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Unit: Electricity
About this unit, electric current & circuit.
- Intro to charge (Opens a modal)
- Unit of charge (Coulombs) (Opens a modal)
- Intro to current (& Amperes) (Opens a modal)
- Finding electric current Get 3 of 4 questions to level up!
Electric potential & potential difference
- Intro to potential difference (& voltage) (Opens a modal)
- Solved example: Potential difference & work done (Opens a modal)
- Voltage and work Get 3 of 4 questions to level up!
Circuits, Ohm's law & resistance
- Introduction to circuits and Ohm's law (Opens a modal)
- Solved example: Ohms law (Opens a modal)
- Ohm's law graph (verifying Ohm's law) (Opens a modal)
- Solved example: (Ohm's law graph) (Opens a modal)
- Ohm's law and resistance Get 3 of 3 questions to level up!
Factors on which resistance of a conductor depends on
- Resistivity and conductivity (Opens a modal)
- Resistance and resistivity Get 5 of 7 questions to level up!
Series and parallel resistors
- Series resistors (Opens a modal)
- Parallel resistors (part 1) (Opens a modal)
- Parallel resistors (part 2) (Opens a modal)
- Parallel resistors (part 3) (Opens a modal)
- Finding equivalent resistance Get 3 of 4 questions to level up!
- Identifying types of resistor combinations Get 3 of 4 questions to level up!
Solving a circuit with series and parallel resistors
- Example: Analyzing a more complex resistor circuit (Opens a modal)
- Solved example: Finding current & voltage in a circuit (Opens a modal)
- Simplifying resistor networks Get 3 of 4 questions to level up!
- Finding currents and voltages (pure circuits) Get 3 of 4 questions to level up!
- Finding currents and voltages (mixed circuits) Get 3 of 4 questions to level up!
Electric power and heating effect of current
- Electric power & energy (Opens a modal)
- Heating effect of current (Opens a modal)
- Solved example - Calculating power & heat dissipated (Opens a modal)
- Electric power (formula) Get 3 of 4 questions to level up!
- Calculating heat dissipated in circuits Get 3 of 4 questions to level up!
Electric circuit with Bulbs
- Solved example: Power dissipated in bulbs (Opens a modal)
- Bulbs connected in series or parallel Get 3 of 4 questions to level up!
Commercial unit of electrical energy
- Commercial unit of electrical energy (Opens a modal)
- Solved example - Cost of operation of electrical device (Opens a modal)
- Finding cost of electrical operation Get 3 of 4 questions to level up!
- Electricity class 10 numerical: CBSE board practice (Opens a modal)
- Electricity class 10: CBSE previous question paper problems (Opens a modal)
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Practice Problems for Electricity Class 10
Last updated at May 26, 2023 by Teachoo
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Download the questions here Chapter 12 Class 10 Electricity.pdf The password to open the file is - teachooisbest
A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly
What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
What is the minimum resistance which can be made using five resistors each of 1/5 Ω?
A cylindrical conductor of length l and uniform area of cross section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed
the shape of the resistor is changed
Explanation - The resistivity of a substance depends on the nature of the material and the physical conditions like temperature. It does not depend on the shape of the resistor.
In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be
In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in parallel to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be
A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams below. The current recorded in the ammeter will be
(a) maximum in (i)
(b) maximum in (ii)
(c) maximum in (iii)
(d) the same in all the cases
Explanation - The current flowing in a circuit does not depend upon the arrangement of the different components, unless they are connected in a different way (Series is changed into parallel, or vice versa.
A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.
Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason.
Find out the following in the electric circuit given in Figure 12.9
(a) Effective resistance of two 8 W resistors in the combination
(b) Current flowing through 4 W resistor
(c) Potential difference across 4 W resistance
(d) Power dissipated in 4 W resistor
(e) Difference in ammeter readings, if any.
The figure below shows three cylindrical copper conductors along with their face areas and lengths. Compare the resistance and the resistivity of the three conductors. Justify your answer
The current flowing through a resistor connected in an electrical circuit and the potential difference developed across its ends are shown in the given ammeter and voltmeter. Find the least count of the voltmeter and ammeter .What is the voltage and the current across the given resistor?
.jpg "how to solve circuit problems in current electricity class 10")
Finding least count
The smallest value that can be measured by a measuring instrument is called its least count.
Numerically,
Least count = Value measure in N divisions/ N
Finding Current through the circuit
The current through the circuit is given by,
Current = Number of division the ammeter needle is pointing x least count In the given figure, The ammeter needle is pointing on the 15th divison
Current = Number of division the ammeter needle is pointing x least count
= 15 × 0.02
= 15 × 2 / 100
= 30 / 100
The ammeter reading is 0.3 A.
Finding potential difference in the circuit
The potential difference in the circuit is given by,
Potential difference = Number of division the voltmeter needle is pointing x least count
In the given figure, The voltmeter needle is pointing on the 21st divison
Potential difference = Number of division the ammeter needle is pointing x least count
= 21 × 0.1
= 21 × 1/10
The voltmeter reading is 2.1V
In a given ammeter, a student sees that needle indicates 17 divisions in ammeter while performing an experiment to verify Ohm’s law. If ammeter has 10 divisions between 0 and 0.5A, then what is the value corresponding to 17 divisions?
Current = Number of division the ammeter needle is pointing × least count
According to the question,
There are 10 divisions between 0 and 0.5A
Hence, 10 divisions measure 0.5A current
Least count = 0.5/10
= 5/ 10 × 10
Least count of ammeter is 0.05 A
According to the question, The ammeter needle is pointing on the 17th divison
= 17 × 0.05
= 17 × 5 / 100
= 85 / 100
The ammeter reading is 0.85 A.
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Extra Question A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly (a) 1020 (b) 1016 (c) 1018 (d) 1023 Current drawn = I = 1 A Time taken = t = 16 s We need to find number of electrons passing First, we need to find the change flown Let the change flown = Q We know that I = 𝑄/𝑡 Q = I t Q = 1 × 16 Q = 16 C We know that Charge on 1 electron = 1.6 × 10−19 C Therefore, 1.6 × 10−19 C = 1 electrons 1 C = 1/(1.6 × 〖10〗^(−19) ) electrons 1 C = 〖10〗^19/1.6 electrons 1 C = 〖10〗^19/16 × 10 electrons 1 C = (〖10〗^20 )/16 electrons Therefore, Number of electrons is 1 C charge = 〖10〗^20/16 electrons Number of electrons is 16 C charge = 16 × 〖10〗^20/16 electrons = 1020 electrons ∴ (a) is correct Extra Question What is the maximum resistance which can be made using five resistors each of 1/5 Ω? (a) 1/5 Ω (b) 10 Ω (c) 5 Ω (d) 1 Ω Resistance of each resistors = 1/5 Ω Number of resistors = 5 We know that, equivalent resistance is maximum when resistors are in series combination We know that, equivalent resistance is given by R = R1 + R2 + R3 + R4 + R5 R = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 R = 5/5 R = 1 Ω Maximum resistance is 1 Ω ∴ (d) is correct Extra Question What is the minimum resistance which can be made using five resistors each of 1/5 Ω? (a) 1/5 Ω (b) 1/25 Ω (c) 1/10 Ω (d) 25 Ω Resistance of each resistors = 1/5 Ω Number of resistors = 5 We know that, equivalent resistance is minimum when resistors are in parallel combination We know that, equivalent resistance is given by R = R1 + R2 + R3 + R4 + R5 1/𝑅 = 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5) 1/𝑅 = 5/1 + 5/1 + 5/1 + 5/1 + 5/1 1/𝑅 = 25 R = 1/25 Ω Minimum resistance is 𝟏/𝟐𝟓 Ω ∴ (b) is correct Extra Question A cylindrical conductor of length l and uniform area of cross section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section (a) A/2 (b) 3A/2 (c) 2 A (d) 3 A Given Length of the conductor = l Area of cross-section = A Resistance of the conductor = R Let the resistivity of the conductor = 𝜌 We know that, R = 𝜌 𝑙/𝐴 ∴ A = 𝝆𝒍/𝑹 New Conductor Given, Length of other conductor = l2 = 2l Resistance of the conductor = R Since the material is same, Resistivity of the conductor = 𝜌 Now, R = 𝜌 𝑙_2/𝐴_2 R = 𝜌 × 2𝑙/𝐴_2 A2 = (𝜌 (2𝑙))/𝑅 A2 = 2((𝜌 𝑙)/𝑅) A2 = 2 A ∴ (c) is correct Extra Question In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be (a) 5 J (b) 10 J (c) 20 J (d) 30 J Given Potential difference = V = 6 V Resistance of R1 = 2 Ω Resistance of R2 = 4 Ω Time taken = t = 5s We need to find heat dissipated by 4 Ω resistor We know that, Heat dissipated = I2 Rt Here, R = R2 I = Current across R2 Since R1 and R2 are in series, Current in R1 and R2 = Current in circuit Finding current in circuit We know that, in series combination equivalent resistance is given by R = R1 + R2 R = 2 + 4 R = 6 Ω By Ohm’s law, V = IR I = 𝑉/𝑅 I = 6/6 I = 1 A Now, Heat dissipated by 4 Ω resistor Heat dissipated = I2R2t = (1)2 (4) (5) = 1 × 4 × 5 = 20 J ∴ (c) is correct Extra Question In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in parallel to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be (a) 45 J (b) 20 J (c) 60 J (d) 65 J Given Potential difference = V = 6 V Resistance of R1 = 2 Ω Resistance of R2 = 4 Ω Time taken = 5s We need to find heat dissipated by 4 Ω resistor Heat dissipated = I2Rt We know that in parallel combination Potential difference across individual resistors = Potential difference across the circuit So, we use Heat formula which has V in it Finding heat dissipated in 4 Ω resistor H = I2 RT H = (𝑉/𝑅)^2R t H = 𝑉^2/𝑅^2 R t H = 𝑉^2/𝑅 t Putting values H = 6^2/4 × 5 H = 36/4 × 5 H = 9 × 5 By Ohm’s law, V = I R I = 𝑉/𝑅 H = 45 J Heat dissipated by 4 Ω resistor is 45 J ∴ (a) is correct Extra Question A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections. Corrections The ammeter should be connected in series. The voltmeter should be connected in parallel. The two ends of the battery should have different signs. Hence, the batteries are not connected properly Corrected figure is Extra Question Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4 Ω resistors? Give reason. The circuit diagram will be Now, the next question is Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4 Ω resistors . We can see that, the parallel combination of two 4Ω resistors give a 2Ω resistor. Hence we can say that, now, two 2Ω resistors are in series. By Ohm's Law, V = IR In a series combination, the current through all the resistors is same. Hence, V = Current × R Since Current is same, and R is same in both the cases. The potential difference across the 2Ω resistor & the parallel combination of 4Ω resistors will be same. Extra Question Find out the following in the electric circuit given in Figure 12.9 (a) Effective resistance of two 8 Ω resistors in the combination (b) Current flowing through 4 Ω resistor (c) Potential difference across 4 Ω resistance (d) Power dissipated in 4 Ω resistor (e) Difference in ammeter readings, if any. Finding equivalent of two 8 Ω resistors in parallel We know that in parallel combination equivalent resistance is given by 1/𝑅 = 1/𝑅_1 + 1/𝑅_2 1/𝑅 = 1/8 + 1/8 1/𝑅 = 2/8 1/𝑅 = 1/4 R = 4 Ω Equivalent resistance is 8 Ω So, our circuit diagram will be Finding current flowing through 4 Ω resistor Since 4 Ω resistor is in series, Current flowing through 4 Ω resistor = Current flowing in the entire circuit Let the current = I Since two 4 Ω resistors are in series Equivalent resistance is R = R1 + R2 R = 4 + 4 R = 8 Ω By Ohm’s law V = I R I = 𝑉/𝑅 I = 8/8 I = 1 A Current flowing = 1 A (c) Potential difference across 4 Ω resistor By Ohm’s law, V = IR V = 1 × 4 V = 4 V (d) Potential dissipated by 4 Ω resistor We know that Power dissipated = V I Putting V = IR = (IR) I = I2 R = (1)2 × 4 = 4 W Power dissipated is 4 W (e) Difference in ammeter readings, if any. Since the two ammeters are connected in series, the current flowing through them would be equal. Hence, there won't be any difference in the current flowing through them. Hence, both the ammeters would show same reading. Extra Question The figure below shows three cylindrical copper conductors along with their face areas and lengths. Compare the resistance and the resistivity of the three conductors. Justify your answer Resistivity We know that the resistivity of a conductor depends on the nature of material of the conductor. Since all the three conductors are made of copper, their nature of material is same. Hence, their resistivity will be same. Resistance We know that R = ρ 𝑙/𝐴 where ρ is the resistivity, l is the length and A is the Area of cross section. Since the material is same, ρ is same in all the three cases. Case I l = L A = A R1 = ρ × 𝐿/𝐴 R1 = ρ 𝐿/𝐴 Case II l = 3L A = A/3 R2 = ρ × 3𝐿/(𝐴/3) R2 = ρ × (3𝐿 × 3)/𝐴 R2 = 9 ρ𝐿/𝐴 Case III I = L/3 l = L/3 A =3A R3 = ρ × (𝐿/3)/3𝐴 R2 = ρ × 𝐿/(3 × 3𝐴) R2 = 1/9 × ρ𝐿/𝐴 Hence, R1 : R2 : R3 = 1 : 9 : 1/9 = 1 × 9 : 9 × 9 : 9 × 1/9 = 9 : 81 : 1
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How to Solve Circuit Problems
Last Updated: December 24, 2022
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 14 people, some anonymous, worked to edit and improve it over time. This article has been viewed 24,064 times. Learn more...
Solving circuits is one of the most challenging tasks for the undergraduate student as it involves numerous theorems, concepts, and processes for solving the circuits. But following a planned problem solving strategy simplifies the problem, making it easier to solve.
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Physics - Current Electricity: Exercises and Example Solved Numerical problems | 12th Physics : Current Electricity
Chapter: 12th physics : current electricity, current electricity: exercises and example solved numerical problems, numerical problems, 1. the following graphs represent the current versus voltage and voltage versus current for the six conductors a,b,c,d,e and f. which conductor has least resistance and which has maximum resistance ans: least: r f = 0.4 ω, maximum r c = 2.5 ω 2. lightning is very good example of natural current. in typical lightning, there is 10 9 j energy transfer across the potential difference of 5 × 10 7 v during a time interval of 0.2 s. using this information, estimate the following quantities (a) total amount of charge transferred between cloud and ground (b) the current in the lightning bolt (c) the power delivered in 0.2 s. ans: charge = 20 c, i = 100 a, p = 5 gw 3. a copper wire of 10-6 m2 area of cross section, carries a current of 2 a. if the number of electrons per cubic meter is 8 × 10 28 , calculate the current density and average drift velocity. ans: j = 2 × 10 6 am −2 ; v d = 15.6 × 10 −5 ms −1 4. the resistance of a nichrome wire at 0º c is 10 ω. if its temperature coefficient of resistance is 0.004/ºc, find its resistance at boiling point of water. comment on the result. ans: r t = 14 ω. as the temperature increases the resistance of the wire also increases. 5. the rod given in the figure is made up of two different materials. both have square cross sections of 3 mm side. the resistivity of the first material is 4 x 10-3 ω.m and it is 25 cm long while second material has resistivity of 5 x 10 -3 ω.m and is of 70 cm long. what is the resistivity of rod between its ends ans: 500 ω 6. three identical lamps each having a resistance r are connected to the battery of emf as shown in the figure. suddenly the switch s is closed. (a) calculate the current in the circuit when s is open and closed (b) what happens to the intensities of the bulbs a,b and c. (c) calculate the voltage across the three bulbs when s is open and closed (d) calculate the power delivered to the circuit when s is opened and closed (e) does the power delivered to the circuit decreases, increases or remain same ans: 7. the current through an element is shown in the figure. determine the total charge that pass through the element at a) t = 0 s, b) t = 2 s, c) t = 5s ans: at t= 0s,dq = 0 c, at t=2 s, dq = 10 c; at t=5 s, dq = 0 c 8. an electronics hobbyist is building a radio which requires 150 ω in her circuit, but she has only 220 ω, 79 ω and 92 ω resistors available. how can she connect the available resistors to get desired value of resistance ans: parallel combination of 220 ω and 79 ω in series with 92 ω 9. a cell supplies a current of 0.9 a through a 2 ω resistor and a current of 0.3 a through a 7 ω resistor. calculate the internal resistance of the cell. ans: 0.5 ω 10. calculate the currents in the following circuit. ans : i 1 = 0.070 a, i 2 = -0.010 a and i 3 = 0.080 a 11. a potentiometer wire has a length of 4 m and resistance of 20 ω. it is connected in series with resistance of 2980 ω and a cell of emf 4 v. calculate the potential along the wire. ans: potential = 0.65 × 10 -2 v m -1 . 12. determine the current flowing through the galvanometer (g) as shown in the figure. 13. two cells each of 5v are connected in series across a 8 ω resistor and three parallel resistors of 4 ω, 6 ω and 12 ω. draw a circuit diagram for the above arrangement. calculate i) the current each resistor ans: the current at 4 ω ,i = 2/4 = 0 .5a, the current at 6 ω, i = 2/6 = 0.33a , the current at 12 ω, i = 2/12 = 0 . 17 a 14. four light bulbs p, q, r, s are connected in a circuit of unknown arrangement. when each bulb is removed one at a time and replaced, the following behavior is observed. draw the circuit diagram for these bulbs. ans 15. in a potentiometer arrangement, a cell of emf 1.25 v gives a balance point at 35 cm length of the wire. if the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell ans: emf of the second cell is 2.25 v, electric current: solved example problems, example 2.1.
Compute the current in the wire if a charge of 120 C is flowing through a copper wire in 1 minute.
The current (rate of flow of charge) in the wire is
I = Q/t = 120/60 = 2A
EXAMPLE 2.2
If an electric field of magnitude 570 N C -1 , is applied in the copper wire, find the acceleration experienced by the electron.
E = 570 N C -1 , e = 1.6 × 10 -19 C, m = 9.11 × 10 -31 kg and a = ?
F = ma = eE
a = eE/ m = 570×1 .6×10 −19 /9 .11×10 -31
= 912 ×10 −19 ×10 31 / 9 .11
= 1.001 × 10 14 m s -2
EXAMPLE 2.3
A copper wire of cross-sectional area 0.5 mm 2 carries a current of 0.2 A. If the free electron density of copper is 8.4 × 10 28 m -3 then compute the drift velocity of free electrons.
The relation between drift velocity of electrons and current in a wire of cross-sectional area A is
v d = I/ ne A
v d = 0.03 x 10 -3 m s -1
EXAMPLE 2.4
Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.
I = 32 A , t = 1 s
Charge of an electron, e = 1.6 × 10 -19 C
The number of electrons flowing per second, n =?
I = q/t = ne / t
n = 32×1 / 1 .6×10 −19 C
n = 20 × 10 19 = 2 × 10 20 electrons
OHM’S LAW: Solved Example Problems
EXAMPLE 2.5
A potential difference across 24 Ω resistor is 12 V. What is the current through the resistor?
V = 12 V and R = 24 Ω
Current, I = ?
From Ohm’s law, I = V/R = 12/24 = 0 .5 A
Resistivity: Solved Example Problems
EXAMPLE 2.6
The resistance of a wire is 20 Ω. What will be new resistance, if it is stretched uniformly 8 times its original length?
R 1 = 20 Ω, R 2 = ?
Let the original length (l 1 ) be l .
The new length, l 2 = 8 l 1 (i.,e) l2 =8 l
The original resistance, R = ρ [ l 1 / A 1 ]
The new resistance
Though the wire is stretched, its volume is unchanged.
Initial volume = Final volume
A 1 l 1 = A 2 l 2 , A 1 l = A 2 8 l
A 1 / A 2 = 8 l / l = 8
By dividing equation R 2 by equation R 1 , we get
Substituting the value of A 1 /A 2 , we get
R 2 / R 1 = 8 ×8 = 64 2
R 2 = 64 × 20=1280 Ω
Hence, stretching the length of the wire has increased its resistance.
EXAMPLE 2.7
Consider a rectangular block of metal of height A, width B and length C as shown in the figure.
If a potential difference of V is applied between the two faces A and B of the block (figure (a)), the current I AB is observed. Find the current that flows if the same potential difference V is applied between the two faces B and C of the block (figure (b)). Give your answers in terms of I AB .
In the first case, the resistance of the block
Resistors in series and parallel: Solved Example Problems
EXAMPLE 2.8
Calculate the equivalent resistance for the circuit which is connected to 24 V battery and also find the potential difference across 4 Ω and 6 Ω resistors in the circuit.
Since the resistors are connected in series, the effective resistance in the circuit
= 4 Ω + 6 Ω = 10 Ω
The Current I in the circuit= V/ R eq = 24/10 = 2 .4 A
Voltage across 4Ω resistor
V 1 = IR 1 = 2 . 4 A× 4 Ω = 9.6V
Voltage across 6 Ω resistor
V 2 = IR 1 = 2 . 4 A× 6 Ω =14 .4V
EXAMPLE 2.9
Calculate the equivalent resistance in the following circuit and also find the current I, I 1 and I 2 in the given circuit.
Since the resistances are connected in parallel, therefore, the equivalent resistance in the circuit is
The resistors are connected in parallel, the potential (voltage) across each resistor is the same.
The current I is the total of the currents in the two branches. Then,
I = I 1 + I 2 = 6 A + 4 A = 10 A
EXAMPLE 2.10
When two resistances connected in series and parallel their equivalent resistances are 15 Ω and 56/15 Ω respectively. Find the individual resistances.
Rs = R 1 + R 2 = 15 Ω (1)
The above equation can be solved using factorisation.
R 1 2 -8 R 1 -7 R 1 + 56 = 0
R 1 (R 1 – 8) – 7 (R 1 – 8) = 0
(R 1 – 8) (R 1 – 7) = 0
If (R 1 = 8 Ω)
using in equation (1)
8 + R 2 = 15
R 2 = 15 – 8 = 7 Ω ,
R 2 = 7 Ω i.e , (when R 1 = 8 Ω ; R 2 = 7 Ω)
If (R 1 = 7 Ω)
Substituting in equation (1)
7 + R 2 = 15
R 2 = 8 Ω , i.e , (when R 1 = 8 Ω ; R 2 = 7 Ω )
EXAMPLE 2.11
Calculate the equivalent resistance between A and B in the given circuit.
EXAMPLE 2.12
Five resistors are connected in the configuration as shown in the figure. Calculate the equivalent resistance between the points a and b.
To find the equivalent resistance between the points a and b, we assume that current is entering the junction a. Since all the resistances in the outside loop are the same (1Ω), the current in the branches ac and ad must be equal. So the electric potential at the point c and d is the same hence no current flows into 5 Ω resistance. It implies that the 5 Ω has no role in determining the equivalent resistance and it can be removed. So the circuit is simplified as shown in the figure.
The equivalent resistance of the circuit between a and b is R eq = 1Ω
Temperature dependence of resistivity: Solved Example Problems
EXAMPLE 2.13
If the resistance of coil is 3 Ω at 20 0C and α = 0.004/0C then determine its resistance at 100 0C.
R 0 = 3 Ω, T = 100ºC, T 0 = 20ºC
α = 0.004/ºC, R T = ?
R T = R0(1 + α(T-T 0 ))
R 100 = 3(1 + 0.004 × 80)
R 100 = 3(1 + 0.32)
R 100 = 3(1.32)
R 100 = 3.96 Ω
EXAMPLE 2.14
Resistance of a material at 10ºC and 40ºC are 45 Ω and 85 Ω respectively. Find its temperature co-efficient of resistance.
T 0 = 10ºC, T = 40ºC, R 0 = 45 Ω , R = 85 Ω
α = 1/R . ΔR /ΔT
α = 0.0296 per ºC
Energy and Power in Electrical Circuits: Solved Example Problems
EXAMPLE 2.15
A battery of voltage V is connected to 30 W bulb and 60 W bulb as shown in the figure. (a) Identify brightest bulb (b) which bulb has greater resistance? (c) Suppose the two bulbs are connected in series, which bulb will glow brighter?
(a) The power delivered by the battery P = VI. Since the bulbs are connected in parallel, the voltage drop across each bulb is the same. If the voltage is kept fixed, then the power is directly proportional to current (P ∝ I). So 60 W bulb draws twice as much as current as 30 W and it will glow brighter then others.
(b) To calculate the resistance of the bulbs, we use the relation P = V 2 / R . In both the bulbs, the voltage drop is the same, so the power is inversely proportional to the resistance or resistance is inversely proportional to the power (R ∝ 1/P) . It implies that, the 30W has twice as much as resistance as 60 W bulb.
(c) When these two bulbs are connected in series, the current passing through each bulb is the same. It is equivalent to two resistors connected in series. The bulb which has higher resistance has higher voltage drop. So 30W bulb will glow brighter than 60W bulb. So the higher power rating does not always imply more brightness and it depends whether bulbs are connected in series or parallel.
EXAMPLE 2.16
Two electric bulbs marked 20 W – 220 V and 100 W – 220 V are connected in series to 440 V supply. Which bulb will be fused?
To check which bulb will be fused, the voltage drop across each bulb has to be calculated.
The resistance of a bulb,
Both the bulbs are connected in series. So the current which passes through both the bulbs are same. The current that passes through the circuit, I = V /R tot .
R tot = ( R 1 + R 2 )
R tot = ( 484 +2420) Ω = 2904Ω
I = 440V/2904Ω ≈ 0.151A
The voltage drop across the 20W bulb is
V 1 = IR 1 = 440/2904 × 2420 ≈ 366.6 V
The voltage drop across the 100W bulb is
V 2 = IR 2 = 440/2904 × 484 ≈ 73.3 V
The 20 W bulb will be fused because its voltage rating is only 220 V and 366.6 V is dropped across it.
EXAMPLE 2.17
A battery has an emf of 12 V and connected to a resistor of 3 Ω. The current in the circuit is 3.93 A. Calculate (a) terminal voltage and the internal resistance of the battery (b) power delivered by the battery and power delivered to the resistor
The given values I = 3.93 A, ξ = 12 V, R = 3 Ω
(a) The terminal voltage of the battery is equal to voltage drop across the resistor
V = IR = 3.93 × 3 = 11.79 V
The internal resistance of the battery,
r = |ξ –V / V| R = | 12 −11 .79 /11 .79 | × 3 = 0.05 Ω
The power delivered by the battery P = Iξ = 3.93 × 12 = 47.1 W
The power delivered to the resistor = I 2 R = 46.3 W
The remaining power = (47.1 – 46.3) P = 0.772 W is delivered to the internal resistance and cannot be used to do useful work. (it is equal to I 2 r).
Cells in series: Solved Example Problems
EXAMPLE 2.18
From the given circuit,
i) Equivalent emf of the combination
ii) Equivalent internal resistance
iii) Total current
iv) Potential difference across external resistance
v) Potential difference across each cell
i) Equivalent emf of the combination ξ eq = nξ = 4 9 = 36 V
ii) Equivalent internal resistance r eq = nr = 4 × 0.1 = 0.4 Ω
iii) Total current I = nξ / R +nr
= [4 ×9] / 10 + ( 4 ×0.1)
= [4 ×9] / [10 +0 .4] = 36 /10.4
iv) Potential difference across external resistance V = IR = 3.46 × 10 = 34.6 V. The remaining 1.4 V is dropped across the internal resistance of cells.
v) Potential difference across each cell V/ n = 34.6/4 = 8 .65V
Cells in parallel: Solved Example Problems
EXAMPLE 2.19
From the given circuit
i) Equivalent emf
iii) Total current (I)
iv) Potential difference across each cell
v) Current from each cell
i) Equivalent emf ξ eq = 5 V
ii) Equivalent internal resistance,
R eq = r/n = 0 .5/4 = 0.125Ω
iii) total current,
iv) Potential difference across each cell V = IR = 0.5 × 10 = 5 V
v) Current from each cell, I ′ = I/n
I ′ = 0.5/4 = 0.125 A
Kirchhoff’s first rule (Current rule or Junction rule): Solved Example Problems
EXAMPLE 2.20
From the given circuit find the value of I.
Applying Kirchoff’s rule to the point P in the circuit,
The arrows pointing towards P are positive and away from P are negative.
Therefore, 0.2A – 0.4A + 0.6A – 0.5A + 0.7A – I = 0
1.5A – 0.9A – I = 0
0.6A – I = 0
Kirchhoff’s Second rule (Voltage rule or Loop rule) : Solved Example Problems
EXAMPLE 2.21
The following figure shows a complex network of conductors which can be divided into two closed loops like ACE and ABC. Apply Kirchoff’s voltage rule.
Thus applying Kirchoff’s second law to the closed loop EACE
I 1 R 1 + I 2 R 2 + I 3 R 3 = ξ
and for the closed loop ABCA
I 4 R 4 + I 5 R 5 - I 2 R 2 = 0
EXAMPLE 2.22
Calculate the current that flows in the 1 Ω resistor in the following circuit.
We can denote the current that flows from 9V battery as I 1 and it splits into I 2 and I 1 – I 2 in the junction according Kirchoff’s current rule (KCR). It is shown below.
Now consider the loop EFCBE and apply KVR, we get
1I 2 + 3I 1 + 2I 1 = 9
5I 1 + I 2 = 9 (1)
Applying KVR to the loop EADFE, we get
3 (I 1 – I 2 ) – 1I 2 = 6
3I 1 – 4I 2 = 6 (2)
Solving equation (1) and (2), we get
I 1 = 1.83 A and I 2 = -0.13 A
It implies that the current in the 1 ohm resistor flows from F to E.
Wheatstone’s bridge : Solved Example Problems
EXAMPLE 2.23
In a Wheatstone’s bridge P = 100 Ω, Q = 1000 Ω and R = 40 Ω. If the galvanometer shows zero deflection, determine the value of S.
EXAMPLE 2.24
What is the value of x when the Wheatstone’s network is balanced?
P = 500 Ω, Q = 800 Ω, R = x + 400, S = 1000 Ω
x + 400 = 0.625 × 1000
x + 400 = 625
x = 625 – 400
Meter bridge : Solved Example Problems
EXAMPLE 2.25
In a meter bridge with a standard resistance of 15 Ω in the right gap, the ratio of balancing length is 3:2. Find the value of the other resistance.
In a meter bridge, the value of resistance in the resistance box is 10 Ω. The balancing length is l 1 = 55 cm. Find the value of unknown resistance.
Q = 10 Ω
Heating Effect of Electric Current, Joule’s law: Solved Example Problems
EXAMPLE 2.27
Find the heat energy produced in a resistance of 10 Ω when 5 A current flows through it for 5 minutes.
R = 10 Ω, I = 5 A, t = 5 minutes = 5 × 60 s
H = I 2 R t
= 5 2 × 10 × 5 × 60
=25 × 10 × 300
=75000 J (or) 75 kJ
EXAMPLE 2.28
An electric heater of resistance 10 Ω connected to 220 V power supply is immersed in the water of 1 kg. How long the electrical heater has to be switched on to increase its temperature from 30°C to 60°C. (The specific heat of water is s = 4200 J kg -1 )
According to Joule’s heating law H = I 2 Rt
The current passed through the electrical heater = 220V/10Ω = 22 A
The heat produced in one second by the electrical heater H = I 2 R
The heat produced in one second H = (22) 2 x 10 = 4840 J = 4.84 k J. In fact the power rating of this electrical heater is 4.84 k W.
The amount of energy to increase the temperature of 1kg water from 30°C to 60°C is
Q = ms ∆T (Refer XI physics vol 2, unit 8)
Here m = 1 kg,
s = 4200 J kg -1 ,
so Q = 1 × 4200 x 30 = 126 kJ
The time required to produce this heat energy t = Q/ I 2 R = 126 ×10 3 / 4840 ≈ 26 .03 s
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- ICSE Class 10
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- ICSE Class 10 Physics Selina Solutions Chapter 8 Current Electricity
Selina Solutions Concise Physics Class 10 Chapter 8 Current Electricity
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Access Answers to Physics Selina Solutions Concise Physics Class 10 Chapter 8 Current Electricity
Exercise 8(A) page no: 186
Question: 1
Define the term current and state its S.I. unit.
The rate of flow of charge is known as current. The S.I. unit of current is Ampere.
Question: 2
Define the term electric potential. State it’s S.I. unit.
The amount of work done per unit charge in bringing a positive test charge from infinity to that point is known as electric potential at a point. The S.I. unit of electric potential is volt.
Question: 3
How is the electric potential difference between the two points defined? State its S.I. unit.
The potential difference between two points can be defined as the potential difference between two points is equal to the work done per unit charge in moving a positive test charge from one point to the other. It’s S.I. unit is volt.
Question: 4
Explain the statement ‘the potential difference between two points is 1 volt’.
The potential difference between two points is said to be 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.
Question: 5
(a) State whether the current is a scalar or vector? What does the direction of current convey?
(b) State whether the potential is a scalar or vector? What does the positive and negative sign of potential convey?
(a) Current is a scalar quantity. The direction of current conveys that the electrons flow opposite to the direction of flow of current.
(b) Potential is a scalar quantity. The positive sign of potential states that work has to be done on the positive test charge against the repulsive force due to the positive charge in bringing it from infinity. Negative sign of potential states that the work done on the negative test charge is due to the attractive force.
Question: 6
Define the term resistance. State its S.I. unit.
The obstruction offered to the flow of current by the conductor is known as its resistance. The S.I. unit of resistance is Ohm.
Question: 7
(a) Name the particles which are responsible for the flow of current in a metallic wire.
(b) Explain the flow of current in a metallic wire on the basis of movement of the particles named by you above in part (a).
(c) What is the cause of resistance offered by the metallic wire in the flow of current through it?
(a) Free electrons are the particles which are responsible for the flow of current in a metallic wire
(b) Free electrons are the moving charges that result in the conduction of electricity in metals. In time ‘t’ if ‘n’ electrons pass through the metallic conductor, then the total charge that has flown is given by
Q (charge) = n × e (charge on an electron)
(c) A metal contain free electrons and fixed positive ions.
When electrons move through a conductor such as a metal wire then an electric current flows. In the metal, the moving electrons can collide with the ions. Thus, this makes it more difficult for the current to flow and therefore causes resistance.
Question: 8
State Ohm’s law and draw a neat labelled circuit diagram containing a battery, a key, a voltmeter, an ammeter, a rheostat and an unknown resistance to verify it.
According to Ohm’s law, the current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant. This is known as Ohm’s law
Question: 9
(a) Name and state the law which relates the potential difference and current in a conductor.
(b) What is the necessary condition for a conductor to obey the law named above in part (a)?
(a) The law is known as Ohm’s law. This law states that the current flowing through the conductor is directly proportional to the potential difference across its ends, provided that the physical conditions and the temperature of the conductor remain constant.
(b) The necessary condition for a conductor to obey Ohm’s law is that the temperature should remain constant.
Question: 10
(a) Draw a V-I graph for a conductor obeying Ohm’s law.
(b) What does the slope of V-I graph for a conductor represent?
(a) V-I graph for a conductor obeying Ohm’s law is given below
(b) Slope of V-I graph for a conductor represents resistance.
Question: 11
Draw a I-V graph for a linear resistor. What does its slope represent?
I-V graph for a linear resistor
Slope of I-V graph: The slope of I-V graph is ΔI / ΔV
ΔI / ΔV is the reciprocal resistance of the conductor i.e.
Slope = ΔI / ΔV
= 1 / resistance of the conductor
Question: 12
What is an ohmic resistor? Give one example of an ohmic resistor. Draw a graph to show its current – voltage relationship. How is the resistance of the resistor determined from this graph?
The conductors which obey the Ohm’s law are known as the ohmic resistors or linear resistances.
Examples: All metallic conductors such as silver, aluminium, copper, iron etc.
Resistance is determined in the form of slope from the above graph
Question: 13
What are non-ohmic resistors? Give one example and draw a graph to show its current-voltage relationship.
The conductors which do not obey the Ohm’s law are known as the non-ohmic resistors or non-linear resistances).
Examples: LED, solar cell, junction diode, etc.
V vs I for non-ohmic conductors
Question: 14
Give two differences between an ohmic and non-ohmic resistor
Question: 15
Fig. below shows the I-V characteristic curves for two resistors. Identify the ohmic and non-ohmic resistors. Give a reason for your answer.
Graph (a) is non-ohmic resistor and Graph (b) is ohmic resistor
The I-V graph for (b) is a straight line or linear while the I-V graph for (a) is a curve
Question: 16
Draw a V – I graph for a conductor at two different temperatures. What conclusion do you draw from your graph for the variation of resistance of conductor with temperature?
In the given graph above T 1 > T 2 . The straight line A is more steeper than the line B because the resistance of conductor is more at high temperature T 1 than at low temperature T 2 . Hence we can conclude that resistance of a conductor increases with the increase in temperature.
Question: 17
(a) How does the resistance of a wire depend on its radius? Explain your answer.
(b) Two copper wires are of same length, but one is thicker than the other. Which will have more resistance?
Resistance of a wire varies inversely as the area of cross section of the wire i.e.,
R ∝ 1 / π 2
Resistance of a wire is directly proportional to its length, i.e., R ∝ l and varies inversely as the area of cross section of the wire i.e., R ∝ 1 / a or R ∝ 1 / πr 2
Since the resistance is inversely proportional to area of cross section, hence thin wire will have more resistance.
Question: 18
How does the resistance of a wire depend on its length? Give a reason of your answer.
Resistance of a wire is directly proportional to the length of the wire.
The resistance of conductor depends on the number of collisions which the electron suffer with the fixed positive ions while moving from one end to other end of the conductor. Therefore in a long conductor, the number of collisions of free electrons with the positive ions will be more when compared to a shorter conductor. Hence a longer conductor offers more resistance.
Question: 19
How does the resistance of a metallic wire depend on its temperature? Explain with reason.
With the increase in temperature of a conductor, the random motion of electrons increases. This makes the number of collisions of electrons with the positive ions to increase. Hence, the resistance of a conductor increases with an increase in its temperature.
The resistance of filament of a bulb is more when it is glowing that is when it is at a high temperature as compared to when it is not glowing that is when it is cold.
Question: 20
Two wires, one of copper and other of iron, are of the same length and same radius. Which will have more resistance? Give reason.
Iron has more resistivity as compared to copper which has less resistivity. So, greater the resistivity, the more the resistance is and the smaller the resistivity, the lesser the resistance. Hence, iron wire has more resistance than copper wire of the same length and same radius.
Question: 21
Name three factors on which the resistance of a wire depends and state how it is affected by the factors stated by you?
The three factors on which the resistance of wire depends are as follows
(i) Resistance of a wire is directly proportional to its length that is
(ii) Resistance of a wire varies inversely as the area of cross section of the wire. The resistance will be less when the area of cross section of the wire is more and vice versa
(iii) Resistance of a conductor increases with an increase in its temperature. As a result the number of collisions increases.
(iv) Resistance depends on the nature of conductor since different substances have different concentration of free electrons. Substances which have a large concentration of electrons, offer less resistance and hence called good conductors and substances which have negligible concentration of free electrons offer very high resistance and are called insulators.
Question: 22
Define the term specific resistance and state its S.I. unit.
Specific resistance of a material is the resistance of a wire of that material of unit length and unit area of cross section. The S.I. unit of specific resistance is ohm × metre
Question: 23
Write an expression connecting the resistance of a wire and specific resistance of its material. State the meaning of symbols used.
The expression is
R = ρ l / a where
ρ = specific resistance of the material of conductor
R = resistance of conductor
l = length of conductor
A = area of cross section of conductor
Question: 24
State the order of specific resistance of (i) a metal, (ii) a semiconductor and (iii) an insulator.
(i) The specific resistance for metals is low, since it allows most of current to pass through it.
(ii) The specific resistance for semiconductor is more than metals
(iii) The specific resistance for insulators is very high, since the current won’t pass through it
Question: 25
(a) Name two factors on which the specific resistance of a wire depends?
(b) Two wires A and B are made of copper. The wire A is long and thin while the wire B is Short and thick. Which will have more specific resistance?
(a) Two factors on which the specific resistance of a wire depends are
(i) Material of the substance and
(ii) Temperature of the substance
(b) Both the wires will have the same specific resistance because the specific resistance depends on the material of the wire and not its dimensions.
Question: 26
Name a substance of which the specific resistance remains almost unchanged by the increase in temperature.
The substance of which the specific resistance remains almost unchanged by the increase in temperature is manganin.
Question: 27
How does specific resistance of a semi-conductor change with the increase in temperature?
With the increase in temperature, specific resistance of a semi-conductor decreases.
Question: 28
How does (a) resistance, and (b) specific resistance of a wire depend on its (i) length, and (ii) radius?
(a) Resistance is directly proportional to length of the wire and inversely proportional to the square of radius of the wire
(b) Specific resistance of a wire do not depend on length and radius of the conductor that is independent of its length and radius.
Question: 29
(a) Name the material used for making the connection wires. Give reason for your answer. (b) Why should a connection wire be thick?
(a) Copper or aluminium are the materials used for making connection wires because they have small specific resistance and hence the wires of these materials possess negligible resistance
(b) The connection wires are made thick to consider their resistance as negligible to the flow of current through the circuit.
Question: 30
Name the material used for making a fuse wire. Give a reason.
Alloy of lead and tin is used for making a fuse wire because it has high resistivity and low melting point.
MULTIPLE CHOICE TYPE
Which of the following is an ohmic resistance?
(b) Junction diode
(c) Filament of a bulb
(d) Nichrome wire
An ohmic resistance is nichrome wire
For which of the following substances, resistance decreases with increase in temperature?
(b) Mercury
(d) Platinum
Resistance decreases with increase in temperature for carbon
In a conductor, 6.25 × 10 16 electrons flow from its end A to B in 2 s. Find the current flowing through the conductor. (e = 1.6 × 10 -19 C)
Number of electrons flowing through the conductor,
n = 6.25 × 10 16 electrons.
Time taken to flow from A to B = 2 s and e = 1.6 × 10 -19 C
Let I be the current flowing through the conductor
Now, I = ne / t
Therefore I = [(6.25 × 10 16 ) (1.6 × 10 -19 )] / 2
I = 5 × 10 -3 A or
Hence, 5 mA current flows from B to A
A current of 1.6 mA flows through a conductor. If charge of an electron is -1.6 x 10 -19 coulomb, find the number of electrons that will pass each second through the cross section of that conductor.
Current, I = 1.6 m A or
I = 1.6 × 10 -3 A
Charge, Q = -1.6 × 10 -19 coulomb
Q = 1.6 × 10 -3 × 1
Number of electrons = 1.6 × 10 -3 / 1.6 × 10 -19
∴ Number of electrons = 10 16
Find the potential difference required to flow a current of 200 mA in a wire of resistance 20 ohm.
Current I = 200 mA
Resistance R = 20 ohm
Potential difference V =?
Using Ohm’s law
V = 0.2 × 20
An electric bulb draws 1.2 A current at 6.0 V. Find the resistance of filament of bulb while glowing.
Current I = 1.2 A
Potential difference or Voltage V = 6.0 V
Resistance R =?
From Ohm’s law
R = 6 / 1.2
A car bulb connected to a 12 volt battery draws 2 A current when glowing. What is the resistance of the filament of the bulb? Will the resistance be more, same or less when the bulb is not glowing.
Potential difference or Voltage V = 12 V
Current I = 2 A
Resistance =?
According to Ohm’s law
Hence, when bulb is not glowing, resistance will be less
Calculate the current flowing through a wire of resistance 5 Ohm connected to a battery of potential difference 3 V.
Potential difference or Voltage V = 3 V
Resistance R = 5 Ohm
In an experiment of verification of Ohm’s law, following observations are obtained.
Draw a characteristic V-I graph and use this graph to find:
(a) potential difference V when the current I is 0.5 A.
(b) current I when the potential difference V is 0.75 V.
(c) resistance in circuit
(a) Potential difference is 1.25 V when the current is 0.5 A
(b) Current is 0.3 A when the potential difference is 0.75 V
(c) The graph is linear and thus resistance can be found from any value of the given table.
If V = 2.5 Volt then
Current I = 1.0 amp
R = 2.5 / 1.0
R = 2.5 Ohm
Two wires of the same material and same length have radii 1 mm and 2 mm respectively. Compare (i) their resistances (ii) their specific resistance.
(i) For wire of radius r 1
R 1 = ρ (l / A 1 )
R 1 = ρ (l / πr 1 2 )
(ii) For wire of radius r 2
R 2 = ρ (l / A 2 )
R 2 = ρ (l / πr 2 2 )
∴ R 1 : R 2 will be
ρ (l / πr 1 2 ) : ρ (l / πr 2 2 )
= r 2 2 : r 1 2
(ii) The resistivities of the two wires will be same because the material of the two wires is same. That is
ρ 1 : ρ 2 = 1: 1
A given wire of resistance 1 Ohm is stretched to double its length. What will be its new resistance?
Let ‘I’ be the length and ‘a’ be the area of cross section of the resistor with resistance, R = 1 Ohm
A given wire is stretched to double its length,
Therefore length l’ = 2l and area of cross section a’ = a / 2
Now Resistance (R’) = ρ l’/ a’
R’ = ρ 2l / a / 2
R’ = 4 ρ (1 / a)
A wire of resistance 3 Ohm and length 10 cm is stretched to length 30 cm. Assuming that it has a uniform cross-section, what will be its new resistance?
Resistance R = 3 Ohm
Length l = 10 cm
The new length l’ = 30 cm = 3 × l
R = ρ (l / A)
New resistance
Stretching length will increase and area of cross section will decrease in same order
R’ = ρ (3l / A / 3)
R’ = 9 ρ (l / A)
R’ = 27 Ohm
A wire of resistance 9 Ohm having length 30 cm is tripled on itself. What is its new resistance?
Resistance R = 9 Ohm
Length l = 30 cm
New length l’ = 3 × l
New resistance R’=?
Area of cross section will also change in same order with change in length
R’ = ρ (l / 3 / 3A)
R’ = 1 / 9 (ρ l / A)
R’ = 1 / 9 R
What length of copper wire of specific resistance 1.7 x 10 -8 ohm m and radius 1 mm is required so that its resistance is 1 ohm.
Resistance R = 1 ohm
Specific resistance = 1.7 × 10 -8 ohm m
Radius r = 1 mm that is 10 -3 m
Length l =?
R = ρ 1 / A
l = Rπr 2 / ρ
l = (1 × π × 10 -6 ) / (1.7 × 10 -8 )
l = (1 × 3.14 × 10 -6 ) / (1.7 × 10 -8 )
l = 1.847 × 10 2 m
l = 184.7 m
The filament of a bulb takes a current 100 mA when potential difference across it is 0.2 V. When the potential difference across it becomes 1.0 V, the current becomes 400 mA. Calculate the resistance of filament in each case and account for the difference.
R 1 = V 1 / I 1
R 1 = 0.2 / 0.1
R 1 = 2 ohm
R 2 = V 2 / I 2
R 2 = 1 / 0.4
R 2 = 2.5 ohm
∴ With increase in temperature resistance of the wire increases. Thus resistance of filament increases with the increase in temperature.
Exercise 8(b) Page no: 200
Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of cell.
e.m.f : When the cell is in open circuit, the potential difference between the terminals of the cell is called its electro-motive force or e.m.f
Terminal voltage: When the cell is in closed circuit, the potential difference between the electrodes of the cell is known as terminal voltage.
Internal resistance: The resistance offered by the electrolyte inside the cell, to the flow of current, is known as the internal resistance of the cell.
State two differences between the e.m.f. and terminal voltage of a cell.
Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.
The factors on which the internal resistance of a cell depends are:
(i) The surface area of the electrodes – larger the surface area of electrodes, less is the internal resistance.
(ii) The distance between the electrodes – more the distance between the electrodes, greater is the internal resistance.
A cell of e.m.f. ε and internal resistance r is used to send current to an external resistance R. Write expressions for (a) the total resistance of circuit, (b) the current drawn from the cell, (c) the p.d. across the cell, and (d) voltage drop inside the cell.
(a) The total resistance of circuit = R + r
(b) The current drawn from the cell
We know that,
= I (R + r)
I = ε / (R + r)
(c) p.d. across the cell: [ε / (R + r)] × R
(d) voltage drop inside the cell: [ε / (R + r)] × r
A cell is used to send current to an external circuit. (a) How does the voltage across its terminal compare with its emf? (b) Under what condition is the emf of the cell equal to its terminal voltage?
(a) When current is drawn from a cell, its terminal voltage V is less than its e.m.f.
∴ Terminal voltage < e.m.f.
(b) When no current is drawn, then the e.m.f. is equal to the terminal voltage.
Explain why the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
The current flows through the circuit when the electric cell is in a closed circuit. There is a fall of potential across the internal resistance of the cell. Hence, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.
Write the expressions for the equivalent resistance R of three resistors R 1 , R 2 and R 3 joined in (a) parallel, (b) series.
(a) Total resistance in parallel is given by
1 / R = 1 / R 1 + 1 / R 2 + 1 / R 3
(b) Total resistance in series is given by
R = R 1 + R 2 + R 3
How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.
If current I is drawn from the battery, the current will also be I through each resistor.
Applying Ohm’s law to the two resistors separately, we get,
V = V 1 + V 2
IR = IR 1 + IR 2
R = R 1 + R 2
Thus total resistance in series is
Show by a diagram how two resistors R 1 and R 2 are joined in parallel. Obtain an expression for the total resistance of the combination.
Applying Ohm’s law separately to the two resistors, we have
I 1 = V / R 1
I 2 = V / R 2
I = I 1 + I 2
V / R = V / R 1 + V / R 2
1 / R = 1 / R 1 + 1 / R 2
State how are the two resistors joined with a battery in each of the following cases when:
(a) same current flows in each resistor
(b) potential difference is same across each resistor.
(c) equivalent resistance is less than either of the two resistances.
(d) equivalent resistance is more than either of the two resistances.
(a) The two resistors are in series
(b) The two resistors are in parallel
(c) The two resistors are in parallel
(d) The two resistors are in series
The V-I graph for a series combination and for a parallel combination of two resistors is shown in fig. Which of the two, A or B, represents the parallel combination? Give a reason for your answer.
Change in V is less for the straight line A than for the straight line B (which means the straight line A is less steeper than B). Hence, the straight line A represents small resistance and the straight line B represents more resistance. The resistance decreases in parallel combination while the resistance increases in series combination. Thus the straight line A represents the parallel combination.
In series combination of resistances:
(a) P.d. is same across each resistance
(b) Total resistance is reduced
(c) Current is same in each resistance
(d) All of the above are true
Current is same in each resistance in series combination of resistances.
In parallel combination of resistances:
(a) P.D. is same across each resistance
(b) Total resistance is increased
P.D is same across each resistance in parallel combination of resistances
Which of the following combinations have the same equivalent resistance between X and Y?
The resistors are connected in parallel in figure (a) between X and Y
Let R’ be their equivalent resistance
Now, 1 / R’ = 1 / 2 + 1 / 2
= 2 / 2 ohm or
A series combination of two 1 ohm resistors is in parallel with another series combination of two 1 ohm resistors in figure (d)
Series resistance of two 1 ohm resistors,
R = (1 + 1) ohm
Therefore we can say that 2 ohm resistors are connected in parallel across X and Y
Let R’ be the net resistance across X and Y.
Then, 1 / R’ = 1 / 2 + 1 / 2
1 / R’ = 2 / 2 ohm or
Thus it is clear from equation [1] and [2] that the figures (a) and (d) have the same equivalent resistance between X and Y
The diagram in figure shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm connected to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammetere and voltmeter when (i) the key K is open, and (ii) the key K is closed
(i) Because of no current, ammeter reading = 0
Voltage V = ε – Ir
V = 2 – 0 × 1
(ii) Ammeter reading
I = 2 / (4 + 1)
I = 0.4 amp
Voltage reading
V = 2 – 0.4 × 1
V = 2 – 0.4
A battery of e.m.f. 3.0 V supplies current through a circuit in which resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery.
r = (e – V) / I
r = (3 – 2.7) / 1.5
r = 0.2 ohm
A cell of emf 1.8 V and internal resistance 2 ohm is connected in series with an ammeter of resistance 0.7 ohm and resistance of 4.5 ohm as shown in figure.
(a) What would be the reading of the ammeter?
(b) What is the potential difference across the terminals of the cell?
(a) ε = 1.8 V
Total resistance = 2 + 4.5 + 0.7
I = ε / R (total resistance)
I = 1.8 / 7.2
Now, excluding internal resistance total resistance = 4.5 + 0.7
V = 0.25 × 5.2
A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm in series. Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.
(a) ε = 15 V
I = 15 / (9 + 3)
I = 15 / 12
External resistance R = 6 + 3
V = 1.25 × 9
V = 11.25 V
A cell of e.m.f. ε and internal resistance r sends current 1.0 A when it is connected to an external resistance 1.9 ohm. But its sends current 0.5 A when it is connected to an external resistance of 3.9 ohm. Calculate the values of e and r.
In first case
I = 1 A, R = 1.9 ohm
ε = I (R + r)
= 1 (1.9 + r)
In second case
I = 0.5 A, R = 3.9 ohm
= 0.5 (3.9 + r)
1.9 + r = 1.95 + 0.5r
r = 0.05 / 0.5
r = 0.1 ohm
Now, substituting the value of r
ε = 1.9 + r
ε = 1.9 + 0.1
Two resistors having resistance 4 ohm and 6 ohm are connected in parallel. Find their equivalent resistance.
Let the equivalent resistance of the 4 ohm and 6 ohm resistors connected in parallel be R’
Then, 1 / R’ = 1 / 4 + 1 / 6
= (3 + 2) / 12
= 5 / 12 ohm or
R’ = 12 / 5
R’ = 2.4 ohm
Four resistors each of resistance 2 ohm are connected in parallel. What is the effective resistance?
R 2 = 2 ohm
R 3 = 2 ohm
R 4 = 2 ohm
1 / R = 1 / R 1 + 1 / R 2 + 1 / R 3 + 1 / R 4
1 / R = 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2
R = 0.5 ohm
You have three resistors of values 2 Ω, 3 Ω and 5 Ω . How will you join them so that the total resistance is less than 1 Ω? Draw diagram and find the total resistance.
To get a total resistance less than 1 Ω, the three resistors should be connected in parallel
Let the total resistance be R’
Then, 1 / R’ = 1 / 2 + 1 / 3 + 1 / 5
1 / R’ = (15 + 10 + 6) / 30
1 / R’ = 31 / 30 Ω or
R’ = 30 / 31
R’ = 0.97 Ω
Three resistors each of 2 W are connected together so that their total resistance is 3 W. Draw a diagram to show this arrangement and check it by calculation.
A parallel combination of two resistors, in series with one resistor.
1 / R’ = 1 / R 1 + 1 / R 2
1 / R’ = 1 / 2 + 1 / 2
R = R’ + R 3
Diagram is shown below
Calculate the equivalent resistance between the points A and B in figure if each resistance is 2.0 Ω
For a parallel resistances
R eff = (R 1 R 2 ) / (R 1 + R 2 )
R eff = (2 × 2) / (2 + 2)
R eff = 4 / 4
R eff = 1 Ω
Hence, total resistance = 2 + 2 + 1
A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.
Resistance of each set:
r 1 = 2 + 2 + 2 = 6 ohm
r 2 = 2 + 2 + 2 = 6 ohm
r 3 = 2 + 2 + 2 = 6 ohm
r 4 = 2 + 2 + 2 = 6 ohm
Now, the above resistances are arranged in parallel
1 / r = 1 / r 1 + 1 / r 2 + 1 / r 3 + 1 / r 4
1 / r = 1 / 6 + 1 / 6 + 1 / 6 + 1 / 6
1 / r = 4 / 6
r = 1.5 ohm
In the circuit shown below in figure, calculate the value of x if the equivalent resistance between the points A and B is 4 ohm
r 1 = 4 ohm
r 2 = 8 ohm
r 3 = x ohm
r 4 = 5 ohm
r’ = r 1 + r 2
r’ = 12 ohm
r’’ = r 3 + r 4
r’’ = (x + 5) ohm
1 / r = 1 / r’ + 1 / r’’
1 / 4 = 1 / 12 + 1 / (5 + x)
1 / 4 – 1 / 12 = 1 / (5 + x)
(3 – 1) / 12 = 1 / (5 + x)
2 / 12 = 1 / (5 + x)
1 / 6 = 1 / (5 + x)
Calculate the effective resistance between the points A and B in the circuit shown in figure
In above figure,
Resistance between XAY = (1 + 1 + 1)
Resistance between XY = 2 ohm
Resistance between XBY = 6 ohm
Let the net resistance between the points X and Y be R’
Then, 1 / R’ = 1 / 2 + 1 / 3 + 1 / 6
1 / R’ = (3 + 2 + 1) / 6
1 / R’ = 6 / 6
1 / R’ = 1 ohm or
Therefore, we can say that three 1 ohm resistors are connected in series between points A and B
Let the net resistance between points A and B be R AB
Then, R AB = (1 + 1 + 1) ohm
R AB = 3 ohm
A uniform wire with a resistance of 27 ohm is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.
Since the wire is divided into three pieces, the new resistance = 27 / 3 = 9
Now, three resistance are joined in parallel
1 / r = 1 / r 1 + 1 / r 2 + 1 / r 3
1 / r = 1 / 9 + 1 / 9 + 1 / 9
1 / r = 3 / 9
1 / r = 1 / 3
A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance of the circuit. Draw a diagram of the arrangement.
1 / r = 1 / 6 + 1 / 3
1 / r = 1 / 2
Calculate the effective resistance between the points A and B in the network shown below in figure.
For parallel resistance
1 / R = 1 / 12 + 1 / 6 + 1 / 4
1 / R = (1 + 2 + 3) / 12
1 / R = 6 / 12
Now, all the resistances are in series
R = 2 + 2 + 5
Calculate the equivalent resistance between the points A and B in figure
R 1 = 3 + 2 = 5 ohm
R 3 = 6 + 4 = 10 ohm
The resistors R 1 , R 2 and R 3 are connected in parallel
1 / R = 1 / 5 + 1 / 30 + 1 / 10
1 / R = (6 + 1 + 3) / 30
1 / R = 10 / 30
1 / R = 1 / 3
In the network shown in adjacent figure, calculate the equivalent resistance between the points (a) A and B (b) A and C
(a) R 1 = 2 + 2 + 2
R 1 = 6 ohm
R 1 and R 2 are connected in parallel
1 / R = 1 / 6 + 1 / 2
1 / R = (1 + 3) / 6
1 / R = 4 / 6
R = 1.5 ohm
(b) R 1 = 2 + 2
R 1 = 4 ohm
R 2 = 2 + 2
R 2 = 4 ohm
The resistors R 1 and R 2 are connected in parallel
1 / R = 1 / 4 + 1 / 4
1 / R = 2 / 4
1 / R = 1 / 2
Five resistors, each of 3 ohm, are connected as shown in figure. Calculate the resistance (a) between the points P and Q, and (b) between the points X and Y.
(a) R 1 = 3 + 3
R 2 = 3 ohm
1 / R = 1 / 6 + 1 / 3
1 / R = (1 + 2) / 6
1 / R = 3 / 6
(b) We know that R = 2 ohm from the above calculation
R 3 = 3 ohm
R 4 = 3 ohm
R’ = R + R 3 + R 4
R’ = 2 + 3 + 3
Two resistors of 2 ohm and 3 ohm are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
Now, I = V / R
I = 1.2 ohm
Here, R 1 and R 2 are connected in parallel
1 / R = 1 / 2 + 1 / 3
1 / R = (3 + 2) / 6
1 / R = 5 / 6
R = 1.2 ohm
I = 6 / 1.2
Therefore in series: 1.2 A and in parallel: 5 A
A resistor of 6 ohm is connected in series with another resistor of 4 ohm. A potential difference of 20 V is applied across the combination. (a) Calculate the current in the circuit and (b) potential difference across the 6 ohm resistor.
(a) To calculate current in the circuit
I = 20 / 10
(b) To calculate the potential difference across the 6 ohm resistor
Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
(a) Draw a labeled diagram of the arrangement
(b) Calculate current in each resistor.
(a) Circuit diagram
(b) Equivalent resistance of the circuit
1 / R = 1 / 4 + 1 / 6
1 / R = (3 + 2) / 12
1 / R = 5 / 12
R = 2.4 ohm
Thus, the e.m.f. of the cell is
V = 0.5 × 2.4
∴ Current through each resistor is
I 4 = V / R 4
I 4 = 1.2 / 4
I 4 = 0.3 A
I 6 = V / R 6
I 6 = 1.2 / 6
I 6 = 0.2 A
Hence, 0.3 A in 4 ohm and 0.2 A in 6 ohm
Calculate the current flowing through each of the resistors A and B in the circuit shown in figure?
For resistor A
For resistor B
Hence, current flowing in resistor A is 2 A and current flowing in resistor B is 1 A
In figure, calculate :
(a) the total resistance of the circuit.
(b) the value of R, and
(c) the current flowing in R
(a) To calculate the total resistance of the circuit
Total resistance R’ =?
R’ = 0.4 / 4
R’ = 10 ohm
(b) To calculate the value of R
R 1 = 20 ohm
1 / R’= 1 / R + 1 / R 1
1 / 10 = 1 / R + 1 / 20
1 / R = 1 / 10 – 1 / 20
1 / R = (2 – 1) / 20
1 / R = 1 / 20
(c) To calculate the current flowing in R
A particular resistance wire has a resistance of 3 ohm per meter. Find:
(a) The total resistance of three lengths of this wire each 1.5 m long, in parallel.
(b) The potential difference of the battery which gives a current of 2 A in each of the 1.5 m length when connected in the parallel to the battery (assume that resistance of the battery is negligible).
(c) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.
(a) Resistance of wire per meter = 3 ohm
So, resistance of three lengths of this wire each 1.5 m long = 3 × 1.5 = 4.5 W
1 / R = 1 / 4.5 + 1 / 4.5 + 1 / 4.5
1 / R = 3 / 4.5
(b) I = 2 A
V = 2 × 4.5
(c) R = 3 ohm for 1 meter wire
Here the area is twice i.e 2 A and Resistance is inversely proportional to area. Thus resistance becomes half
R = 7.5 ohm
A cell supplies a current of 1.2 A through two 2 ohm resistors connected in parallel. When resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.
In parallel R = 1 / 2 + 1 / 2 = 1 ohm
ε = 1.2 (1 + r)
ε = 1.2 + 1.2r
ε = 0.4 (4 + r)
ε = 1.6 + 0.4r
This means:
1.2 + 1.2r = 1.6 + 0.4r
1.2r – 0.4r = 1.6 – 1.2
r = 0.4 / 0.8
r = 0.5 ohm
(i) Internal resistance r = 0.5 ohm
(ii) ε = I(R + r)
ε = 1.2 (1 + 0.5)
A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors 3 ohm and 6 ohm connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 ohm resistor (d) the current in 6 ohm resistor.
(a) In parallel
1 / R = 1 / 3 + 1 / 6
1 / R = (2 + 1) / 6
15 = I (2 + 3)
(c) V = 6 V
(d) R = 6 ohm
The circuit diagram in figure shows three resistors 2 ohm, 4 ohm and R ohm connected to a battery of e.m.f. 2 V and internal resistance 3 ohm. If main current of 0.25 A flows through the circuit, find:
(a) the p.d. across the 4 ohm resistor
(b) the p.d. across the internal resistance of the cell,
(c) the p.d. across the R ohm or 2 ohm resistor, and
(d) the value of R.
(a) To calculate the p.d. across the 4 ohm resistor
V = 0.25 × 4
(b) To calculate the p.d. across the internal resistance of the cell
Internal resistance r = 3 ohm
V = 0.25 × 3
(c) To calculate the p.d. across the 2 ohm resistor
Effective resistance of parallel combination of 2 ohm resistances = 1 ohm
V = 0.25 / 1
(d) To calculate the value of R
ε = I (R’ + r)
2 = 0.25 (R’ + 3)
Three resistors of 6.0 ohm, 2.0 ohm and 4.0 ohm are joined to an ammeter A and a cell of emf 6.0 V as shown in figure. Calculate:
(a) the efective resistance of the circuit.
(b) the reading of ammeter
(a) R 1 = 6 W
R’ = R 2 + R 3
R 1 and R’ are connected in parallel
1 / R = 1 / R 1 + 1 / R’
1 / R = 1 / 6 + 1 / 6
1 / R = 2 / 6
(b) R = 3 ohm
The diagram below in Fig., shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8 V. Calculate:
a. The total resistance of the circuit
b. The reading of ammeter A.
(a) In the above figure,
Let R XY be the resistance between X and Y
Then, 1 / R XY = 1 / 10 + 1 / 40
1 / R XY = (4 + 1) / 40
1 / R XY = 5 / 40 ohm
Or R XY = 8 ohm
Then, 1 / R AB = 10 ohm
∴ The total resistance of the circuit = 8 ohm + 10 ohm
(b) Current I = Voltage / Total resistance
I = 1.8 / 18 A
Hence, the reading of ammeter is 0.1 A
Question: 31
A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in fig.
(a) The reading of the ammeter,
(b) The potential difference across the terminals of the cells, and
(c) The potential difference across the 4.5 ohm resistor.
The total resistance of the circuit is
R eq = R cell + R ammeter + R 1 || R 2
∴ R eq = 1.2 + 0.8 + (R 1 R 2 ) / R 1 + R 2
∴R eq = 2 + (4.5 × 9) / 4.5 + 9
R eq = 2 + 40.5 / 13.5
∴R eq = 5 ohm
(a) The current through the ammeter is
I = E cell / R eq
(b) The potential difference across the ends of the cells is
V cell = E cell – IR cell
V cell = 2 – 0.4 × 1.2
V cell = 2 – 0.48
∴V cell = 1.52 V
(c) The potential difference across the 4.5 ohm resistor is
V 4.5 = V cell – V ammeter
V 4.5 = 1.52 – 0.4 × 0.8
V 4.5 = 1.52 – 0.32
∴ V 4.5 = 1.2 V
Exercise 8(C) Page No: 211
Write an expression for the electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time.
The expression for the electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time is
Electrical energy, W = I 2 Rt joule
Write an expression for the electrical power spent in flow of current through a conductor in terms of (a) resistance and potential difference, (b) current and resistance.
(a) Expression for electrical power spent in flow of current through a conductor in terms of resistance and potential difference is
Electrical Power, P = V 2 / R
(b) Expression for electrical power spent in flow of current through a conductor in terms of current and resistance is
Electrical Power, P = I 2 R
Electrical power P is given by the expression P = (Q × V) ÷ time
(a)What do the symbols Q and V represent?
(b)Express the power P in terms of current and resistance explaining the meanings of symbols used there in.
(a) The symbol Q represents Charge and the symbol V represents Voltage
(b) Electrical Power, P = I 2 R
-where I represents current and R represents resistance
Name the S.I. unit of electrical energy. How is it related to Wh?
Joule is the S.I. unit of electrical energy. It is related to Wh as
1 Wh = 3600 J
Explain the meaning of the statement ‘the power of an appliance is 100 W’.
The power of an appliance is 100 W. This means that 100 J of electrical energy is consumed by the appliance in 1 second.
State the S.I. unit of electrical power.
Watt is the S.I. unit of electrical power.
(i)State and define the household unit of electricity.
(ii)What is the voltage of the electricity that is generally supplied to a house?
(iii) What is consumed while using different electrical appliances, for which electricity bills are paid?
(i) Kilowatt hour (kWh) is the household unit of electricity. The electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour is One kilowatt hour (kWh)
(ii) The voltage of the electricity that is generally supplied to a house is 220 volt.
(iii) The electrical energy is consumed while using different electrical appliances, for which electricity bills are paid
Name the physical quantity which is measured in (i) kW, (ii) kWh. (iii) Wh
(i) The quantity which is measured in kW is electrical power
(ii) The quantity which is measured in kWh is electrical energy
Define the term kilowatt – hour and state its value in S.I. unit.
The electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour is one kilowatt hour (kWh). Its value in S.I. unit is 1kWh = 3.6 × 10 6 J
How do kilowatt and kilowatt-hour differ?
Kilowatt- hour is the unit of electrical energy whereas the kilowatt is the unit of electrical power.
Complete the following:
(a) 1 kWh = (1 volt × 1 ampere × ……..) / 1000
1 kWh= ________ J
(a) 1 kWh = (1 volt × 1 ampere × 1 hour) / 1000
(b) 1 kWh = 3.6 × 10 6 J
What do you mean by power rating of an electrical appliance? How do you use it to calculate (a) the resistance of the appliance and (b) the safe limit of the current in it, while in use?
The power rating is an electrical appliance such as electric bulb, geyser etc. is rated with power (P) and voltage (V). If an electric bulb is rated as 50 W – 220 V, it means that when the bulb is lighted on a 220 V supply, 50 W electrical power is consumed by it.
(a) To calculate the resistance of the appliance, the expression is:
Resistance, R = V 2 / P
(b) The safe limit of current in it, while in use is
An electric bulb is rated ‘100 W, 250 V’. What information does this convey?
An electric bulb is rated ‘100 W, 250 V’ means that if the bulb is lighted on a 250 V supply, it consumes 100 W electric power (which means 100 J of electrical energy is converted into the light and heat energy in 1 second in the filament).
List the names of three electrical gadgets used in your house. Write their power, voltage rating and approximate time for which each one is used in a day. Hence find the electrical energy consumed by each in a month of 30 days.
Two lamps, one rated 220 V, 50 W and the other rated 220 V, 100 W, are connected in series with mains of voltage 220 V. Explain why does the 50 W lamp consume more power.
Resistance of 220 V, 50 W lamp is
R 1 = V 2 / P 1
R 1 = (220) 2 / 100
R 1 = 968 ohm
Resistance of 220 V, 100 W lamp is
R 2 = V 2 / P 2
R 2 = (220) 2 / 100
R 2 = 484 ohm
Same current I passes through each lamp because the two lamps are connected in series.
Power consumed in 220 V, 50 W lamp is
P 1 = I 2 R 1
Power consumed in 220 V, 100 W lamp is
P 2 = I 2 R 2
Here, R 1 > R 2
So, P 1 > P 2
Thus 50 W lamp consumes more power
Name the factors on which the heat produced in a wire depends when current is passed in it, and state how does it depend on the factors stated by you.
The amount of heat produced in a wire on passing current through it, depends on the following three factors.
(i) The amount of current passing through the wire
(ii) The resistance of wire and
(iii) The time for which current is passed in the wire
(i) Dependence of heat produced on the current in wire: The amount of heat H produced in the wire is directly proportional to the square of current I passing through the wire, i.e., H ∝ I 2
(ii) Dependence of heat produced on the resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R
(iii) Dependence of heat produced on the time: The amount of heat H produced in a wire is directly proportional to the time t for which current is passed in the wire i.e., H ∝ t
When a current I flows through a resistance R for time t , the electrical energy spent is:
(d) I 2 R / t
The electrical energy spent when a current I flows through a resistance R for time t is I 2 Rt
An electrical appliance has a rating 100 W, 120 V. The resistance of element of appliance when in use is:
(a) 1.2 ohm
(b) 144 ohm
(c) 120 ohm
(d) 100 ohm
The resistance of element of appliance when in use is 144 ohm
An electric bulb of resistance 500 ohm draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at its end.
Resistance of electric bulb (R) = 500 ohm
Current drawn from the source (I) = 0.4 A
(a) Power of the bulb (P) = VI
V = 0.4 × 500
(b) The potential difference at its end is 200 V
Power (P) = VI
P = 200 × 0.4
Hence, the power of the bulb is 80 Watt
A current of 2 A is passed through a coil of resistance 75 Ω for 2 minutes. (a) How much heat energy is produced? (b) How much charge is passed through the resistance?
Current (I) = 2 A
Resistance, R = 75 Ω
Time, t = 2 min or 120 s
(a) Heat produced, H = I 2 Rt or
H = (2) 2 (75) (120) J = 36000 J
(b) Charge passed through the resistance, Q = It or
Q = (2) (120) C
Calculate the current through a 60 W lamp rated for 250 V. If the line voltage falls to 200 V, how is power consumed by the lamp affected?
Power, P = 60 W
Voltage, V = 250 V
Power, P = VI
I = 60 / 250
Resistance of lamp R = V 2 / P
R = (250) 2 / 60
R = 1041.6 ohm
If voltage falls to 200 V, then the power consumed will be
P = V 2 / R
P = (200) 2 / 1041.6
Hence, consumed reduces to 38.4 W
An electric bulb is rated ‘100 W, 250 V’. How much current will the bulb draw if connected to a 250 V supply?
Power, P = 100 W
I = 100 / 250
An electric bulb is rated at 220 V, 100 W. (a) What is its resistance? (b) What safe current can be passed through it?
Voltage, V = 220 V
Power, P = V 2 / R
R = (220) 2 / 100
R = 484 ohm
(b) The safe limit of current that can be passed through it is
I = 100 / 220
A bulb of power 40 W is used for 12.5 h each day for 30 days. Calculate the electrical energy consumed.
Energy consumed for each day, E = P × t
E = 40 × 12.5
Energy consumed for 30 days
E = 500 × 30
E = 15000 Wh
An electric press is rated ‘750 W, 230 V’. Calculate the electrical energy consumed by the press in 16 hours
Energy, E = Power × time
E = 750 × 16
E = 12000 Wh
An electrical appliance having a resistance of 200 ohm is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes (i) in joule, (ii) in kWh
Resistance, R = 200 ohm
Voltage, V = 200 V
Time, t = 5 minutes
t = 5 × 60 sec
t = 300 sec
As we know,
Energy, E = V 2 t / R
(i) In joules
E = [(200) 2 × 300] / 200
E = 60000 J
(ii) In kWh
As 1 kWh = 3.6 × 10 6 J
1 J = 1 / 3.6 × 10 6 kWh
60000 J = 60000 / 3.6 × 10 6
60000 J = 0.0167 kWh
A bulb rated 12 V, 24 W operates on a 12 volt battery for 20 minutes. Calculate:
(i) the current flowing through it, and
(ii) the energy consumed.
Power, P = 24 W
Voltage, V = 12 V
Current, I =?
(i) The current flowing through it is
I = 24 / 12
(ii) Energy, E = P × t
E = 24 × 20 × 60
E = 28,800 J
A current of 0.2 A flows through a wire whose ends are at a potential difference of 15 V. Calculate:
(i)The resistance of the wire, and
(ii)The heat energy produced in 1 minute.
Current, I = 0.2 A
Potential difference, V = 15 V
Time, t = 60 sec
(a) To calculate the resistance of the wire
R = 15 / 0.2
(b) To calculate the heat energy produced in 1 minute
Heat energy, H = I 2 Rt
H = (0.2) 2 × 75 × 60
What is the resistance, under normal working conditions, of an electric lamp rated at ‘240 v’, 60 W? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.
Voltage, V = 240 V
R = V 2 / P
R = (240) 2 / 60
R = 960 ohm
I = 60 / 240
Thus when one lamp is connected across the mains, it draws 0.25 A current. If two such lamps are connected in series across the mains, current through each bulb becomes
Two bulbs are rated 60 W, 220 V and 60 W, 110 V, respectively. Calculate the ratio of their resistances.
Voltage, V 1 = 220 V
V 2 = 110 V
Power, P 1 = P 2 = P = 60 W
We know that, R = V 2 / P
R 1 = V 1 2 / P
R 1 = (220) 2 / 60
R 2 = V 2 2 / P
R 2 = (110) 2 / 60
Now, dividing R 1 and R 2 we get,
R 1 / R 2 = 4 / 1
R 1 : R 2 = 4: 1
An electric bulb is rated 250 W, 230 V.
(i) the energy consumed in one hour, and
(ii) the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains?
Power, P = 250 W
Voltage, V = 230 V
(i) Energy, E = P × t
Time, t = 1 × 60 × 60
t = 3600 sec
E = 250 × 3600
E = 9 × 10 5 J
(ii) 1000 Wh = 250 × t
time, t = 1000 / 250
t = 4 hours
Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. Calculate:
(i)The total current taken from the supply,
(ii)The resistance of each heater, and
(iii)The energy supplied in kWh to the three heaters in 5 hours.
Voltage, V = 100 V
(i) Current through each heater, I =?
I = 250 / 100
∴ Current taken for the three heaters
(ii) Resistance for each heater, R = V / I
R = 100 / 2.5
(iii) Time for which energy is supplied, t = 5 h
Energy, E = P × t
E = 250 × 5
E = 1250 Wh
E = 1.25 kWh
Energy for three heaters = 3 × 1.25
A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 ohm. A steady current of 0.5 A flows through the circuit. Calculate:
(i)The total energy supplied by the battery in 10 minutes,
(ii)The resistance of the bulb, and
(iii)The energy dissipated in the bulb in 10 minutes.
Voltage, V = 4 V
Resistance of the battery, R B = 2.5 ohm
Current, I = 0.5 A
(i) Energy supplied by the battery, E = V 2 t / R
t = 10 × 60
t = 600 sec
R = 4 / 0.5
E = [(4) 2 × 600] / 8
(ii) Total resistance, R = 8 ohm
Resistance of the bulb, R b = 8 – 2.5 ohm
R b = 5.5 ohm
(iii) Energy dissipated in the bulb in 10 min, E = I 2 Rt
E = (0.5) 2 × 5.5 × 600
Two resistors A and B of 4 ohm and 6 ohm, respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, (ii) the power dissipated in each resistor.
Resistance, R A = 4 ohm
Resistance, R B = 6 ohm
Voltage, V = 6 V
(i) Since the resistances are connected in parallel
Equivalent Resistance, 1 / R = 1 / R A + 1 / R B
1 / R = 10 / 24
P = (6) 2 / 2.4
(ii) Power dissipation across each resistor, P = VI
Current across resistor R A , I A = V / R A
I A = 6 / 4 = 1.5 A
Power dissipation across resistor R A ,
P = 6 × 1.5
(iii) Current across resistor R B , I B = V / R B
I B = 6 / 6
Power dissipation across resistor R B ,
A battery of e.m.f. 15 V and internal resistance 2 ohm is connected to two resistors of resistances 4 ohm and 6 ohm joined in series. Find the electrical energy spent per minute in 6 ohm resistor.
e.m.f. of battery, V = 15 V
Internal resistance of battery, R B = 2 ohm
Resistances given in circuit,
R 1 = 4 ohm and R 2 = 6 ohm
(i) When resistors are connected in series,
Equivalent resistance, R = R B + R 1 + R 2
Current in the circuit, I = 15 / 12
Voltage across resistor R 2 , V 2 = IR = 1.25 × 6
V 2 = 7.50 V
Time, t = 1 min = 60 sec
Energy across R 2 , E = V 2 t / R
E = [(7.5) 2 × 60] / 6
E = 562.5 J
Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point. How long will it take if the supply had been of 200 V?
Heat gained = (V 2 / R) × t
(V 1 2 / R) × t 1 = (V 2 2 / R) × t 2
t 2 = (V 1 / V 2 ) 2 × t 1
t 2 = (220 / 200) 2 × 300
t 2 = 363 s
t 2 = 6.05 min
An electric toaster draws current 8 A in a circuit with source of voltage 220 V. It is used for 2 h. Find the cost of operating the toaster if the cost of electrical energy is ₹ 4.50 per kWh
Current, I = 8 A
Time, t = 2 h
Energy, E = VIt
E = 220 × 8 × 2
E = 3520 Wh
E = 3.52 kWh
Cost of energy = ₹ 4.50 per kWh
∴ Cost of 3.52 kWh of energy = ₹ 4.50 × 3.52 kWh
An electric kettle is rated 2.5 kW, 250 V’. Find the cost of running the kettle for two hours at ₹ 5.40 per unit.
Power of kettle, P = 2.5 kW
As, Energy, E = P × t
E = 2.5 × 2
Cost per unit of energy = ₹ 5.40
Cost for 5 kWh of energy = 5.40 × 5 = ₹ 27
A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. Calculate:
(i)The current drawn,
(ii)The energy consumed in 50 hours, and
(iii)The cost of energy consumed at ₹ 4.20 per kWh.
Power of geyser, P = 1500 W
(i) Current, I = P / V
I = 1500 / 250
(ii) Time, t = 50 h
E = 1500 × 50
E = 75000 Wh
(iii) Cost per unit of energy = ₹ 4.20
Cost for 75 kWh of energy = 4.20 × 75 = ₹ 315
Selina Solutions Concise Physics Class 10 Chapter 8 Current Electricity discusses different types of sources of Current. Some of the important points covered in this chapter are concept of charge, concept of current, concept of potential and potential difference, concept of resistance, ohm’s law. Students are provided questions, numerical problems at the end of the chapter in order to assess their knowledge
List of subtopics covered in Selina Solutions Concise Physics Class 10 Chapter 8 Current Electricity
List of Exercises
Key Features of Selina Solutions Concise Physics Class 10 Chapter 8 Current Electricity
- Selina Solutions are readily accessible for students
- Use of graphs and tables, wherever necessary
- Numerical problems are solved with precision
- After extensive research, Selina Solutions have been solved by our subject experts
The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2023-24 will be updated soon.
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Solved Examples of Comlicated Circuits
Illustration:.
Let us analyse a simple circuit shown in the figure alongside. Assume current values (I 1 , I 2 & I 3 ) at random directions.
Alt txt: simple circuit
Solutions
Þ All through the branch gfdab current in I 1
All through the branch geb current is I 3
All through the branch ghcb current is I 2
Applying KCL at b Þ I 1 + I 2 + I 3 = 0 …… (i)
(Note: We will get the same eqn. at node g)
The voltage drops across the circuit elements in loop fdbegf,
V d – V f = 10V (constant voltage source)
V d – V a = 1 × I 1 (drop in the direction of current flow)
V e – V b = 2 × I 3 (drop in the direction of current flow)
V g – V e = 15V (constant voltage source)
Moving anti clockwise we write,
KVL: (V f – V d )+(V d – V a )+(V a – V b )+(V b – V e )+(V e – V g )+(V g – V f ) = 0
Þ (–10) + (I 1 ) + (0) + (–2I 3 ) + (–15) + 0 = 0
Þ I 1 – 2 I 3 = 25 …… (ii)
Consider loop bchgeb. The voltage drop across the elements is,
V h – V c = 4I2
V e – V b = 2I3
V g – V e = 15
KVL: (V b – V c ) + (V c – V h ) + (V h – V g ) + (V g – V e ) + (V e – V b ) = 0
Þ (0) + (–4I 2 ) + (0) + (15) + 2I 3 = 0
Þ 2I 3 – 4I 2 = –15 …… (iii)
Now solve (i), (ii), & (iv) simultaneously, you shall arrive at,
I 1 = 120/14, I 2 = 5 – 5/14, I 3 = –115/14
Illustration:
What is the potential difference between the points M and N for the circuits shown in the figures, for case l and case ll?
Alt txt : Potential-difference-between-points
Solution:
l = E 1 – E 2 / r 2 + r 1 = 12 – 6 / 3 + 2 = 1.2 A
Alt txt: circuit-1
For cell E 1 : v A – E 1 + lr 1 = v B
i.e. v A – v B = E 1 – lr 1
= 12 – 1.2 × 3 = 8.4 V
For cell E 2 , v C – E 2 – lr 2 = v D
i.e. v C – v D = 6 + 1.2 × 2 = 8.4 V
Hence, v C – v D = v A – v B = v M – v N = 8.4 V
l = E 1 + E 2 / r 1 + r 2
= 12 + 6 / 3 + 2 = 3.6 A
Alt txt: Circuit-2
For cell E 1 :
v A – E 1 + lr 1 = v B ,
i.e. v A – v B = E 1 – lr 1 = 12 – 3.6 × 3 = 1.2 V
For cell E 2 :
v C + E 2 – lr 2 = v D
i.e. v C – v D = –E 2 + lr 2 = – 6 + 3.6 × 2 = 1.2 V
Hence, v A – v B = v C – v D = v M – v N = 1.2 V
In the adjacent circuit, find the effective resistance between the points A and B.
Alt txt: effective resistance between two points
Resistors AF and FE are in series with each other. Therefore, network AEF reduces to a parallel combination of two resistors of 6 W each.
R eq = 6 × 6 / 6 + 6 = 3 W.
Similarly, the resistance between A and D is given by 6 × 6 / 6 + 6 = 3 W.
Now, resistor AC is in parallel with the series combination of AD and DC. Therefore, resistance between A and C is 6 × 6 / 6 + 6 = 3 W. Now, the combination of (AC + CB) is in parallel to AB. Therefore, since AC and CB are in series, their combined resistance = 3 + 3 = 6 W. Resistance between A and B is given by, 1 / R = 1/6 + 1/3 = 3/6 or R AB = 2W.
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