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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!
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- 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

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

(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

## 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 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?

## Finding least count

The smallest value that can be measured by a measuring instrument is called its least count.

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

## 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

## 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

There are 10 divisions between 0 and 0.5A

Hence, 10 divisions measure 0.5A current

Least count of ammeter is 0.05 A

According to the question, The ammeter needle is pointing on the 17th divison

The ammeter reading is 0.85 A.

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## How to Solve Circuit Problems

Last Updated: December 24, 2022

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## Physics - Current Electricity: Exercises and Example Solved Numerical problems | 12th Physics : Current Electricity

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

## EXAMPLE 2.2

E = 570 N C -1 , e = 1.6 × 10 -19 C, m = 9.11 × 10 -31 kg and a = ?

a = eE/ m = 570×1 .6×10 −19 /9 .11×10 -31

## EXAMPLE 2.3

The relation between drift velocity of electrons and current in a wire of cross-sectional area A is

## EXAMPLE 2.4

Charge of an electron, e = 1.6 × 10 -19 C

The number of electrons flowing per second, n =?

n = 20 × 10 19 = 2 × 10 20 electrons

## OHM’S LAW: Solved Example Problems

A potential difference across 24 Ω resistor is 12 V. What is the current through the resistor?

From Ohm’s law, I = V/R = 12/24 = 0 .5 A

## Resistivity: Solved Example Problems

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 ]

Though the wire is stretched, its volume is unchanged.

A 1 l 1 = A 2 l 2 , A 1 l = A 2 8 l

By dividing equation R 2 by equation R 1 , we get

Substituting the value of A 1 /A 2 , we get

Hence, stretching the length of the wire has increased its resistance.

Consider a rectangular block of metal of height A, width B and length C as shown in the figure.

In the first case, the resistance of the block

## Resistors in series and parallel: Solved Example Problems

Since the resistors are connected in series, the effective resistance in the circuit

The Current I in the circuit= V/ R eq = 24/10 = 2 .4 A

V 1 = IR 1 = 2 . 4 A× 4 Ω = 9.6V

V 2 = IR 1 = 2 . 4 A× 6 Ω =14 .4V

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

The above equation can be solved using factorisation.

R 1 (R 1 – 8) – 7 (R 1 – 8) = 0

R 2 = 7 Ω i.e , (when R 1 = 8 Ω ; R 2 = 7 Ω)

R 2 = 8 Ω , i.e , (when R 1 = 8 Ω ; R 2 = 7 Ω )

Calculate the equivalent resistance between A and B in the given circuit.

The equivalent resistance of the circuit between a and b is R eq = 1Ω

## Temperature dependence of resistivity: Solved Example Problems

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

T 0 = 10ºC, T = 40ºC, R 0 = 45 Ω , R = 85 Ω

## Energy and Power in Electrical Circuits: Solved Example Problems

To check which bulb will be fused, the voltage drop across each bulb has to be calculated.

R tot = ( 484 +2420) Ω = 2904Ω

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 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

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

## Cells in series: Solved Example Problems

i) Equivalent emf of the combination

ii) Equivalent internal resistance

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 +0 .4] = 36 /10.4

v) Potential difference across each cell V/ n = 34.6/4 = 8 .65V

## Cells in parallel: Solved Example Problems

iv) Potential difference across each cell

ii) Equivalent internal resistance,

iv) Potential difference across each cell V = IR = 0.5 × 10 = 5 V

v) Current from each cell, I ′ = I/n

## Kirchhoff’s first rule (Current rule or Junction rule): Solved Example Problems

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

## Kirchhoff’s Second rule (Voltage rule or Loop rule) : Solved Example Problems

Thus applying Kirchoff’s second law to the closed loop EACE

I 1 R 1 + I 2 R 2 + I 3 R 3 = ξ

I 4 R 4 + I 5 R 5 - I 2 R 2 = 0

Calculate the current that flows in the 1 Ω resistor in the following circuit.

Now consider the loop EFCBE and apply KVR, we get

Applying KVR to the loop EADFE, we get

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

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

## Meter bridge : Solved Example Problems

## Heating Effect of Electric Current, Joule’s law: Solved Example Problems

R = 10 Ω, I = 5 A, t = 5 minutes = 5 × 60 s

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 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)

The time required to produce this heat energy t = Q/ I 2 R = 126 ×10 3 / 4840 ≈ 26 .03 s

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## 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

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.

Define the term electric potential. State it’s S.I. unit.

How is the electric potential difference between the two points defined? State its S.I. unit.

Explain the statement ‘the potential difference between two points is 1 volt’.

(a) State whether the current is a scalar or vector? What does the direction of current convey?

Define the term resistance. State its S.I. unit.

(a) Name the particles which are responsible for the flow of current in a metallic wire.

(c) What is the cause of resistance offered by the metallic wire in the flow of current through it?

Q (charge) = n × e (charge on an electron)

(c) A metal contain free electrons and fixed positive ions.

(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) 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.

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.

= 1 / resistance of the conductor

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

Examples: LED, solar cell, junction diode, etc.

V vs I for non-ohmic conductors

Give two differences between an ohmic and non-ohmic resistor

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

(a) How does the resistance of a wire depend on its radius? Explain your answer.

Resistance of a wire varies inversely as the area of cross section of the wire i.e.,

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.

How does the resistance of a metallic wire depend on its temperature? Explain with reason.

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

Define the term specific resistance and state its S.I. unit.

ρ = specific resistance of the material of conductor

A = area of cross section of conductor

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

(a) Name two factors on which the specific resistance of a wire depends?

(a) Two factors on which the specific resistance of a wire depends are

(i) Material of the substance and

(ii) Temperature of the substance

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.

Name the material used for making a fuse wire. Give a reason.

Which of the following is an ohmic resistance?

An ohmic resistance is nichrome wire

For which of the following substances, resistance decreases with increase in temperature?

Resistance decreases with increase in temperature for carbon

Number of electrons flowing through the conductor,

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

Therefore I = [(6.25 × 10 16 ) (1.6 × 10 -19 )] / 2

Hence, 5 mA current flows from B to A

Charge, Q = -1.6 × 10 -19 coulomb

Number of electrons = 1.6 × 10 -3 / 1.6 × 10 -19

Find the potential difference required to flow a current of 200 mA in a wire of resistance 20 ohm.

Potential difference or Voltage V = 6.0 V

Potential difference or Voltage V = 12 V

Hence, when bulb is not glowing, resistance will be less

Potential difference or Voltage V = 3 V

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.

(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.

ρ (l / πr 1 2 ) : ρ (l / πr 2 2 )

A given wire of resistance 1 Ohm is stretched to double its length. What will be its new resistance?

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’

The new length l’ = 30 cm = 3 × l

Stretching length will increase and area of cross section will decrease in same order

A wire of resistance 9 Ohm having length 30 cm is tripled on itself. What is its new resistance?

Area of cross section will also change in same order with change in length

Specific resistance = 1.7 × 10 -8 ohm m

Radius r = 1 mm that is 10 -3 m

l = (1 × π × 10 -6 ) / (1.7 × 10 -8 )

l = (1 × 3.14 × 10 -6 ) / (1.7 × 10 -8 )

Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of cell.

State two differences between the e.m.f. and terminal voltage of a cell.

The factors on which the internal resistance of a cell depends are:

(a) The total resistance of circuit = R + r

(b) The current drawn from the cell

(c) p.d. across the cell: [ε / (R + r)] × R

(d) voltage drop inside the cell: [ε / (R + r)] × r

(a) When current is drawn from a cell, its terminal voltage V is less than its e.m.f.

(b) When no current is drawn, then the e.m.f. is equal to the terminal voltage.

(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

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,

Thus total resistance in series is

Applying Ohm’s law separately to the two resistors, we have

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

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

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

Series resistance of two 1 ohm resistors,

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.

(i) Because of no current, ammeter reading = 0

(a) What would be the reading of the ammeter?

(b) What is the potential difference across the terminals of the cell?

Total resistance = 2 + 4.5 + 0.7

Now, excluding internal resistance total resistance = 4.5 + 0.7

(a) the current through the battery

(b) the p.d. between the terminals of the battery.

Now, substituting the value of r

Let the equivalent resistance of the 4 ohm and 6 ohm resistors connected in parallel be R’

Four resistors each of resistance 2 ohm are connected in parallel. What is the effective resistance?

1 / R = 1 / R 1 + 1 / R 2 + 1 / R 3 + 1 / R 4

1 / R = 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2

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

A parallel combination of two resistors, in series with one resistor.

Calculate the equivalent resistance between the points A and B in figure if each resistance is 2.0 Ω

R eff = (R 1 R 2 ) / (R 1 + R 2 )

Hence, total resistance = 2 + 2 + 1

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

Calculate the effective resistance between the points A and B in the circuit shown in figure

Resistance between XAY = (1 + 1 + 1)

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

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

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

Calculate the effective resistance between the points A and B in the network shown below in figure.

1 / R = 1 / 12 + 1 / 6 + 1 / 4

Now, all the resistances are in series

Calculate the equivalent resistance between the points A and B in figure

The resistors R 1 , R 2 and R 3 are connected in parallel

1 / R = 1 / 5 + 1 / 30 + 1 / 10

R 1 and R 2 are connected in parallel

The resistors R 1 and R 2 are connected in parallel

(b) We know that R = 2 ohm from the above calculation

Here, R 1 and R 2 are connected in parallel

Therefore in series: 1.2 A and in parallel: 5 A

(a) To calculate current in the circuit

(b) To calculate the potential difference across the 6 ohm resistor

(a) Draw a labeled diagram of the arrangement

(b) Calculate current in each resistor.

(b) Equivalent resistance of the circuit

Thus, the e.m.f. of the cell is

∴ Current through each resistor is

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?

Hence, current flowing in resistor A is 2 A and current flowing in resistor B is 1 A

(a) the total resistance of the circuit.

(a) To calculate the total resistance of the circuit

(b) To calculate the value of R

(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.

(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

(c) R = 3 ohm for 1 meter wire

In parallel R = 1 / 2 + 1 / 2 = 1 ohm

(i) Internal resistance r = 0.5 ohm

(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

(a) To calculate the p.d. across the 4 ohm resistor

(b) To calculate the p.d. across the internal resistance of the cell

(c) To calculate the p.d. across the 2 ohm resistor

Effective resistance of parallel combination of 2 ohm resistances = 1 ohm

(d) To calculate the value of R

(a) the efective resistance of the circuit.

R 1 and R’ are connected in parallel

a. The total resistance of the circuit

Let R XY be the resistance between X and Y

Then, 1 / R XY = 1 / 10 + 1 / 40

∴ The total resistance of the circuit = 8 ohm + 10 ohm

(b) Current I = Voltage / Total resistance

Hence, the reading of ammeter is 0.1 A

(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

(a) The current through the ammeter is

(b) The potential difference across the ends of the cells is

(c) The potential difference across the 4.5 ohm resistor is

Electrical energy, W = I 2 Rt joule

Electrical power P is given by the expression P = (Q × V) ÷ time

(a)What do the symbols Q and V represent?

(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

Explain the meaning of the statement ‘the power of an appliance is 100 W’.

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?

(ii) The voltage of the electricity that is generally supplied to a house is 220 volt.

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.

How do kilowatt and kilowatt-hour differ?

(a) 1 kWh = (1 volt × 1 ampere × ……..) / 1000

(a) 1 kWh = (1 volt × 1 ampere × 1 hour) / 1000

(a) To calculate the resistance of the appliance, the expression is:

(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?

Resistance of 220 V, 50 W lamp is

Resistance of 220 V, 100 W lamp is

Same current I passes through each lamp because the two lamps are connected in series.

Power consumed in 220 V, 50 W lamp is

Power consumed in 220 V, 100 W lamp is

Thus 50 W lamp consumes more power

(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

When a current I flows through a resistance R for time t , the electrical energy spent is:

The electrical energy spent when a current I flows through a resistance R for time t is I 2 Rt

The resistance of element of appliance when in use is 144 ohm

Resistance of electric bulb (R) = 500 ohm

Current drawn from the source (I) = 0.4 A

(a) Power of the bulb (P) = VI

(b) The potential difference at its end is 200 V

Hence, the power of the bulb is 80 Watt

(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

Resistance of lamp R = V 2 / P

If voltage falls to 200 V, then the power consumed will be

Hence, consumed reduces to 38.4 W

(b) The safe limit of current that can be passed through it is

Energy consumed for each day, E = P × t

A bulb rated 12 V, 24 W operates on a 12 volt battery for 20 minutes. Calculate:

(i) the current flowing through it, and

(i) The current flowing through it is

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.

Potential difference, V = 15 V

(a) To calculate the resistance of the wire

(b) To calculate the heat energy produced in 1 minute

Now, dividing R 1 and R 2 we get,

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?

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.

(i) Current through each heater, I =?

∴ Current taken for the three heaters

(ii) Resistance for each heater, R = V / I

(iii) Time for which energy is supplied, t = 5 h

Energy for three heaters = 3 × 1.25

(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.

Resistance of the battery, R B = 2.5 ohm

(i) Energy supplied by the battery, E = V 2 t / R

(ii) Total resistance, R = 8 ohm

Resistance of the bulb, R b = 8 – 2.5 ohm

(iii) Energy dissipated in the bulb in 10 min, E = I 2 Rt

(i) Since the resistances are connected in parallel

Equivalent Resistance, 1 / R = 1 / R A + 1 / R B

(ii) Power dissipation across each resistor, P = VI

Current across resistor R A , I A = V / R A

Power dissipation across resistor R A ,

(iii) Current across resistor R B , I B = V / R B

Power dissipation across resistor R B ,

Internal resistance of battery, R B = 2 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

Energy across R 2 , E = V 2 t / R

(V 1 2 / R) × t 1 = (V 2 2 / R) × t 2

Cost of energy = ₹ 4.50 per kWh

∴ Cost of 3.52 kWh of energy = ₹ 4.50 × 3.52 kWh

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:

(ii)The energy consumed in 50 hours, and

(iii)The cost of energy consumed at ₹ 4.20 per kWh.

(iii) Cost per unit of energy = ₹ 4.20

Cost for 75 kWh of energy = 4.20 × 75 = ₹ 315

List of subtopics covered in Selina Solutions Concise Physics Class 10 Chapter 8 Current Electricity

## Key Features of Selina Solutions Concise Physics Class 10 Chapter 8 Current Electricity

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## Solved Examples of Comlicated Circuits

## 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

Consider loop bchgeb. The voltage drop across the elements is,

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

Now solve (i), (ii), & (iv) simultaneously, you shall arrive at,

I 1 = 120/14, I 2 = 5 – 5/14, I 3 = –115/14

## Illustration:

Alt txt : Potential-difference-between-points

## Solution:

l = E 1 – E 2 / r 2 + r 1 = 12 – 6 / 3 + 2 = 1.2 A

For cell E 1 : v A – E 1 + lr 1 = v B

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

i.e. v A – v B = E 1 – lr 1 = 12 – 3.6 × 3 = 1.2 V

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

Similarly, the resistance between A and D is given by 6 × 6 / 6 + 6 = 3 W.

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Solving Electric Circuit Problems. When tackling a circuit problem you may need to figure out the equivalent resistance of the circuit, voltage drops across resistors, total current coming out of the battery or current through specific resistors, power dissipated by resistors or provided by the battery, relative brightness of light bulbs in a circuit, the effect of a shorted resistor, or a ...

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4. Write down all the givens, organize the information and think about the best way to attack the problem. 5. Work out the possible solutions on paper. Normally, there are many ways to solve a circuit but there are only one or two easy ways to the solution. 6. Keep the overview of the problem in sight. Remember, while solving circuits, to ...

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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, ∆T = 30, so Q = 1 × 4200 x 30 = 126 kJ. The time required to produce this heat energy t = Q/ I2R = 126 ×103 / 4840 ≈ 26 .03 s. Prev Page.

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The lightbulb filament violates Ohm's Law. Ohm's Law Statement: Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperature, remain constant. Ohm's Law Equation: V = IR, where V is the voltage across the conductor, I is the current ...

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. Solution: Given, Voltage, V = 220 V. 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 ...

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.