Home | | Physics 12th Std | Current Electricity: Exercises and Example Solved Numerical problems

# Current Electricity: Exercises and Example Solved Numerical problems

Physics : Current Electricity: Solved Example Numerical problems with Answers, Solution and Explanation

## Numerical 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.

Solution

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.

### Solution:

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 ×1031 / 9 .11

= 1.001 × 1014 m s-2

### EXAMPLE 2.3

A copper wire of cross-sectional area 0.5 mm2 carries a current of 0.2 A. If the free electron density of copper is 8.4 × 1028 m-3 then compute the drift velocity of free electrons.

### Solution

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

vd = I/ ne A vd = 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.

### Solution

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 = It/e

n = 32×1 / 1 .6×10−19 C

n = 20 × 1019 = 2 × 1020 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?

Solution 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?

Solution

R1 = 20 Ω, R2= ?

Let the original length (l1) be l.

The new length, l2 = 8l1 (i.,e) l2 =8l

The original resistance, R = ρ [ l1 / A1]

The new resistance

R2 = Though the wire is stretched, its volume is unchanged.

Initial volume = Final volume

A1l1= A2l2 , A1l = A28l

A1 / A= 8l / l = 8

By dividing equation R2 by equation R1, we get Substituting the value of A1/A2, we get

R2 / R = 8 ×8 = 64 2

R2 = 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 IAB 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 IAB.

Solution

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. Solution

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/ Req = 24/10 = 2 .4 A

Voltage across 4Ω resistor

V1 = IR1 = 2 . 4 A× 4 Ω = 9.6V

Voltage across 6 Ω resistor

V2 = IR1 = 2 . 4 A× 6 Ω =14 .4V

EXAMPLE 2.9

Calculate the equivalent resistance in the following circuit and also find the current I, I1 and I2 in the given circuit. Solution

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 = I1 + I2= 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.

Solution

Rs = R1 + R2 = 15 Ω (1) The above equation can be solved using factorisation.

R12-8 R1-7 R1+ 56 = 0

R1 (R1– 8) – 7 (R1– 8) = 0

(R1– 8) (R1– 7) = 0

If (R1= 8 Ω)

using in equation (1)

8 + R2 = 15

R2 = 15 – 8 = 7 Ω ,

R2 = 7 Ω i.e , (when R1 = 8 Ω ; R2 = 7 Ω)

If (R1= 7 Ω)

Substituting in equation (1)

7 + R2 = 15

R2 = 8 Ω , i.e , (when R1 = 8 Ω ; R2 = 7 Ω )

EXAMPLE 2.11

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

Case (a)

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 Req = 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.

Solution

R0= 3 Ω,  T = 100ºC,  T0 = 20ºC

α = 0.004/ºC,  RT = ?

RT= R0(1 + α(T-T0))

R100 = 3(1 + 0.004 × 80)

R100 = 3(1 + 0.32)

R100 = 3(1.32)

R100 = 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.

Solution

T0 = 10ºC, T = 40ºC, R0= 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? Solution

(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 = V2 / 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 (R1/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?

Solution

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 /Rtot.

Rtot = ( R1 + R2 )

Rtot = ( 484 +2420) Ω = 2904Ω

I = 440V/2904Ω ≈ 0.151A

The voltage drop across the 20W bulb is

V1 = IR1 = 440/2904 × 2420 ≈ 366.6 V

The voltage drop across the 100W bulb is

V2 = IR2 = 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.

Determination of internal resistance: Solved Example Problems

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

Solution

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 = I2 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 I2 r).

## Cells in series: Solved Example Problems

EXAMPLE 2.18

From the given circuit, Find

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

Solution

i) Equivalent emf of the combination ξeq = nξ = 4 9 = 36 V

ii) Equivalent internal resistance req = 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

I = 3.46 A

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 Find

i) Equivalent emf

ii) Equivalent internal resistance

iii) Total current (I)

iv) Potential difference across each cell

v) Current from each cell

Solution

i) Equivalent emf ξeq = 5 V

ii) Equivalent internal resistance,

Req = r/n = 0 .5/4 = 0.125Ω

iii) total current, I ≈ 0.5 A

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. Solution

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

I = 0.6 A

## 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. Solution

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

I1R1 + I2R2 + I3R3 = ξ

and for the closed loop ABCA

I4R4 + I5R5-I2R2= 0

EXAMPLE 2.22

Calculate the current that flows in the 1 Ω resistor in the following circuit. Solution We can denote the current that flows from 9V battery as I1 and it splits into I2 and I1  – I2 in the junction according Kirchoff’s current rule (KCR). It is shown below.

Now consider the loop EFCBE and apply KVR, we get

1I2 + 3I1 + 2I1 = 9

5I1 + I2 = 9 (1)

Applying KVR to the loop EADFE, we get

3 (I1 – I2 ) – 1I2 = 6

3I1 – 4I2 = 6 (2)

Solving equation (1) and (2), we get

I1 = 1.83 A and I2 = -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.

Solution 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 Solution

P/Q = R/S x + 400 = 0.625 × 1000

x + 400 = 625

x = 625 – 400

x = 225 Ω

## 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.

Solution EXAMPLE 2.25

In a meter bridge, the value of resistance in the resistance box is 10 Ω. The balancing length is l1 = 55 cm. Find the value of unknown resistance.

Solution

Q = 10 ## 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.

Solution

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

H = I2 R t

= 52 × 10 × 5 × 60

=25 × 10 × 300

=25 × 3000

=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)

Solution

According to Joule’s heating law H = I2 Rt

The current passed through the electrical heater = 220V/10Ω = 22 A

The heat produced in one second by the electrical heater H = I2 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,

∆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

Tags : Physics , 12th Physics : Current Electricity
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Physics : Current Electricity : Current Electricity: Exercises and Example Solved Numerical problems | Physics

Related Topics