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Jee Numericals of Current electricity class-12 CBSE GSEB

Current Electricity

Question 3.1:

Car has a storage battery with an emf of 12 V. 0.4Ω is the internal resistance of the battery, what is the maximum current that can be drawn from the battery?

Answer 3.1:

Given that Emf of the battery, E = 12 V

Battery has an Internal resistance of R = 0.4 Ω

The Maximum current drawn from the battery is given by  = I

According to Ohm’s law,

E = IR

I=ER

I=120.4

= 30 A

The maximum current drawn from the given battery is 30 A.

Question 3.2:

A battery has an emf of 10 V and internal resistance is observed to be 3 Ω and is connected to a resistor. If the current flowing in the circuit is 0.5 A, calculate the resistance of the resistor? Also calculate the terminal voltage of the battery when the circuit is closed.

Answer 3.2:

Given Data :

Given that Emf of the battery, E = 10 V

Battery has an Internal resistance of R = 3 Ω

Current flowing in the circuit , I = 0.5 A

Let the Resistance of the resistor be  = R

The relation for current using Ohm’s law is,

I=ER+r

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR

V = 0.5 × 17

V = 8.5 V

Therefore, the resistance of the resistor calculated is 17 Ω and the terminal voltage is found to be

8.5 V.

Question 3.3 :

a) A series combination of three resistors with the following resistance r 1 = 1 Ω , r 2= 2 Ω and r 3 = 3 Ω is made. Calculate the total resistance of the combination.

b) Calculate the potential drop across each resistors given above , when this combination is connected to a battery of emf 12 V with negligible internal resistance.

Answer 3.3 :

1. a) Resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3= 3 Ω are combined in series

Total resistance of the above series combination can be calculated by the algebraic sum of individual resistances

Therefore, total resistance is given by :

R = 1 Ω + 2 Ω + 3 Ω = 6 Ω

Thus calculated R = 6 Ω

b ) I is the current flowing the given circuit

Also,

Given that emf of the battery , E = 12 V

Total resistance of the circuit ( calculated above ) = R = 6 Ω

Using Ohm’s law , relation for current can be obtained

I=ER

I=126

= 2 A

Therefore, the current calculated is 2 A

Let the Potential drop across 1 Ω resistor = V 1

The value of V 1 can be obtained from Ohm’s law as :

V 1 = 2 x 1 = 2 V

Let the Potential drop across 2 Ω resistor = V 2

The value of V 2 can be obtained from Ohm’s law as :

V 2 = 2 x 2 = 4 V

Let the Potential drop across 3 Ω resistor = V 3

The value of V 3 can be obtained from Ohm’s law as :

V 3 = 2 x 3 = 6 V

Therefore, the potential drops across the given resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3= 3 Ω are calculated to be

V 1 = 2 x 1 = 2 V

V 1 = 2 x 1 = 4 V

V 1 = 2 x 1 = 6 V

Question 3.4 :

a) A parallel combination of three resistors with the following resistance r 1 = 2 Ω , r 2= 4 Ω and r 3 = 5 Ω is made. Calculate the total resistance of the combination.

b) Calculate the current drawn from the battery when the above combination of resistors is given is connected to a battery of emf 20 V with negligible internal resistance.

Answer 3.4 :

A ) Resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are combined in parallel

Hence the total resistance of the above circuit can be calculated  by the following formula :

1R=1R1+1R2+1R3

1R=12+14+15

1R=10+5+420

1R=1920

Therefore, total resistance of the parallel combination given above is given by :

R =

1920

B ) Given that emf of the battery , E = 20 V

Let the current flowing through resistor R 1 be I 1

I 1 is given by :

I1=VR1

I1=202

I1=10A

Let the current flowing through resistor R 2 be I 2

I 2 is given by :

I2=VR2

I2=204

I2=5A

Let the current flowing through resistor R 3 be I 3

I 3 is given by :

I3=VR1

I3=205

I3=4A

Therefore , the total current can be found by the following formula :

I = I 1 + I 2 + I 3 = 10 + 5 + 4 = 19 A

therefore the current flowing through each resistors is calculated to be :

I1=10A

I2=5A

I3=4A

and the total current is calculated to be, I = 19 A

Question 3.5 :

Resistance of a heating element is observed to be 100 Ω at a room temperature of ( 27 ° C ) . Calculate the resistance of this element if the resistance is 117 Ω, given that the temperature coefficient  of the material used for the element is 1.70 x 10 – 4 C – 1

Answer 3.5 :

Given that the room temperature , T = 27 ° C

The heating element has a resistance of , R = 100 Ω

Let the increased temperature of the filament be T 1

At T 1 , the resistance of the heating element is  R 1 = 117 Ω

Temperature coefficient of the material used for the element is 1.70 x 10 – 4 C – 1

α = 1.70 x 10 – 4 C – 1

α is given by the relation ,

α=R1–RR(T1–T)

T1–T=R1–RRα

T1–27=117–100100(1.7x10−4)

T1–27=1000

T 1 = 1027 ° C

Therefore, the resistance of the element is 117 Ω at T 1 = 1027 ° C

Question 3.6 :

The length of a wire is 15 m and uniform cross  – section is 6.0 x 10 – 7 m 2.  Negligibly small current is passed through it with a resistance of 5.0 Ω. Calculate the resistivity of the material at the temperature at which the experiment is conducted.

Answer 3.6 :

Given that the length of the wire , L = 15 m

Area of cross – section is given as , a = 6.0 x 10 – 7 m 2

Let the resistance of the material of the wire be , R , ie. , R = 5.0 Ω

Resistivity of the material is given as ρ

R=ρLA

ρ=RxAL=5×6×10–715=2×10–7

Therefore, the resistivity of the material is calculated to be

2×10–7

Question 3.7 :

A silver wire is observed to have a resistance of 2.1 Ω at a temperature of 27.5 ° C and a resistance of 2.7 Ω at a temperature of 100 ° C. Calculate the temperature coefficient of resistivity of silver.

Answer 3.7 :

Given :

Given that temperature T 1 = 27.5 ° C

Resistance R 1 at temperature T 1 is given as :

R 1 =  2.1 Ω ( at T 1 )

Given that temperature T 2 = 100 ° C

Resistance R 2 at temperature T 2 is given as :

R 2 =  2.7 Ω ( at T 2 )

Temperature coefficient of resistivity of silver = α

α=R2–R1R1(T2–T1)

α=2.7–2.12.1(100–27.5)=0.0039∘C–1

Therefore, the temperature coefficient of resistivity of silver is

0.0039∘C–1

Question 3.8 :

A heating element made up of nichrome is connected to a 230 V supply draws an initial current of 3.2 A  which after a few seconds, settles to a steady value of 2.8 A. calculate the steady temperature of the heating element at a room temperature of 27.0 ° C. The temperature coefficient of nichrome ( material used to make the above  heating device ) averaged over the temperature range involved is 1.70 x 10 – 4° C  – 1   

Answer 3.8 :

Given that the supply voltage , V = 230 V

Initial current drawn is observed to be , I 1= 3.2 A

Let the initial resistance  be R 1 , which can be found by the relation :

R1=VI

R1=2303.2

= 71.87 Ω

Value of current at steady state , I 2 = 2.8 A

Value of resistance at steady state = R 2

R 2 can be calculated by the following equation :

R2=2302.8

R2=82.14Ω

The temperature coefficient of nichrome ( material used to make the above  heating device ) averaged over the temperature range involved is 1.70 x 10 – 4 ° C – 1

Value of initial temperature of nichrome , T 1 = 27.0 ° C

Value of steady state temperature reached by nichrome = T 2

This temperature T 2 can be obtained by the following formula :

α=R2–R1R1(T2–T1)

T2–27=82.14–71.8771.87x(1.7x10−4)

T2–27=840.5

T 2 = 840.5 + 27 = 867.5 ° C

Hence, the steady temperature of the heating element is 867.5 ° C