*Question 3.1:*

*The 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

\(I = \frac{ E }{ R }\)\(I = \frac{ 12 }{ 0.4 }\) = 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,

\(I = \frac{ E }{ R + r }\)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 resistor given above, when this combination is connected to a battery of emf 12 V with negligible internal resistance.*

*Answer 3.3 :*

- 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

\(I = \frac{ E }{ R }\)\(I = \frac{ 12 }{ 6 }\) = 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 :

\(\frac{ 1 }{ R } = \frac{ 1 }{ R1 } + \frac{ 1 }{ R2 } + \frac{ 1 }{ R3 }\) \(\frac{ 1 }{ R } = \frac{ 1 }{ 2 } + \frac{ 1 }{ 4 } + \frac{ 1 }{ 5 }\) \(\frac{ 1 }{ R } = \frac{ 10 + 5 + 4 }{ 20 }\) \(\frac{ 1 }{ R } = \frac{ 19 }{ 20 }\)Therefore, total resistance of the parallel combination given above is given by :

R = \(\frac{ 19 }{ 20 }\)

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 :

Let the current flowing through resistor R _{2} be I _{2}

I _{2} is given by :

Let the current flowing through resistor R _{3} be I _{3}

I _{3} is given by :

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 :

\(I_{ 1 } = 10 A\) \(I_{ 2 } = 5 A\) \(I_{ 3 } = 4 A\)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 ,

\(\alpha = \frac{ R_{ 1 } – R }{ R \left ( T _{ 1 } – T \right )}\) \(T _{ 1 } – T = \frac{ R _{ 1 } – R }{ R \alpha }\) \(T _{ 1 } – 27 = \frac{ 117 – 100 }{ 100 ( 1.7 x 10_{-4}) }\) \(T _{ 1 } – 27 = 1000\)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

\(R = \rho \frac{ L }{ A }\) \(\rho = \frac{ R x A }{ L } = \frac{ 5 \times 6 \times 10 ^{ – 7 }}{ 15 } = 2 \times 10 ^{ – 7 }\)Therefore, the resistivity of the material is calculated to be \(2 \times 10 ^{ – 7 }\)

*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

\(\alpha = \frac{ R _{ 2 } – R _{ 1 }}{ R _{ 1 } \left ( T _{ 2 } – T _{ 1 }\right )}\) \(\alpha = \frac{ 2.7 – 2.1 }{ 2.1 \left ( 100 – 27.5 \right )} = 0.0039 ^{\circ} C ^{ – 1 }\)Therefore, the temperature coefficient of resistivity of silver is \(0.0039 ^{\circ} C ^{ – 1 }\)

*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 :

\(R _{ 1 } = \frac{ 230 }{ 3.2 }\) = 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 :

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 :

T _{2} = 840.5 + 27 = 867.5 ° C

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

*Question 3.9 :*

*Calculate the current in each branch of the network shown in the figure shown below :*

*Answer 3.9 :*

The current flowing through various branches of the network is shown in the figure given below :

Let I _{1} be the current flowing through the outer circuit

Let I _{2} be the current flowing through AB branch

Let I _{3} be the current flowing through AD branch

Let I _{2} – I _{4} be the current flowing through branch BC

Let I _{3} + I _{4} be the current flowing through branch BD

Let us take closed circuit ABDA into consideration, we know that potential is zero.

i.e , 10 I _{2} + 5 I _{4} – 5 I _{3} = 0

2 I _{2} + I _{4} – I _{3} = 0

I _{3} = 2 I _{2} + I _{4 } . . . . . . . . . . . . . . . . . . . . . . eq ( 1 )

Let us take closed circuit BCDB into consideration , we know that potential is zero.

i.e , 5 ( I _{2} – I _{4} ) – 10 ( I _{3} + I _{4} ) – 5 I _{4} = 0

5 I _{2} + 5 I _{4} – 10 I _{3} – 10 I _{4} – 5 I _{4} = 0

5 I _{2} – 10 I _{3} – 20 I _{4} = 0

I _{2} = 2 I _{3} – 4 I _{4} . . . . . . . . . . . . . . . . . . . . . . eq ( 2 )

Let us take closed circuit ABCFEA into consideration , we know that potential is zero.

i.e , – 10 + 10 ( I _{1} ) + 10 ( I _{2} ) + 5 ( I _{2} – I _{4} ) = 0

10 = 15 I _{2} + 10 I _{1} – 5 I _{4}

3 I _{2} + 2 I _{2} – I _{4} = 2 . . . . . . . . . . . . . . . . . . . . . . eq ( 3 )

From equation ( 1 ) and ( 2 ) , we have :

I _{3} = 2 ( 2 I _{3} + 4 I _{4} ) + I _{4}

I _{3} = 4 I _{3} + 8 I _{4} + I _{4}

– 3 I _{3} = 9 I _{4}

– 3 I _{4} = + I _{3} . . . . . . . . . . . . . . . . . . . . . . eq ( 4 )

Putting equation ( 4 ) in equation ( 1 ) , we have :

I _{3} = 2 I _{2} + I _{4}

– 4 I _{4} = 2 I _{2}

I _{2} = – 2 I _{4} . . . . . . . . . . . . . . . . . . . . . . eq ( 5 )

From the above equation , we infer that :

I_{ 1} = I _{3} + I _{2 } . . . . . . . . . . . . . . . . . . . . . . eq ( 6 )

Putting equation ( 4 ) in equation ( 1 ) , we obtain

3 I _{2} + 2 ( I _{3} + I _{2} ) – I _{4} = 2

5 I _{2 }+ 2 I _{3} – I _{4} = 2 . . . . . . . . . . . . . . . . . . . . . . eq ( 7 )

Putting equations ( 4 ) and ( 5 ) in equation ( 7 ) , we obtain

5 ( – 2 I _{4} ) + 2 ( – 3 I _{4} ) – I _{4} = 2

– 10 I _{4} – 6 I _{4} – I _{4} = 2

17 I _{4} = – 2

Equation ( 4 ) reduces to

I _{3} = – 3 ( I _{4} )

I _{2} = – 2 ( I _{4} )

Therefore, current in each branch is given as :

In branch \(AB = \frac{ 4 }{ 17 } A\)

In branch \(BC = \frac{ 6 }{ 17 } A\)

In branch \(CD = \frac{ – 4 }{ 17 } A\)

In branch \(AD = \frac{ 6 }{ 17 } A\)

In branch \(BD = \frac{ – 2 }{ 17 } A\)

Total current \(= \frac{ 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 2 }{ 17 } = \frac{ 10 }{ 17 } A\)

*Question 3. 10 :*

*A ) In the metre bridge shown below in the figure, 39.5 cm is found to be the balance point from the end A with N resistor having a value of 12.5 Ω. Calculate the value of the resistor M. Also, state the reason behind making the connections between the resistors in a Wheatstone bridge or meter bridge of thick copper strips.*

*B ) Calculate the new balance point of the bridge above if M and N are interchanged.*

*C ) Would the galvanometer show any deflection if the cell and galvanometer are interchanged at the balance point of the Wheatstone bridge? *

*Answer 3. 10 :*

A meter bridge with resistors M and N are shown in the figure.

( a ) Let L _{1} be the balance point from end A ,

Given that , L _{1} = 39.5 cm

Given that resistance of the resistor N = 12.5 Ω

We know that , condition for the balance is given by the equation :

\(\frac{ M }{ N } = \frac{ 100 – L _{ 1 } }{ L _{ 1 } }\) \(M = \frac{ 100 – 39.5 }{ 39.5 } \times 12.5 = 8.2 \Omega\)Thus calculated the resistance of the resistor M , M = \(8.2 \Omega\)

*Question 3.11 :*

*A storage battery has an emf of 8.0 V and an internal resistance of 0.5 Ω is allowed by charged by a 120 V, DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?*

* *

*Answer 3.11 :*

Given :

Emf of the given storage battery, E = 8.0 V

Given that the Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Effective voltage in the circuit = V^{ 1}

R is connected to the storage battery in series.

Hence, it can be written as

V ^{1} = V – E

V ^{1} = 120 – 8 = 112 V

Current flowing in the circuit = I , which is given by the relation ,

\(I = \frac{ V ^{ 1 }}{ R + r }\) \(I = \frac{ 112 }{ 15.5 + 5 }\) \(I = \frac{ 112 }{ 16 }\) \(I = 7 A\)We know that Voltage across a resistor R given by the product,

I x R = 7 × 15.5 = 108.5 V

We know that ,

DC supply voltage = Terminal voltage + voltage drop across R

Terminal voltage of battery = 120 – 108.5 = 11.5 V

A series resistor when connected in a charging circuit limits the current drawn from the external source.

The current will become extremely high in its absence. This is extremely dangerous.

** Question 3.12 **:

*In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm*

*length of the wire. If the cell is replaced by another cell and the balance point shifts to*

*63.0 cm, what is the emf of the second cell?*

*Answer 3.12 :*

Emf of the cell, E _{1} = 1.25 V

Balance point of the potentiometer, l _{1} = 35 cm

The cell is replaced by another cell of emf E _{2}.

New balance point of the potentiometer, l _{2} = 63 cm

The balance condition is given by the relation ,

\(\frac{ E _{ 1 } }{ E _{ 2 } } = \frac{ I _{ 1 } }{ I _{ 2 } }\) \(E _{ 2 } = E _{ 1 } \times \frac{ I_{ 2 } }{ I _{ 1 } }\) \(E _{ 2 } = 1.25 \times \frac{ 63 }{ 35 } = 2.25 V\)* *

*Question 3.13 :*

*A copper conductor has a number density of free electrons estimated in Example 3.1 of*

*8.5 × 10 ^{28} m ^{– 3}. How long does an electron take to drift from one end of a wire 3.0 m*

*long to its other end? The area of the cross-section of the wire is 2.0 × 10 ^{– 6} m ^{2} and it is*

*carrying a current of 3.0 A. *

* **Answer 3.13 :*

Given that Number density of free electrons in a copper conductor , n = 8.5 x 10 ^{28} m ^{– 3}

Let the Length of the copper wire be l

Given , l = 3.0 m

Let the area of cross – section of the wire be A = 2.0 x 10 ^{– 6} m ^{2}

Value of the current carried by the wire , I = 3.0 A , which is given by the equation ,

I = n A e V _{d}

Where,

e = electric charge = 1.6 x 10 ^{– 19} C