Q1. The Storage battery of a car has an emf of 12 V . If the internal resistance of the battery is 0.4 Ω , what is the maximum current that can be drawn from the battery ?

Sol. Given , E = 12 V , r = 0.4 Ω , I_{max = ?
}

By Formula ,

$ \displaystyle I = \frac{E}{R+r} $

For maximum Current , R= 0

So , $ \displaystyle I_{max} = \frac{E}{r} = \frac{12}{0.4} $

I_{max} = 30 A

Q2. A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor . If the current in the circuit is 0.5 A , what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?

Sol. Given , E = 10 V , r = 3 Ω , I = 0.5 A , R = ? , V = ?

By Formula , $ \displaystyle I = \frac{E}{R+r} $

$ \displaystyle R = \frac{E}{I}-r $

R = (10/0.5)−3

= 20 −3 =17Ω

By relation V = IR

V = 0.5×17 = 8.5 V

Q3. (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination ?

(b) If the combination is connected to a battery of emf 12 V and

negligible internal resistance, obtain the potential drop across

each resistor.

Sol. (a)Given R_{1} = 1 Ω , R_{2} = 2 Ω , R_{3} = 3 Ω

By formula , Rs = R_{1 }+ R_{2 }+ R_{3}

Rs = 1 + 2 + 3 = 6 Ω

(b)E = 12 V

By relation , I = E/R = 12/6 = 2 A

V_{1} = I R_{1} = 1 × 2 = 2V

V_{2} = I R_{2}= 2 × 2 = 4V

V_{3} = IR_{3}= 2 × 3 = 6V

Q4. (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What

is the total resistance of the combination ?

(b) If the combination is connected to a battery of emf 20 V and

negligible internal resistance, determine the current through

each resistor, and the total current drawn from the battery.

Sol. (a) By Formula ,

$ \displaystyle \frac{1}{R} = \frac{1}{2}+\frac{1}{4}+\frac{1}{5} $

On solving we get ,

R = 20/19 Ω

(b) By formula ,

I_{1} = V/R_{1} = 20/2 = 10 A

I_{2} = V/R_{2} = 20/4 = 5 A

I_{3} = V/R_{3} = 20/5 = 4 A

Total current I = I_{1} + I_{2 }+ I_{3}

I = 10 + 5 + 4 = 19 A

Q5. At room temperature (27.0 °C) the resistance of a heating element

is 100 Ω. What is the temperature of the element if the resistance is

found to be 117 Ω, given that the temperature coefficient of the

material of the resistor is 1.70 × 10^{–4} °C^{–1}.

Sol. R_{1} = 100 Ω , R_{2} = 117 Ω

α = 1.70 × 10^{–4} °C^{–1}

By formula ,

R_{2} = R_{1} [1 + α(t_{2} − t_{1} )]

R_{2} − R_{1} = R_{1} α(t_{2} − t_{1} )

117 − 100 = 100 × 1.70 × 10^{–4}( t_{2} − 27 )

On Solving , t_{2} = 1027 °C

Q6. A negligibly small current is passed through a wire of length 15 m

and uniform cross-section 6.0 × 10^{–7} m^{2}, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment ?

Sol. By formula ,

$ \displaystyle R = \rho \frac{l}{A} $

$ \displaystyle \rho = \frac{R A}{l} $

$ \displaystyle \rho = \frac{5\times 6\times 10^{-7}}{15} $

= 2 × 10^{–7} ohm-m

Q7. A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance

of 2.7 Ω at 100 °C. Determine the temperature coefficient of

resistivity of silver.

Sol.

By formula ,

R_{2} = R_{1} [1 + α(t_{2} − t_{1} )]

R_{2} − R_{1} = R_{1} α(t_{2} − t_{1} )

2.7 − 2.1 = 2.1α(100−27.5 )

α = 0.6/(2.1×72.5)

= 3.941×10^{–3 }°C^{–1}

Q.8. A heating element using nichrome connected to a 230 V supply

draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C ?

Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^{–4} °C^{–1}

Sol.

R_{27} = 230/3.2 ohm = 2300/32 ohm ,

R_{t} = 230/2.8 ohm = 2300/28 ohm

α = 1.70 × 10^{–4} °C^{–1}

Using formula ,

$ \displaystyle \alpha = \frac{R_t – R_{27}}{R_{27}(t – 27)} $

$ \displaystyle (t – 27) = \frac{R_t – R_{27}}{R_{27}\times \alpha} $

$\displaystyle t = \frac{R_t – R_{27}}{R_{27}\times \alpha} + 27 $

t = 867.2°c

Q: 9. Determine the current in each branch of the following network : Fig.

Sol. The current through the various arms of the circuit have been shown in Fig

According to Kirchhoff’s second law; In a closed circuit ABCEA ;

10(i_{1}+i_{2}) + 10i_{1} + 5(i_{1}-i_{3}) -10 = 0

25 i_{1} + 10 i_{2}-5i_{3} = 10

5i_{1} + 2i_{2}-i_{3} = 2 …(i)

In a closed circuit ABDA ;

10 i_{1} + 5i_{3}-5i_{2} = 0

2 i_{1} + i_{3}-i_{2} = 0

i_{2} = 2 i_{1} + i_{3} ….(ii)

In a closed circuit BCDB ;

5 (i_{1} -i_{3})-10 (i_{2} + i_{3} )-5i_{3}=0

5i_{1} – 10i_{2}-20i_{3} = 0

i_{1} = 2i_{2} + 4i_{3} …(iii)

On solving ;

i_{1} =-2 (-2⁄(17))=4⁄17 A ;

i_{2} =-3 (-2⁄17)= 6⁄17 A

i_{1}+i_{2} =(4⁄17)+(6⁄17) = 10⁄17 A

i_{1}-i_{3}=4⁄17-(-2⁄17)= 6⁄17 A

i_{2}+i_{3} = 6⁄(17+(-2⁄(17) = 4⁄17 A