**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 ,****For maximum Current , R= 0**

**So , **

**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 , **

**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 ,**

**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 ,**

**= 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 ,**

**t = 867.2°c**