**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 , I = E/(R+r)

For maximum Current , R= 0

So , I_{max =} E/r

= 12/0.4 = 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 , I = E/(R+r)

R = (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 x 2 = 2V

V_{2} = I R_{2}= 2 x 2 = 4V

V_{3} = IR_{3}= 2 x 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 ,

1/R = 1/2 + 1/4 + 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