A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor.

Solution:

Emf of the battery, E = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Resistance of the resistor = R

The relation for current using Ohm’s law is,

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

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

$=\frac{10}{0.5}=20 \Omega$

$\therefore R=20-3=17 \Omega$

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR

= 0.5 × 17

= 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

8.5 V.