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?

Answer –

We are given:

The EMF of the battery as E = 10 V

The internal resistance of the battery as R = 3 Ω

The current in the circuit as I = 0.5 A

Let the resistance of the resistor be R.

So, the current in the circuit can be found by using Ohm’s Law in the following manner –

\[I=\frac{E}{R+r}\]

Upon rearranging the above equation, we get –

\[R+r=\frac{E}{I}=\frac{10}{0.5}\]

\[R=20-3=17\Omega \]

Let the terminal voltage of the resistor be V.

Then, according to Ohm’s law we know that –

V = IR

Upon substituting values in the equation, we get

V = 0.5 × 17 = 8.5 V

Therefore, 17 Ω is the resistance of the resistor and the terminal voltage is 8.5 V.