Question

A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).

Hint:

To solve such problems, use Ohm’s law to relate voltage, current, and resistance, and apply the concept of internal resistance of the battery.

Answer 1

We use the relation between emf, current, potential difference, and internal resistance: \[ E = I_1 (r + R_1) \quad \text{and} \quad E = I_2 (r + R_2) \] Given: - \( I_1 = 2A \), potential difference \( V_1 = 5V \), - \( I_2 = 4A \), potential difference \( V_2 = 4V \). For the first case: \[ V_1 = I_1 R_1 \quad \Rightarrow \quad R_1 = \frac{V_1}{I_1} = \frac{5}{2} = 2.5 \, \Omega \] For the second case: \[ V_2 = I_2 R_2 \quad \Rightarrow \quad R_2 = \frac{V_2}{I_2} = \frac{4}{4} = 1 \, \Omega \] The internal resistance \( r \) is the same for both cases, and can be found by solving the system of equations. The solution gives: \[ E = 6 \, \text{V}, \quad r = 1 \, \Omega \] Thus, the value of \( E \) is 6V and \( r \) is 1Ω.