Question

A potentiometer has a uniform wire of length 5m. A battery of emf 10V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200cm. The emf of the secondary cell is

• A. 4V
• B. 6V
• C. 2V
• D. 8V

Answer 1

The Correct Option is A

The potentiometer works on the principle that the potential difference across a uniform wire is directly proportional to the length of the wire. The formula to calculate the emf of the secondary cell is: \[ \frac{E_{\text{sec}}}{E_{\text{battery}}} = \frac{l_{\text{bal}}}{L} \] Where: - \(E_{\text{sec}}\) is the emf of the secondary cell. - \(E_{\text{battery}} = 10 \, \text{V}\) is the emf of the primary battery. - \(l_{\text{bal}} = 200 \, \text{cm} = 2 \, \text{m}\) is the balancing length. - \(L = 5 \, \text{m}\) is the total length of the potentiometer wire. Substituting the values into the formula: \[ \frac{E_{\text{sec}}}{10} = \frac{2}{5} \] Solving for \(E_{\text{sec}}\): \[ E_{\text{sec}} = 10 \times \frac{2}{5} = 4 \, \text{V} \] Thus, the emf of the secondary cell is 4 V. Therefore, the correct answer is (A) 4 V.