Question

A potentiometer wire of length \(10\) m and resistance \(9\,\Omega\) is connected to a cell of emf \(E\) and internal resistance \(1\,\Omega\). If \(V\) is the maximum potential difference that can be measured using the potentiometer, then the condition for the potentiometer to work is

• A. \(E=V\)
• B. \(9E\ge10V\)
• C. \(E<V\)
• D. \(9V=10E\)

Hint:

For a potentiometer, \[ \boxed{ V_{\max}=IR_{\text{wire}}. } \] The unknown potential difference must always satisfy \[ \boxed{ V\le V_{\max}. } \]

Answer 1

The Correct Option is B

Step 1: Find the current through the potentiometer wire. The potentiometer wire of resistance \[ 9\,\Omega \] is connected to a cell of emf \[ E \] having internal resistance \[ 1\,\Omega. \] Hence, the current through the circuit is \[ I = \frac{E}{9+1} = \frac{E}{10}. \]

Step 2: Find the maximum potential difference. The maximum potential drop across the potentiometer wire is \[ V_{\max} = IR = \frac{E}{10}\times9 = \frac{9E}{10}. \] For the potentiometer to measure a potential difference \[ V, \] it must satisfy \[ V\le V_{\max}. \] Therefore, \[ V\le\frac{9E}{10}. \] or \[ \boxed{9E\ge10V.} \]

Step 3: Write the answer. Hence, \[ \boxed{9E\ge10V.} \] Thus, \[ \boxed{(B)} \] is the correct answer.