Question

A potentiometer of length \(4\,\mathrm{m}\) and resistance \(20\,\Omega\) is connected in series to a resistor of resistance \(28\,\Omega\). If the emf of the cell is \(5\,\mathrm{V}\) and its internal resistance is \(2\,\Omega\), then the maximum potential difference that the potentiometer measures is

• A. \(4\,\mathrm{V}\)
• B. \(3\,\mathrm{V}\)
• C. \(2\,\mathrm{V}\)
• D. \(5\,\mathrm{V}\)

Hint:

The maximum measurable potential difference of a potentiometer equals the potential drop across its wire: \[ \boxed{ V_{\max}=IR_{\text{wire}} } \] where \[ I=\frac{E}{R_{\text{external}}+R_{\text{wire}}+r}. \]

Answer 1

The Correct Option is C