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

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The maximum measurable potential difference of a potentiometer equals the potential drop across its wire: \[ \boxed{ V_{\max}=IR_{\text{wire}} } \] where \[ I=\frac{E}{R_{\text{external}}+R_{\text{wire}}+r}. \]
Updated On: Jul 15, 2026
  • \(4\,\mathrm{V}\)
  • \(3\,\mathrm{V}\)
  • \(2\,\mathrm{V}\)
  • \(5\,\mathrm{V}\)
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The Correct Option is C

Solution and Explanation

Step 1: Calculate the current in the circuit. The total resistance is \[ R_{\text{total}} = 20+28+2 = 50\,\Omega. \] Hence, \[ I = \frac{E}{R_{\text{total}}} = \frac{5}{50} = 0.1\,\text{A}. \]

Step 2:
Find the potential drop across the potentiometer wire. The potentiometer resistance is \[ 20\,\Omega. \] Therefore, \[ V = IR = 0.1\times20 = 2\,\text{V}. \] This is the maximum potential difference measurable by the potentiometer. Hence, \[ \boxed{2\,\text{V}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.
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