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JEE Mains · Physics · STD 12 - 3. current electricity

In the potentiometer, when the cell in the secondary circuit is shunted with 4 \(\Omega\) resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12 \(\Omega\) resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ___________ \(\Omega\).

  1. A 3
  2. B 4
  3. C 12
  4. D 6
Verified Solution

Answer & Solution

Correct Answer

(B) 4

Step-by-step Solution

Detailed explanation

Let E is emf and r is internal resistance of cell. \(\frac{E \cdot 4}{r+4} = 120K\) \(\frac{E \cdot 12}{r+12} = 180K\) \(\Rightarrow \frac{1}{3} \frac{ r +12}{ r +4}=\frac{2}{3}\) \(r+12=2(r+4)\) \(\Rightarrow r=4\)
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