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 ___________ \(\Omega\).
- A 3
- B 4
- C 12
- D 6
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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