JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2022
- Medium
- Numerical
A cell, shunted by a \(8 \; \Omega\) resistance, is balanced across a potentiometer wire of length \(3 \; m\). The balancing length is \(2 \; m\) when the cell is shunted by \(4 \; \Omega\) resistance. The value of internal resistance of the cell will be \(\dots \; \Omega .\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\(\frac{V_{1}}{V_{2}}=\frac{3}{2}=\frac{E-i_{1} r}{E-i_{2} r}\)
\(=\frac{E-\frac{E}{8+r} \times r}{E-\frac{E}{4+r} \times r}\)
\(\frac{3}{2}=\frac{8(4+r)}{4(8+r)}\)
\(24+3 r=16+4 r\)
\(r=8 \; \Omega\)
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