MHT CET · Physics · Current Electricity
A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by \(5 \Omega\). When a resistance of \(15 \Omega\) is used for shunting, null point moves to 300 cm . The internal resistance of the cell is
- A \(3 \Omega\)
- B \(4 \Omega\)
- C \(5 \Omega\)
- D \(6 \Omega\)
Answer & Solution
Correct Answer
(C) \(5 \Omega\)
Step-by-step Solution
Detailed explanation
\(\frac{E R_1}{R_1 + r} = k L_1 \implies \frac{E(5)}{5 + r} = k(200)\) (1) \(\frac{E R_2}{R_2 + r} = k L_2 \implies \frac{E(15)}{15 + r} = k(300)\) (2)
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