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JEE Mains · Physics · STD 12 - 3. current electricity
To find the resistance of a galvanometer by the half deflection method the following circuit is used with resistances \(R_1 = 9970\,\Omega,\) \(R_2 = 30\,\Omega\) and \(R_3 = 0\,\Omega.\) The deflection in the galvanometer is \(d\). With \(R_3 = 107\,\Omega\) the deflection changed to \(\frac {d}{2}\)The galvanometer resistance is approximately ............... \(\Omega\)

- A \(107\)
- B \(137\)
- C \(53.5\)
- D \(77\)
Answer & Solution
Correct Answer
(D) \(77\)
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