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A galvanometer is shunted with a 4 ohm resistance, which reduces its deflection to one-fifth of the original. If the galvanometer is then additionally shunted with a 2 ohm wire (in parallel with the 4 ohm shunt), the new deflection will be (the main current remains the same):
- 5/9 of the deflection when shunted with 4 ohm only
- 5/13 of the deflection when shunted with 4 ohm only
- 1/13 of the original deflection
- 1/9 of the original deflection
Correct answer: 5/13 of the deflection when shunted with 4 ohm only
Solution
From shunt condition with S = 4 ohm: I_g / I = S / (G + S) = 1/5 => G = 4S = 16 ohm. When a 2 ohm is added in parallel with 4 ohm: S_eff = (4 * 2)/(4 + 2) = 4/3 ohm. New I_g: I_g * G = (I - I_g) * S_eff => 16 * I_g = (I - I_g) * 4/3 => 48 I_g = 4I - 4 I_g => 52 I_g = 4I => I_g = I/13. Deflection ratio to '4 ohm shunt only' case (I_g was I/5): ratio = (I/13) / (I/5) = 5/13.
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