Question:
The galvanometer deflection, when key \(K_1\) is closed but \(K_2\) is open, equals \(\theta_0\) (see figure). On closing \(K_2\) also and adjusting \(R_2\) to \(5\,\Omega\), the deflection in galvanometer becomes \(\frac{\theta_0}{5}\). The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :
The galvanometer deflection, when key \(K_1\) is closed but \(K_2\) is open, equals \(\theta_0\) (see figure). On closing \(K_2\) also and adjusting \(R_2\) to \(5\,\Omega\), the deflection in galvanometer becomes \(\frac{\theta_0}{5}\). The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :
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Deflection is proportional to galvanometer current; with \(K_2\) closed, \(R_2\) acts as a shunt across G, all in series with \(R_1 = 220\,\Omega\).
Updated On: Nov 5, 2026
- \(5\,\Omega\)
- \(25\,\Omega\)
- \(12\,\Omega\)
- \(22\,\Omega\)
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The Correct Option is D
Solution and Explanation
Let the battery emf be E and the galvanometer resistance G. Deflection \(\propto\) current through G.
\(K_2\) open: \(I_0 = \frac{E}{220 + G}\), deflection \(\theta_0\).
\(K_2\) closed: G in parallel with \(R_2 = 5\,\Omega\); \(R_p = \frac{5G}{5+G}\). Total current \(I = \frac{E}{220 + R_p}\). Current through G: \(I_G = I\cdot\frac{5}{5+G} = \frac{5E}{220(5+G) + 5G}\).
Condition \(I_G = \frac{I_0}{5}\): \(\frac{5E}{1100 + 225G} = \frac{E}{5(220+G)}\).
\(25(220 + G) = 1100 + 225G \Rightarrow 5500 + 25G = 1100 + 225G \Rightarrow 200G = 4400 \Rightarrow G = 22\,\Omega\).
Answer: (4) \(22\,\Omega\)
\(K_2\) open: \(I_0 = \frac{E}{220 + G}\), deflection \(\theta_0\).
\(K_2\) closed: G in parallel with \(R_2 = 5\,\Omega\); \(R_p = \frac{5G}{5+G}\). Total current \(I = \frac{E}{220 + R_p}\). Current through G: \(I_G = I\cdot\frac{5}{5+G} = \frac{5E}{220(5+G) + 5G}\).
Condition \(I_G = \frac{I_0}{5}\): \(\frac{5E}{1100 + 225G} = \frac{E}{5(220+G)}\).
\(25(220 + G) = 1100 + 225G \Rightarrow 5500 + 25G = 1100 + 225G \Rightarrow 200G = 4400 \Rightarrow G = 22\,\Omega\).
Answer: (4) \(22\,\Omega\)
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