To find the resistance of a galvanometer by the half deflection method, the following circuit is used with resistances R₁ = 9970 Ω, R₂ = 30 Ω and R₃ = 0. The deflection in the galvanometer is d. With R₃ = 107 Ω, the deflection changed to d/2. The galvanometer resistance is approximately: (a) 107 Ω (b) 137 Ω (c) 107/2 Ω (d) 77 Ω

Answer: (d). R₁ is very large, so the main current i stays almost constant. G + R₃ is in parallel with R₂: Ig = i(R₂)/(R₂ + G + R₃) Half deflection: R₂ + G + 107 = 2(R₂ + G) ⇒ G = 107 - 30 = 77 Ω So the answer is (d) 77 Ω.

To find the resistance of a galvanometer by the half deflection… | JEE Main 2013 (22 Apr, Shift 1)

Q188JEE Main 2013 · 22 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

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 $R_3 = 0$ . The deflection in the galvanometer is d. With $R_3 = 107\ \Omega$ , the deflection changed to $\frac{d}{2}$ . The galvanometer resistance is approximately:

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Answer: (d)

$R_1$ is very large, so the main current $i$ stays almost constant.
$G + R_3$ is in parallel with $R_2$ : $I_g = i\frac{R_2}{R_2 + G + R_3}$
Half deflection: $R_2 + G + 107 = 2(R_2 + G) \Rightarrow G = 107 - 30 = 77\ \Omega$
So the answer is (d) $77\ \Omega$ .

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