θ0∝iG⇒θ0C=iG
I-case Cθ0=220+RV ......(i)
II-case C5θ0=(220+5+R5R)V×5+R5 ....(ii)
From (i) and (ii) R=22Ω
The galvanometer deflection, when key K1 is closed but K2 is open, equals θ0 (see figure). On closing K2 also and adjusting R2 to 5Ω, the deflection in galvanometer becomes 5θ0. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

Held on 12 Jan 2019 · Verified 6 Jul 2026.
12Ω
22Ω
5Ω
25Ω
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