A galvanometer of resistance $G\,\Omega$ is converted into an ammet...

Question

A galvanometer of resistance $G\,\Omega$ is converted into an ammeter of range 0 to 1 A. If the current through the galvanometer is 0.1% of 1 A, the resistance of the ammeter is:

Accepted Answer

Step 1: $I_g = 0.1\% $ of 1 A $= 0.001$ A. Shunt current $I_s = 1 - 0.001 = 0.999$ A. Step 2: Since $G$ and shunt $S$ are in parallel, $I_gG = I_sS \Rightarrow S = \dfrac{0.001\,G}{0.999} = \dfrac{G}{999}$. Step 3: Ammeter resistance $= \dfrac{GS}{G+S} = \dfrac{G\cdot G/999}{G + G/999} = \dfrac{G}{1000}$. Answer: $\dfrac{G}{1000}\,\Omega$ (Option B).

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