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In the circuit shown in figure resistance of galvanometer coil is $1 Ω$ . Then

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Current through the galvanometer G is zero

Reading of ammeter is 0.5 A

charged stored in the 5 $μ F$ capacitor is 30 $μ C$

Current through the galvanometer is 0.2 A

answer is C.

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Detailed Solution

In steady state, $I_{g} = \frac{10}{4 + 1 + 5} A = 1 A$

Potential drop across the 5 $μ F$ capacitor = $(5 + 1) \times 1 v o l t = 6 v o l t . T h e r e f o r e c h a r g e s t o r e d = 6 \times 5 μ C = 30 μ C$

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