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Questions  

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

a
Current through the galvanometer G is zero
b
Reading of ammeter is 0.5 A
c
charged stored in the 5μF capacitor is 30 μC
d
Current through the galvanometer is 0.2 A

detailed solution

Correct option is C

In steady state, Ig=104+1+5A=1 APotential drop across the 5 μF capacitor = (5+1)×1 volt =6 volt . Therefore charge stored = 6×5 μC =30 μC

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