Why does a galvanometer show a momentary deflection at the time of charging or discharging a capacitor? Write the necessary expression to explain this observation.
Solution
As the capacitor is connected to the battery, the electrons start moving towards the plate connected to the negative terminal of the battery and electrons leave from the plate connected to the positive terminal of the battery. This happens until the potential of the capacitor becomes equal to that of the battery. As this happens very quickly, the charging current produces a deflection. The reverse process is repeated for discharging that is the charge is lost. As the galvanometer can be considered as a resistance, the circuit behaves like a R-C circuit having time constant equal to RC. Hence the expression governing this phenomenon is
`q=q_0(1-e^(-t/(RC)))`
