An electric lamp connected in series with a capacitor and an a.c source is glowing with certain brightness. On reducing the frequency of source the brightness of the lamp

Step-by-step solution

The impedance of the series RC circuit is given by Z = R^2 + X_C^2 , where X_C = 1 2 f C . When the frequency f of the AC source is reduced, the capacitive reactance X_C increases. As a result, the total impedance Z of the circuit increases. The current in the circuit is I = V Z . Since Z increases, the current I decreases. The brightness of the lamp is proportional to the power dissipated, P = I^2 R . With a decrease in current, the power dissipated decreases, and hence the brightness of the lamp is reduced. Answe