A bulb and a capacitor are connected in series to an a.c. source of varialbe frequency. How will the brightness of the bulb change on increasing the frequency of a.c. source ?

Explanation

The bulb acts as a resistor. When it is connected to an a.c. source through the capacitor, the circuit is RC circuit, whose impedance,
`Z = sqrt(R^(2) + X_(C )^(2)) = sqrt(R^(2) + (1)/(omega^(2) C^(2)))`
As frequency `v = (omega)/(2 pi)` of a.c. source is
increased, Z decreases. The current through the bulb `I_(v) = E_(v)//Z` increases. Therefore, brightness of the bulb (which corresponds to `I_(v)^(2)`) increases.