Question

A lamp is connected in series with a capacitor and an AC source. What happens if the capacity of the capacitor is reduced ?
(A) The lamp shines more brightly
(B) The lamp shines less brightly
(C) There is no change in the brightness of the lamp
(D) Brightness may increase or decrease depending on the frequency of the AC

Answer 1

Hint:In order to solve this problem, first remember that when lamp are connected in series with AC source and capacitor then impedance Z of circuit is the combination of resistance of bulb R and reactance of capacitor ${X_C}$. Now, see the effect of capacitance C on the reactance ${X_C}$ using the above formula and connect it with circuit theory we will get a desired solution.

Formula used:
Reactance of capacitor i.e.,
${X_C} = \dfrac{1}{{\omega C}}$ ………………(1)
Where
$\omega = $ Frequency of AC source
C $ = $ Capacitance of capacitor

Complete step by step answer:
We know that the impedance Z of the given circuit in question depends on the resistance of lamp R and the reactance of the capacitor ${X_C}$. Where ${X_C}$ is given by equation number (1).Given that capacity C of capacitance is reduced. So, according to the above expression of ${X_C}$, we can conclude that if C is reduced then ${X_C}$ will be increased.Hence, due to increment of ${X_C}$, impedance Z will also be increased.So, the current in the circuit will decrease.If current decreases then according to the formula of power i.e., $P = {I^2}R$, power will also decrease.

Hence, the lamp shines less brightly. So, option B is the correct answer.

Note: Many times students may get confused between the AC and DC flow through capacitance. Capacitor has a property which allows AC to flow from it but blocks DC. Because according to the expression of reactance i.e., ${X_C} = \dfrac{1}{{\omega C}}$. AC is high frequency i.e., $\omega \uparrow $.So, the reactance becomes low i.e., ${X_C} \downarrow $ for AC. Hence, AC can easily flow through the capacitor but DC is low frequency i.e., $\omega \downarrow $. So, the reactance becomes high i.e., ${X_C} \uparrow $ for DC. Hence, capacitor block DC.