A bulb is connected in series with a variable capacitor and an AC source as shown. What happens to the brightness of the bulb when the key is plugged in and capacitance of the capacitor is gradually reduced?
Advertisement Remove all ads
Solution
The capacitance of the capacitor is gradually reduced. Therefore, the capacitive reactance `X_c=1/(2pifC)` increases.
Therefore, essentially, the overall resistance of the circuit increases. This causes a reduction in the amount of current flowing through the circuit. Therefore, the brightness of the bulb reduces
Concept: Capacitors and Capacitance
Is there an error in this question or solution?
APPEARS IN
Advertisement Remove all ads
