Question

A light bulb connected in series with a capacitor and an a.c.source is glowing with certain brightness. On reducing the value of capacitance and frequency respectively, the brightness of the bulb

• A. is reduced, is reduced
• B. is more, is more
• C. is more, is reduced
• D. is reduced, is more

Hint:

Capacitive reactance is 1/(2 pi f C), so reducing C or f increases it and lowers the current.

Answer 1

The Correct Option is A

Step 1: Think of the capacitor as a blocker of low-frequency current:
A capacitor passes high frequency easily and blocks low frequency.

Step 2: Apply to both changes:
A smaller capacitor stores less charge per cycle, so it passes less current. A lower frequency means fewer charge flows per second, so the current is lower.

Step 3: Pick:
Less current means a dimmer bulb in both cases, option A.

Final Answer:
Reducing C or f raises capacitive reactance, which lowers the current. \[ \boxed{\text{(A) }\text{is reduced, is reduced}} \]