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: Understanding the Concept:
In a series circuit with a bulb (resistance \(R\)) and a capacitor, the current is \(I=\dfrac{V}{\sqrt{R^2+X_C^2}}\) with \(X_C=\dfrac{1}{2\pi fC}\).

Step 2: Reduce the capacitance:
A smaller \(C\) gives a larger \(X_C\). So the impedance rises and the current falls. The bulb gets dimmer.

Step 3: Reduce the frequency:
A smaller \(f\) also gives a larger \(X_C\). So again the current falls and the bulb gets dimmer.

Step 4: Conclusion:
The brightness is reduced in both cases. Option A.

Final Answer:
The brightness is reduced both times. \[ \boxed{\text{(A) }\text{is reduced, is reduced}} \]