Question:
An electric lamp connected in series with a capacitor and an a.c source is glowing with certain brightness. On reducing the frequency of source the brightness of the lamp
An electric lamp connected in series with a capacitor and an a.c source is glowing with certain brightness. On reducing the frequency of source the brightness of the lamp
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Capacitive reactance is 1/(omega C), which rises when frequency falls.
Updated On: Oct 1, 2026
- is increased
- is reduced
- remain the same
- becomes zero
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
In an a.c. circuit with a lamp (resistance R) and a capacitor in series, the current is \(I = \dfrac{V}{\sqrt{R^2 + X_C^2}}\) with \(X_C = \dfrac{1}{2\pi fC}\).
Step 2: Effect of lower frequency
When f decreases, \(X_C\) increases, so the impedance rises and the current falls. A smaller current means the lamp glows less brightly.
The lamp does not become fully dark (D), because the circuit still carries a smaller a.c. current.
Final Answer:
The brightness is reduced, option (B). \[ \boxed{\text{Reduced}} \]
In an a.c. circuit with a lamp (resistance R) and a capacitor in series, the current is \(I = \dfrac{V}{\sqrt{R^2 + X_C^2}}\) with \(X_C = \dfrac{1}{2\pi fC}\).
Step 2: Effect of lower frequency
When f decreases, \(X_C\) increases, so the impedance rises and the current falls. A smaller current means the lamp glows less brightly.
The lamp does not become fully dark (D), because the circuit still carries a smaller a.c. current.
Final Answer:
The brightness is reduced, option (B). \[ \boxed{\text{Reduced}} \]
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