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Question
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?
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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
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