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प्रश्न
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?
उत्तर
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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संबंधित प्रश्न
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(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
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Y − disconnect the battery.
Z − reconnect the battery with polarity reversed.
W − insert a dielectric slab in the capacitor.
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(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.
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