a lamp is connected in series with a capacitor predict your observation for Ac and do. what happen when capacitance is decrease?
Sauhard1:
sorry it's not do it's dc
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With DC source, the lamp will glow for a moment and remain off. This is so because a fully discharged capacitor behaves like a short circuit, charges up and is an open circuit then on. Decreasing the capacitor has virtually no change in the behavior.
In case of AC source, the the lamp will remain lit at less than its normal brightness depending upon the impedance the capacitor offers. Decreasing the capacitance will further diminish the brightness of the lamp
In case of AC source, the the lamp will remain lit at less than its normal brightness depending upon the impedance the capacitor offers. Decreasing the capacitance will further diminish the brightness of the lamp
can you give me a expression for this question
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