Solution:
As frequency increases, capacitive reactance \(X_C = \frac{1}{2\pi f C}\) decreases. This decreases the total impedance \(Z = \sqrt{R^2 + X_C^2}\), thereby increasing the current, causing the bulb to glow more brightly.
A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then:
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