Question

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….

Answer 1

Hint: To find whether the bulb will glow more or less, we have to use the concept of impedance and reactance of the capacitor in which Impedance extends the concept of resistance to alternating current (AC) circuits, and possesses both magnitude and phase, unlike resistance, which has only magnitude and reactance is the opposition of a circuit element to the flow of current due to that element's inductance or capacitance.

Complete step by step answer:
The capacitive reactance of the circuit is given by
${\chi _c} = \dfrac{1}{f}$
And the impedance of the circuit is given by
$Z = \sqrt {{R^2} + \chi _c^2} $
Where, $f$ - frequency of the AC source and $R$ - resistance in the circuit.
Here, a bulb and a capacitor are connected in series with an AC source.

Now, when we increase the frequency of the voltage source, the capacitive reactance of the circuit decreases which in turns decreases the impedance of the circuit. When the impedance of the circuit decreases, more current will flow through the circuit and hence, the bulb will glow brighter.

Note: We should know the formula for impedance and reactance of the capacitor in the circuit. The higher the frequency of the AC power source, the higher the current and the brightness. On a DC source the current will flow only for a limited short time and the bulb will run low for a short time.