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

Capacitor and capacitive reactance

1. A capacitor is a two-terminal electronic device, that stores the electrical energy from charge.
2. Capacitive reactance is a property of a capacitor, which resists the current flow through it.
3. The capacitive reactance is defined by the form, $X_c=\frac{1}{2\mathrm{\pi fc}}$, where, f and c are the frequency of alternating current and capacitance of the capacitor, $X_c$which is the capacitive reactance.
4. We know the impedance of the circuit, $Z=\sqrt{\left(R^2+{X_c}^2\right)}$, where, R is the resistance and Z is the impedance of the circuit.

Explanation

1. The bulb and the capacitor are connected in a series combination and the voltage of the alternating current source is constant.
2. We know that the capacitive reactance, $X_c=\frac{1}{2\mathrm{\pi fc}}$ .
3. The capacitive reactance is inversely proportional to the frequency of the alternating current. If we increase the frequency, then the capacitive reactance will decrease.
4. As we know the impedance of the circuit, $Z=\sqrt{\left(R^2+{X_c}^2\right)}$. When the capacitive reactance decrease, the impedance will decrease. So, the current flow in the circuit will increase, that's why the light bulb will blow brighter.

Diagram

Therefore, if the frequency of the alternating current increase, then the bulb in the circuit will blow brighter.