A circuit consists of a light bulb, a capacitor, and an inductor connected in series to an ac source. What happens to the brightness of the bulb when the inductor is omitted? When the inductor is left in the circuit but the capacitor is omitted? Explain

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. When an inductor is attached to an AC supply, the resistance produced by it is called inductive reactance (X_L).

Resistance is measure of opposition to the flow of current in a closed electrical circuit. It is measured in Ohm (Ω).

Impedance is defined as the effective resistance of an electric circuit to the flow of current due to the combined effect of resistance (offered by resistor) and reactance (offered by capacitor and inductor).

A capacitor is a device consisting of two conductors in close proximity that are used to store electrical energy. These conductors are insulated from each other. When a capacitor is attached to an AC supply, the resistance produced by it is called capacitive reactance (X_C).

Power dissipated in the bulb is given by

$P=I_{rms}{R}^2$

If Z is the impedance of the circuit, then current in the circuit is given by

$$\begin{gathered} Z=\sqrt{R^2+{X_C}^2+{X_L}^2}=\sqrt{R^2+{\left(1/C\omega \right)}^2+{\left(\omega L\right)}^2} \\ {I}_{rms}=V/Z \end{gathered}$$

If the inductance is omitted, the impedance will decrease, and thus rms current will increase due to which current consumed by bulb increases and power dissipated by bulb increases which means bulb will be brighter.

If the capacitor is omitted, the impedance will increase as L increases, and thus rms current will decrease due to which current consumed by bulb decreases and power dissipated by bulb decreases which means bulb will be less bright.

Therefore, when inductor is omitted, bulb will shine brighter, and when capacitor is omitted, bulb will shine less bright.