Question

A circuit consists of a light bulb, a capacitor, and an inductor connected in series to an ac source. Is it possible for both the capacitor and the inductor to be removed and the brightness of the bulb to remain the same? Explain.

Short answer

The brightness of the bulb will remain the same when the capacitor and the inductor are removed.

Step-by-step solution

Step-1: Formulas used  

The power dissipated in the bulb is given by$\mathbf{P}\mathbf{=}{{\mathbf{i}}_{\mathbf{rms}}}^{\mathbf{2}}\mathbf{R}$.

Use impedance to get the current in the circuit by${\mathbf{i}}_{\mathbf{rms}}\mathbf{=}\frac{\mathbf{V}}{\mathbf{Z}}$.

The impedance is given by$\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left(X_L-X_{C)}\right)}^{\mathbf{2}}}$.

Step-2: Calculations

When the capacitor and the inductor are removed,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{C}}={\mathrm{X}}_{\mathrm{L}}=0 \\ \mathrm{Z}=\sqrt{{\mathrm{R}}^2+0^2+0^2} \\ =\sqrt{{\mathrm{R}}^2} \\ =\mathrm{R} \end{gathered}$$

As shown, the impedance is same for the resistance R and hence the resistance is same. Therefore, the power consumed by the bulb is the same, and hence, the brightness of the bulb will remain the same.

Therefore, the brightness of the bulb will remain the same.