Question

A capacitor, an inductor and an electric bulb are connected in series to an a.c. supply of variable frequency. As the frequency of the supply is increased gradually, then the electric bulb is found to

• A. increase in brightness.
• B. decrease in brightness.
• C. increase, reach a maximum and then decrease in brightness.
• D. show no change in brightness.

Hint:

Logic Tip: The current in a series LCR circuit always follows a bell-curve shape when plotted against frequency, peaking precisely at the resonant frequency $\nu_0 = \frac{1}{2\pi\sqrt{LC$.

Answer 1

The Correct Option is C

Concept:
The circuit described is an LCR series circuit, where the electric bulb acts as the resistor $R$. The brightness of the bulb depends entirely on the power dissipated by it ($P = I_{rms}^2 R$), which in turn depends on the RMS current $I_{rms}$ flowing through the circuit.
Step 1: Analyze the impedance of the LCR circuit.
The current in the circuit is governed by its impedance $Z$: $$I_{rms} = \frac{V_{rms{Z} = \frac{V_{rms{\sqrt{R^2 + (X_L - X_C)^2$$ where $X_L = \omega L$ (inductive reactance) and $X_C = \frac{1}{\omega C}$ (capacitive reactance).
Step 2: Determine the effect of varying the frequency ($\omega$).
At very low frequencies ($\omega \to 0$), $X_C$ is extremely high and $X_L$ is very low. The impedance $Z$ is high, so the current is low (dim bulb). As the frequency gradually increases, $X_L$ increases and $X_C$ decreases. The difference $(X_L - X_C)$ becomes smaller, which decreases the total impedance $Z$. Consequently, the current increases, and the bulb grows brighter.
Step 3: Analyze behavior at and beyond resonance.
At a specific resonant frequency ($\omega = \frac{1}{\sqrt{LC$), $X_L = X_C$. The impedance reaches its minimum possible value ($Z_{min} = R$), causing the current to reach its maximum value. The bulb achieves maximum brightness here. If the frequency is increased further, $X_L$ becomes larger than $X_C$, and their difference $(X_L - X_C)$ starts increasing again. This increases $Z$, decreases the current, and causes the bulb's brightness to decrease. Therefore, the brightness increases, hits a maximum at resonance, and then decreases.