Question

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:

• A. increases
• B. remains same
• C. becomes zero
• D. decreases

Answer 1

Answer: (A)

increases

Answer 2

Step 1: Understanding the circuit The capacitor is in series with the bulb in an AC circuit. The impedance Z of the capacitor is given by:

$Z_C = \frac{1}{\omega C},$

where:

• $\omega$ is the angular frequency of the AC supply,
• $C$ is the capacitance of the capacitor.

Step 2: Effect of placing a dielectric Placing a dielectric between the plates of the capacitor increases the capacitance $C$, as:

$C' = \kappa C,$

where $\kappa > 1$ is the dielectric constant.

Step 3: Impedance of the capacitor Since $Z_C \propto \frac{1}{C}$, increasing $C$ reduces the capacitive impedance $Z_C$.

Step 4: Impact on the bulb The total impedance of the circuit decreases, leading to an increase in the current through the circuit. As the current increases, the glow of the bulb increases.

Final Answer: The glow of the bulb increases.