QJEEM24Ph18f5: JEE Main 2024 Physics - Class 12 JEEM24Ph18f5 Solution: A bulb and a capacitor are connected in series across an ac…

To understand the impact of placing a dielectric between the plates of the capacitor on the glow of the bulb, we need to consider the properties and behavior of capacitors in an AC circuit.

When a capacitor is connected in series with a bulb in an AC circuit, the impedance of the capacitor plays a significant role in determining the current through the circuit. The impedance $Z_C$ of a capacitor in an AC circuit is given by:

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

where:

• $Z_C$ is the capacitive reactance (impedance of the capacitor).
• $\omega$ is the angular frequency of the AC supply ( $\omega = 2 \pi f$ , where $f$ is the frequency).
• $C$ is the capacitance of the capacitor.

When we place a dielectric between the plates of the capacitor, the capacitance $C$ increases. The capacitance with a dielectric can be described as:

$C' = \kappa C$

where:

• $C'$ is the new capacitance with the dielectric present.
• $\kappa$ is the dielectric constant ( \kappa > 1 ).

Since C' > C , the new capacitive reactance $Z_C'$ can be given as:

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

Because $C' = \kappa C$ , we have:

$Z_C' = \frac{1}{\omega \kappa C} = \frac{1}{\kappa} \left( \frac{1}{\omega C} \right) = \frac{Z_C}{\kappa}$

Since \kappa > 1 , Z_C' < Z_C . This means the impedance of the capacitor decreases when a dielectric is placed between its plates. In a series circuit, the overall impedance decreases when the impedance of one component decreases, leading to an increase in the current through the circuit.

Thus, with an increase in current, the bulb will glow brighter. Therefore, the correct option is:

Option C: increases