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 :

<p>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.</p> <p>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:</p> <p>$Z_C = \frac{1}{\omega C}$</p> <p>where:</p> <ul> <li>$ Z_C $ is the capacitive reactance (impedance of the capacitor).</li> <li>$ \omega $ is the angular frequency of the AC supply ( $ \omega = 2 \pi f $ , where $ f $ is the frequency).</li> <li>$ C $ is the capacitance of the capacitor.</li> </ul> <p>When we place a dielectric between the plates of the capacitor, the capacitance $ C $ increases. The capacitance with a dielectric can be described as:</p> <p>$C' = \kappa C$</p> <p>where:</p> <ul> <li>$ C' $ is the new capacitance with the dielectric present.</li> <li>$ \kappa $ is the dielectric constant ( $ \kappa > 1 $ ).</li> </ul> <p>Since $ C' > C $, the new capacitive reactance $ Z_C' $ can be given as:</p> <p>$Z_C' = \frac{1}{\omega C'}$</p> <p>Because $ C' = \kappa C $, we have:</p> <p>$$ Z_C' = \frac{1}{\omega \kappa C} = \frac{1}{\kappa} \left( \frac{1}{\omega C} \right) = \frac{Z_C}{\kappa} $$</p> <p>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.</p> <p>Thus, with an increase in current, the bulb will glow brighter. Therefore, the correct option is:</p> <p><strong>Option C: increases</strong></p>