1. Series circuit in AC

A bulb behaves approximately like a resistor $R$, and the capacitor has capacitive reactance

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

So the total impedance of the series circuit is

$Z = \sqrt{R^2 + X_C^2}$

and the current is

$I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}}$

2. Effect of inserting a dielectric

When a dielectric is placed between the plates of the capacitor, its capacitance increases:

$C' = K C \quad (K>1)$

Therefore the capacitive reactance decreases:

$X_C' = \frac{1}{\omega C'} = \frac{1}{\omega K C} = \frac{X_C}{K}$

So,

$X_C' < X_C$

3. Effect on total impedance and current

Since $X_C$ decreases, the total impedance becomes

$Z' = \sqrt{R^2 + {X_C'}^2}$

which is smaller than before:

$Z' < Z$

Hence the current increases:

$I' = \frac{V}{Z'} > \frac{V}{Z} = I$

4. Glow of the bulb

The brightness of the bulb depends on the power dissipated in it:

$P = I^2 R$

Since $R$ is unchanged and $I$ increases, the power increases. Therefore the bulb glows more brightly.

5. Option check

• A: becomes zero — false
• B: remains same — false
• C: increases — true
• D: decreases — false

Therefore, the correct answer is:

$\boxed{\text{C: increases}}$