A capacitor 'C', a variable resistor 'R', and a bulb 'B' are connected in series to the AC mains. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R the same; (ii) the resistance R is increased, keeping the capacitance the same?

(i) As the dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V = Q/C). Since the charge Q remains constant in a series circuit, an increase in capacitance C leads to a decrease in voltage V across the capacitor. As a result, the potential drop across the bulb will increase (since both are connected in series). The total voltage across the series combination remains constant (AC mains voltage). Therefore, if the voltage across the capacitor decreases, the voltage across the bulb must increase. So, its brightness will increase.

(ii) As the resistance (R) is increased, the potential drop across the resistor will increase (V = IR, where I is the current, which remains constant in a series circuit). As a result, the potential drop across the bulb will decrease (since both are connected in series). Again, because the total voltage is constant, an increase in the voltage drop across the resistor must result in a decrease in the voltage drop across the bulb. So its brightness will decrease.