A capacitor, an inductor and an electric bulb are connected in series to an AC supply of variable frequency. As the frequency of supply is increased gradually, then the electric bulb is found toIncrease in brightnessDecrease in brightnessIncrease in brightness, reach a maximum and then decrease in brightnessOf same brightness

The brightness of the bulb in a series LCR circuit (inductor, capacitor, resistor/bulb) connected to an AC supply depends on the impedance of the circuit. The impedance (Z) of a series LCR circuit is given by Z = √[(R^2) + (XL - XC)^2], where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

1. At low frequencies, the capacitive reactance XC is high and the inductive reactance XL is low. This makes the overall impedance Z high, so the current in the circuit is low, and the bulb is dim.

2. As the frequency increases, XC decreases and XL increases. However, XC decreases faster than XL increases, so the overall impedance Z decreases, and the current in the circuit increases. Therefore, the bulb gets brighter.

3. At the resonant frequency, XC equals XL, so the impedance Z is at its minimum (equal to R), and the current is at its maximum. Therefore, the bulb is at its brightest.

4. If the frequency continues to increase beyond the resonant frequency, XC continues to decrease, but now XL is larger than XC, so the overall impedance Z starts to increase again, and the current decreases. Therefore, the bulb gets dimmer.

So, the correct answer is: The brightness of the bulb will increase, reach a maximum at the resonant frequency, and then decrease as the frequency of the AC supply is increased gradually.