When bulbs are connected in series, the same amount of current flows through each bulb. However, the voltage drop across each bulb is proportional to its wattage rating. Therefore, the bulb with the highest wattage rating will have the highest voltage drop and will glow the brightest.



To calculate the voltage drop across each bulb, we can use Ohm's Law:

V = I * R

where V is the voltage drop, I is the current, and R is the resistance.

Assuming each bulb has a resistance of R, the total resistance of the circuit is:

R_total = 3R

Since the supply voltage is 220V and the current is the same for each bulb, we can calculate the voltage drop across each bulb as:

V_drop = V_supply / 3 = 220V / 3 = 73.33V

Using Ohm's Law again, we can calculate the current through each bulb as:

I = V_drop / R

For a 40W bulb:

R = V^2 / P = 73.33V^2 / 40W = 133.33 ohms
I = 73.33V / 133.33 ohms = 0.55A

For a 60W bulb:

R = V^2 / P = 73.33V^2 / 60W = 89.63 ohms
I = 73.33V / 89.63 ohms = 0.82A

For a 100W bulb:

R = V^2 / P = 73.33V^2 / 100W = 53.33 ohms
I = 73.33V / 53.33 ohms = 1.38A



Based on the calculations, the 100W bulb will glow the brightest when connected in series to a 220V supply. It has the highest voltage drop and therefore the highest brightness. However, it is important to note that connecting bulbs in series is not recommended as if one bulb fails, the circuit will be broken and none of the bulbs will light up.

I m not sure bt .i think it will be 100 W.coz we know V=P/I..and I will be same so the one with more power will have more V .so it should be 100am i correct ???