Two bulbs of 60W and 100W are connected
in series with an AC power supply of 100V.
Which bulb will glow brighter and why?
When two bulbs are connected in series, the same current flows through both bulbs. The voltage across each bulb depends on its resistance.
The bulb with higher resistance will have a higher voltage drop across it, and hence will glow brighter ¹². The power of a bulb is given by the formula P = V^2/R, where P is the power in watts, V is the voltage across the bulb, and R is the resistance of the bulb.
The resistance of a 60W bulb can be calculated as R = V^2/P = (100V)^2/60W = 166.67 ohms. Similarly, the resistance of a 100W bulb can be calculated as R = V^2/P = (100V)^2/100W = 100 ohms.When these two bulbs are connected in series, their total resistance is R_total = R_60W + R_100W = 266.67 ohms.
The current flowing through the circuit can be calculated as I = V/R_total = 100V/266.67 ohms = 0.375 A.
Using the formula P = V^2/R, we can calculate the power dissipated by each bulb:- Power dissipated by the 60W bulb: P_60W = (100V)^2/166.67 ohms = 60W- Power dissipated by the 100W bulb: P_100W = (100V)^2/100 ohms = 100WTherefore, both bulbs will glow with their rated wattage, but the **100W bulb** will glow brighter than the **60W bulb** because it has a lower resistance and hence a higher voltage drop across it.