Two bulbs of 60W and 100W are connected
in series with an AC power supply of 100V.
Which bulb will glow brighter and why?
Answer:
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.