When bulbs are connected in series, the same current flows through each bulb. The power dissipation in a bulb is given by:
P=‌where:
P is the power
V is the voltage across the bulb
R is the resistance of the bulb
Given that the power ratings of the bulbs are
40W,60W, and 100 W , and they are all rated for a common voltage source (typically each bulb would be rated for 220 V ), we can use the power rating to deduce the resistance of each bulb:
Rb=‌Let's calculate the resistance for each bulb:
For the 40 W bulb:
R40=‌=1210ΩFor the 60 W bulb:
R60=‌=806.67ΩFor the 100 W bulb:
R100=‌=484Ω When bulbs are connected in series, the total resistance
R‌total ‌ is the sum of the individual resistances
‌R‌total ‌=R40+R60+R100‌R‌total ‌=1210+806.67+484=2500.67ΩThe total current in the series circuit is given by Ohm's Law:
‌I=‌| V‌source ‌ |
| R‌total ‌ |
‌I=‌=0.088A The voltage drop across each bulb can be found using Ohm's Law again:
Voltage drop across the 40 W bulb:
V40=I×R40=0.088×1210=106.48VVoltage drop across the 60 W bulb:
V60=I×R60=0.088×806.67=71.79VVoltage drop across the 100 W bulb:
V100=I×R100=0.088×484=42.59VThe power dissipated by each bulb in this series arrangement is:
Power in the 40 W bulb:
P40=I2×R40=(0.088)2×1210=9.38WPower in the 60 W bulb:
P60=I2×R60=(0.088)2×806.67=6.30WPower in the 100 W bulb:
P100=I2×R100=(0.088)2×484=3.74WFrom these calculations, it is clear that the 40 W bulb dissipates the most power when connected in series and thus will glow the brightest.
Therefore, the answer is:
Option A: 40 W