Question

Two lamps, one rated 220 V, 50 W and the other rated 220 V, 100 W are connected in series with mains of 220 V. Choose the correct one.

• A. 100 W lamp consume more power
• B. 50 W lamp consume more power
• C. 50 W lamp consume less power
• D. both consume same power.

Answer 1

The correct option is B 50 W lamp consume more power

Assuming the bulbs were manufactured to operate at 220 volts, according to energy dissipation law I=P/V, the current and according to the ohm's law R=V/I, the resistance is determined.
For 100 W bulb:
The current I=P/V, I=100/220, that is, I[100]=0.45 amps for the 100 watt bulb.The resistance R=V/I, R=220/0.45, that is, R[100]=489 ohms. The 100 watt bulb has 489 ohms resistance when burning.
For 50 W bulb:
The current I=P/V, I=50/220, that is, I[50]=0.23 amps for the 50 watt bulb.The resistance R=V/I, R=220/0.23, that is, R[50]=957 ohms. The 50 watt bulb has 957 ohms resistance when burning.
As these 2 bulbs are connected in series, the series resistances add directly. So, the total resistance is 489 ohms + 957 ohms = 1446 ohms.
So in our 220 volt circuit, I=V/R, I=220/1446, I=0.152 amps.
Hence, the total circuit current is 0.152 amps.
All points in a series circuit see the same current. So, volt drop at the 50 watt bulb is V=IR, that is, $0.152\times 957=114.8$ volts of drop and volt drop at the 100 watt bulb is V=IR, that is, $0.152\times 489=74.3$ volts of drop.
At the 50 watt bulb; $P=voltagedropV\times I$; $P=114.8\times 0.152$; P= 17.44 watts.
At the 100 watt bulb; $P=voltagedropV\times I$; $P=74.3\times 0.152$; P = 11.29 watts.

The more watts burning means brighter light from the 50 watt bulb. Hence, the 50 W bulb consumes more power.