Question 44: Current Electricity

Question

Three bulbs of $$40 \mathrm{~W}, 60 \mathrm{~W}$$, and $$100 \mathrm{~W}$$ are arranged in series with a $$220 \mathrm{~V}$$ source. The maximum light is obtained from

Accepted Answer

When bulbs are connected in series, the same current flows through each bulb. The power dissipation in a bulb is given by:
$$ P = \frac{V^2}{R} $$
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 $$40 \mathrm{~W}, 60 \mathrm{~W},$$ and $$100 \mathrm{~W}$$, 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:
$$ R_b = \frac{V_b^2}{P} $$
Let's calculate the resistance for each bulb:
For the $$40 \, \mathrm{W}$$ bulb:
$$ R_{40} = \frac{220^2}{40} = 1210 \, \Omega $$
For the $$60 \, \mathrm{W}$$ bulb:
$$ R_{60} = \frac{220^2}{60} = 806.67 \, \Omega $$
For the $$100 \, \mathrm{W}$$ bulb:
$$ R_{100} = \frac{220^2}{100} = 484 \, \Omega $$
When bulbs are connected in series, the total resistance $$ R_{total} $$ is the sum of the individual resistances:
$$ R_{total} = R_{40} + R_{60} + R_{100} $$
$$ R_{total} = 1210 + 806.67 + 484 = 2500.67 \, \Omega $$
The total current in the series circuit is given by Ohm's Law:
$$ I = \frac{V_{source}}{R_{total}} $$
$$ I = \frac{220}{2500.67} = 0.088 \, \mathrm{A} $$
The voltage drop across each bulb can be found using Ohm's Law again:
Voltage drop across the $$40 \mathrm{~W}$$ bulb: $$ V_{40} = I 	imes R_{40} = 0.088 	imes 1210 = 106.48 \, \mathrm{V} $$
Voltage drop across the $$60 \mathrm{~W}$$ bulb: $$ V_{60} = I 	imes R_{60} = 0.088 	imes 806.67 = 71.79 \, \mathrm{V} $$
Voltage drop across the $$100 \mathrm{~W}$$ bulb: $$ V_{100} = I 	imes R_{100} = 0.088 	imes 484 = 42.59 \, \mathrm{V} $$
The power dissipated by each bulb in this series arrangement is:
Power in the $$40 \mathrm{~W}$$ bulb: $$ P_{40} = I^2 	imes R_{40} = (0.088)^2 	imes 1210 = 9.38 \, \mathrm{W} $$
Power in the $$60 \mathrm{~W}$$ bulb: $$ P_{60} = I^2 	imes R_{60} = (0.088)^2 	imes 806.67 = 6.30 \, \mathrm{W} $$
Power in the $$100 \mathrm{~W}$$ bulb: $$ P_{100} = I^2 	imes R_{100} = (0.088)^2 	imes 484 = 3.74 \, \mathrm{W} $$
From these calculations, it is clear that the $$40 \mathrm{~W}$$ bulb dissipates the most power when connected in series and thus will glow the brightest.
Therefore, the answer is:
Option A: 40 W

Answer

40 W

Answer

60 W

Answer

All give the same light

Answer

100 W

Current Electricity

Three bulbs of $40 \mathrm{~W}, 60 \mathrm{~W}, and 100 \mathrm{~W} are arranged in series with a 220 \mathrm{~V}$ source. The maximum light is obtained from

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