Q166 / 180 · COMEDK 2024 (Evening Shift) · Physics
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
- A100 W
- B60 W
- Correct answerC40 W
- DAll give the same light
Solution
The power consumed by a bulb is given by $P = I^2 R$, where $I$ is the current flowing through the bulb and $R$ is its resistance. Since the bulbs are connected in series, the same current $I$ flows through all of them.
The resistance $R$ of a bulb is related to its rated power $P_{rated}$ and rated voltage $V$ by the formula $R = \dfrac{V^2}{P_{rated}}$.
For the given bulbs:
$R_{40} = \dfrac{V^2}{40}$
$R_{60} = \dfrac{V^2}{60}$
$R_{100} = \dfrac{V^2}{100}$
Comparing the resistances, we find $R_{40} > R_{60} > R_{100}$.
Since $P = I^2 R$ and $I$ is constant for all bulbs in series, the power consumed by each bulb is directly proportional to its resistance. Therefore, the bulb with the highest resistance will consume the most power and emit the most light.
Thus, the $40 \mathrm{~W}$ bulb has the highest resistance and will glow the brightest.
Answer: 40 W
The resistance $R$ of a bulb is related to its rated power $P_{rated}$ and rated voltage $V$ by the formula $R = \dfrac{V^2}{P_{rated}}$.
For the given bulbs:
$R_{40} = \dfrac{V^2}{40}$
$R_{60} = \dfrac{V^2}{60}$
$R_{100} = \dfrac{V^2}{100}$
Comparing the resistances, we find $R_{40} > R_{60} > R_{100}$.
Since $P = I^2 R$ and $I$ is constant for all bulbs in series, the power consumed by each bulb is directly proportional to its resistance. Therefore, the bulb with the highest resistance will consume the most power and emit the most light.
Thus, the $40 \mathrm{~W}$ bulb has the highest resistance and will glow the brightest.
Answer: 40 W
Hence, the correct answer is option: 'C'.