Asked by violet
A circuit contains two lamps connected in series. What would happen to the brightness of each lamp if two more lamps were added? Explain.
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GPT-5 mini
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Assume the lamps are identical and the supply is a fixed voltage source. In series the same current flows through every lamp, so adding two more in series increases the total resistance and reduces the current.
Let each lamp have resistance R and the battery voltage be V.
- With 2 lamps: R_total = 2R, I2 = V/(2R). Power per lamp P2 = I2^2R = V^2/(4R).
- With 4 lamps: R_total = 4R, I4 = V/(4R) = I2/2. Power per lamp P4 = I4^2R = V^2/(16R) = P2/4.
So each lamp’s current is halved and the electrical power (hence brightness) of each lamp falls to one quarter of what it was with two lamps. (If the source or lamp resistance is nonideal the numerical factor may differ, but the lamps will still become dimmer.)
Let each lamp have resistance R and the battery voltage be V.
- With 2 lamps: R_total = 2R, I2 = V/(2R). Power per lamp P2 = I2^2R = V^2/(4R).
- With 4 lamps: R_total = 4R, I4 = V/(4R) = I2/2. Power per lamp P4 = I4^2R = V^2/(16R) = P2/4.
So each lamp’s current is halved and the electrical power (hence brightness) of each lamp falls to one quarter of what it was with two lamps. (If the source or lamp resistance is nonideal the numerical factor may differ, but the lamps will still become dimmer.)
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