Three incandescent bulbs of $100\, W$ each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same wattage are connected in parallel to the same source. Will the bulbs in the two circuits glow with the same brightness? Justify your answer.
(N/A) No, the bulbs will not glow with the same brightness.
$1$. In a series circuit, the total resistance is $R_{series} = R + R + R = 3R$. Since the voltage $V$ is constant, the current in each bulb is $I_{series} = V / (3R) = (1/3) \times (V/R)$.
$2$. In a parallel circuit, each bulb is connected directly to the source voltage $V$. Therefore, the current in each bulb is $I_{parallel} = V/R$.
$3$. Since the power dissipated by a bulb is $P = I^2R$, the power in the series circuit is $P_{series} = (V/3R)^2 \times R = V^2 / (9R)$, while in the parallel circuit, it is $P_{parallel} = (V/R)^2 \times R = V^2 / R$.
$4$. Because $P_{parallel} > P_{series}$, the bulbs in the parallel circuit will glow much more brightly than those in the series circuit.
$1$. In a series circuit, the total resistance is $R_{series} = R + R + R = 3R$. Since the voltage $V$ is constant, the current in each bulb is $I_{series} = V / (3R) = (1/3) \times (V/R)$.
$2$. In a parallel circuit, each bulb is connected directly to the source voltage $V$. Therefore, the current in each bulb is $I_{parallel} = V/R$.
$3$. Since the power dissipated by a bulb is $P = I^2R$, the power in the series circuit is $P_{series} = (V/3R)^2 \times R = V^2 / (9R)$, while in the parallel circuit, it is $P_{parallel} = (V/R)^2 \times R = V^2 / R$.
$4$. Because $P_{parallel} > P_{series}$, the bulbs in the parallel circuit will glow much more brightly than those in the series circuit.
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