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nrqed

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It's actually the other way around.

In series the current is the same in all three so the best equation to use for the power is P = R I^2 which shows that the largest resistance will dissipate the most power.

In parallel they all have the same voltage so it's better to use P = V^2/R which shows that the largest resistance will dissipate the smallest power.

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In series, the total resistance is 6 ohms. So the current is V/6. The power in each bulb is the same: PB1,PB2,PB3 = 1/6 V^2, so they are equally bright. This is because each drops a different amount of voltage. VB1 = V/6, VB2 = V/3, VB3 = V/2.

In parallel, the voltage drop for each bulb is the same: V. The current is different. IB1 = V/1, IB2 = V/2, IB3 = V/3. So the power in each bulb is PB1 = V^2, PB2 = 1/2 V^2, PB3 = 1/3 V^2.

So, in series all the bulbs are the same brightness. In parallel, B3 is the dimmest of the 3. But its still brighter than in series.

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In series, the total resistance is 6 ohms. So the current is V/6. CORRECTION: each bulb drops a different amount of voltage. VB1 = V/6, VB2 = V/3, VB3 = V/2. So, the power in each bulb is PB1 = 1/36*V^2, PB2 = 1/18*V^2, PB3 = 1/12*V^2.

In parallel, the voltage drop for each bulb is the same: V. The current is different. IB1 = V/1, IB2 = V/2, IB3 = V/3. So the power in each bulb is PB1 = V^2, PB2 = 1/2 V^2, PB3 = 1/3 V^2.

So, in series B3 is brightest. In parallel, B3 is the dimmest of the 3. But its still brighter than in series.

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series the current is constant ,parallel he voltage is constant

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