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Two heaters of same length and material but of different thickness are connected in series across a power supply. The power dissipated:

(a) will be same in both.

(b) will be more in thinner wire

(c) will be more in thicker wire

(d) cannot be predicted

My answer is (b) as thinner wire has high resistance and power is directly proportional to current squared and resistance. The answer key states the correct answer to be option (c). Why is it so?

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    – Yashas
    Apr 22, 2017 at 12:15
  • $\begingroup$ Sir please give me answer .its urgent $\endgroup$ Apr 22, 2017 at 12:23
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3 Answers 3

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A thicker wire always has lesser resistance as electrons are able to flow more easily, rather than when they have to squeeze through a narrow cross-section, which offers a lot of obstruction to their movement.

And, as the wires are in series, both resistors receive the same current. Thus, power is proportional only to resistance and the power dissipated in the thinner wire is MORE.

Perhaps your answer key is wrong. Are you sure they're in series?

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The answer key is wrong.

Imagine that the resistance of the heater with thicker wire is so less that practically there is no resistance. This means that we can assume that the second heater is not even there.

In which case, obviously the first heater will consume more power, because the other one is not even there!

So, the answer to this question is definitely not (c).

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You have a choice of formulae for the power $P$ dissipated in a resistor $R$.

$P=IV=I^2R=\frac{V^2}{R}$

In this example you are told that the resistors are in series so the current $I$ through each of the resistors is the same. So in this case use $P=I^2R$ to make the comparison.

If the resistors had been parallel then the voltage across the resistors would be the same, so use $P=\frac{V^2}{R}$ to make the comparison.

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