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I've been working on this problem, but I can't reach the answer which is B, 0.5W, for this question.

Using

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

As the period is 0.2s, frequency is 5Hz

$f = 5Hz$

$R=20\Omega$

Now consider one cycle,

for the first 0.1 seconds, the PSU pushes 4V out in the positive direction

then for the remaining 0.1 seconds of the cycle, the resistor has a potential difference of 2V in the other negative direction.

My reasoning therefore is that the net voltage per cycle is +2V.

Or should it be the average which is $V_{avg} = \frac{4V-2V}{2} = 1V$

But using my original equation does not then give 0.5W

Do I need to consider time or frequency? Or should I use $P = I^2R$ instead, however, the current is unknown so it would need to be calculated fist.

The question was taken from a paper that does not require any knowledge of AC circuits so there should be a fairly simple solution as this question is supposed to take about 2 minutes to solve.

Thanks

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closed as off-topic by John Rennie, Kyle Kanos, Jon Custer, Chris, Emilio Pisanty Feb 5 '18 at 13:31

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  • $\begingroup$ Suggest you go back to the definition:$$\text{average power}=\frac{\text{energy transferred}}{\text{time taken}}.$$ You must decide on a suitable time interval. $\endgroup$ – Philip Wood Feb 4 '18 at 11:33
  • $\begingroup$ The average power is the average of the instantaneous power, that is, you have to average the power dissipated over the two half-periods. $\endgroup$ – Massimo Ortolano Feb 4 '18 at 11:34
  • $\begingroup$ Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Feb 4 '18 at 12:32
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I would work this out by splitting the problem into two parts: work out the power when at $+4V$ and when at $-2V$

Lets start with $4V$. Over a second you can work out that the resistor is at $4V$ for half the time. The energy dissipated in a second when at $4V$ is therefore: $$\frac{4^2}{20} \times 0.5seconds = 0.4J$$

Using the same idea but substituting $-2V$ you find: $$\frac{(-2)^2}{20} \times 0.5 = 0.1J$$

Add the two energies together gives you the energy dissipated in 1 second which is the power - $0.5W$

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  • $\begingroup$ Sorry but… The $power$ dissipated in the first 0.1 s is $\frac{4^2}{20} \text{W}=0.64 \text{W}$, not half of this. Similar comments for the second 0.1 s. Neither is it logical in general to add power over consecutive time intervals. However, we can say: $energy$ dissipated in first 0.1 s = 0.8 W $\times$ 0.1 s = 0.08 J, and energy dissipated in second 0.1 s =0.02 J, so total energy dissipated over cycle = 0.10 J. So average power =$\frac{0.10 \text{J}}{0.20 \text{s}}$ = 0.50 W. $\endgroup$ – Philip Wood Feb 4 '18 at 13:19
  • $\begingroup$ Correction. It should be 0.8 W, not 0.64 W in first line of above comment. $\endgroup$ – Philip Wood Feb 4 '18 at 13:36
  • $\begingroup$ Edited my answer to talk in terms of the energy - you are correct that my poor use of the wording was technically incorrect. $\endgroup$ – Tech Feb 4 '18 at 13:39
  • $\begingroup$ It's 0.1 s, not 0.5 s. Sorry to be such a bore. $\endgroup$ – Philip Wood Feb 4 '18 at 13:45

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