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The instantaneous current $i$ and the instantaneous voltage $v$ in a pure resistance a.c. circuit is given by:

$$i=i_{\text{max}} \sin \omega t$$ $$ \text{and} \space v= v_{\text{max}} \sin \omega t$$

Since the power follows:

$$P = iv =i^2 R$$

Show that an equation for instantaneous power is;

$$P=i_{\text{max}}^2 R[1-\cos 2(\omega t)]/2$$

I have been revising basic compound angles and I am struggling to understand the following question from the examples I have previously studied on such topic.

I cannot see how the compound angle formulae relates to this question (if I am correct in thinking that it is relevant for a question of this nature).

A first step, or point of direction/suggestion would be brilliant. Thank You.

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1 Answer 1

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If you're question is "how can I turn $(\sin(\omega t))^2$ into a double-angle-type form", then you should look at your double angle formulas.

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  • $\begingroup$ Would that be be the first part in the process to consider the double angled formula for instantaneous current and voltage to prove the equation? $\endgroup$
    – Andre
    Feb 2, 2014 at 17:46
  • $\begingroup$ @Andre I don't think I know what your question is asking, because I don't know what process you could possibly have in mind. As far as I can tell, the $I^2$ in your formula leads to a sin-squared term, and all you're trying to do now is simplify that term. $\endgroup$
    – tabstop
    Feb 2, 2014 at 17:48
  • $\begingroup$ Sorry for the confusion. I am in the process of completing a past mathematics paper as revision in preparation for an exam in two weeks from which I came across the question in subject. I have to show the equation for instantaneous power is as the above, where I know the voltage, current and power. I interpreted it in the form of 'compound angles', which I am no longer sure is correct? So any possible solutions given would be amazing!! $\endgroup$
    – Andre
    Feb 2, 2014 at 18:00
  • $\begingroup$ If you know that $\cos 2x=1-2\sin^2 x$, which you had better, then you can rearrange this formula to "solve" for $\sin^2 x$ and use that to replace the $\sin^2 x$ in your equation above. $\endgroup$
    – tabstop
    Feb 2, 2014 at 18:04
  • $\begingroup$ Thanks. So regarding sin, is it then a case of substitution to prove? $\endgroup$
    – Andre
    Feb 2, 2014 at 21:28

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