# Where does the power absorbed of this capacitor come from?

In my circuits book, the voltage across a capacitor is given as $v(t)=100\cos({2\pi60t})$. Because $i_{ab}=C\frac{d}{dt}v_{ab}$, we naturally compute the current as $i(t)=-120\pi\sin({2\pi60t})$. The book concurs. It is easy to understand that for any given two-terminal component, $p=vi$, or in this case, $p(t)=v(t)i(t)$. However, the book gives the answer to $p(t)$ as $-18,850\sin({2\pi120t})$. However, WolframAlpha gives the link as $-6000\pi^2sin(240\pi t)$.

I have no idea where $\pi$ dissapears off to. I have no idea where $18,850$ comes from. I would be most grateful to anyone who could get me out of this jam.

## 1 Answer

You have

$$p(t)=v(t)i(t)=-100\cdot120\pi\sin(2\pi 60t)\cos(2\pi 60t)= -50\cdot 120\pi\sin(4\pi 60t)=\\=-6000\pi\sin(240\pi t)=-18850\sin(240\pi t)$$

I think you entered one factor of $\pi$ too many in WolframAlpha. Note that I used $2\sin(x)\cos(x)=2\sin(2x)$.

• Insert Picard Facepalm here So its rounding! – user1833028 Sep 29 '14 at 8:44