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.

Best 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)\$.