Circuit Analysis - series RC Circuit with AC supply If anyone is familiar with Horowitz and Hill... its exercise 1.19
Show that all the average power delivered to the preceding circuit winds up in the resistor. Do this by computing the value of $V^2/R$. What is the power, in watts for a series circuit of a 1$\mu F$ capacitor and a 1k resistor placed across the 110 Volt RMS, 60 Hz powerline?
The circuit in question is an AC supply with a cap and resistor in series. Simple.
I've been pouring myself into this complex algebra problem and have made no progress over several hours. I'm desperate to understand what they want but cannot get there without some help.
 A: (I am not sure what part is throwing you, please ask for elaboration in the comments and I will do my best to explain.)
The usual convention is to use $j = \sqrt{-1}$ to avoid confusion with the  symbol for current, $i$.
If the supply voltage is $v(t) = \sqrt{2} \operatorname{Re} V_{\text{sup}} e^{j \omega t}$ and the steady state current is $i(t) = \sqrt{2} \operatorname{Re} I e^{j \omega t}$, then the instantaneous power delivered to the circuit is (as is usual convention, $V_{\text{sup}}, I \in \mathbb{C}$ are RMS quantities, also note that $\operatorname{Re} Z = \frac{1}{2}(Z+\overline{Z})$):
\begin{eqnarray}
p(t) &=& v(t)i(t) \\
&=&  2(\operatorname{Re} V_{\text{sup}} e^{j \omega t})(\operatorname{Re} I e^{j \omega t}) \\
&=&  2 \frac{1}{2}(V_{\text{sup}} e^{j \omega t}+ \overline{V_{\text{sup}}} e^{-j \omega t}) \frac{1}{2}(I e^{j \omega t}+ \overline{I} e^{-j \omega t}) \\
&=& \frac{1}{2}( V_{\text{sup}} I e^{j 2 \omega t} +
\overline{V_{\text{sup}}} I +
V_{\text{sup}} \overline{I} +
 \overline{V_{\text{sup}}} \overline{I} e^{-j 2 \omega t}) \\
 &=& (\operatorname{Re} V_{\text{sup}} I e^{j 2 \omega t}) + (\operatorname{Re} \overline{V_{\text{sup}}} I)
 \end{eqnarray}
 from which we see that the average power (integrating $t \mapsto e^{j 2 \omega t}$ over a cycle results in zero) is given by $P = \operatorname{Re} \overline{V_{\text{sup}}} I$.
Since the circuit is linear (lumped, time invariant, etc.), the voltage and current are related by $V_{\text{sup}} = I Z(j\omega)$.
  In this case, $Z(j\omega) = R + \frac{1}{j \omega C}$, so we obtain the average power delivered to the circuit is
  \begin{eqnarray}
  P &=& \operatorname{Re} \overline{V_{\text{sup}}} I \\ 
  &=&  \operatorname{Re} \frac{|V_{\text{sup}}|^2 }{Z(j\omega)} \\
  &=&  |V_{\text{sup}}|^2 \operatorname{Re}( \frac{1}{R} \frac{(\omega R C)^2 + j \omega R C}{1+(\omega R C)^2}) \\
  &=& \frac{|V_{\text{sup}}|^2}{R} \frac{(\omega R C)^2}{1+(\omega R C)^2}
  \end{eqnarray}
Now consider the average power delivered to the resistor: $P_R = \operatorname{Re} \overline{V_{\text{R}}} I$. Since the capacitor and resistor form a voltage divider, we have $V_{\text{R}} = V_{\text{sup}} \frac{R}{Z(j \omega)}$, from which we get $P_R = \operatorname{Re} \overline{V_{\text{sup}} \frac{R}{Z(j \omega)}} \frac{V_{\text{sup}}}{Z(j \omega)} = R \frac{|V_{\text{sup}}|^2}{|Z(j \omega)|^2} = P $. Hence the average power delivered to the circuit is the same as the average power delivered to the resistor.
For the computational part, we use $P= \frac{|V_{\text{sup}}|^2}{R} \frac{(\omega R C)^2}{1+(\omega R C)^2} = \frac{|V_{\text{sup}}|^2}{R} \frac{1}{1+\frac{1}{(\omega R C)^2}}$, with $|V_{\text{sup}}| = 110$, $\omega = 120 \pi$, $C=1 \mu F$, $R=1k$ to get $P \approx 1.5057 W$.
(Contrast this with the $> 12 W$ power that would be delivered if the capacitor was removed.)
A: This is a simple series $RC$ across a $110$ volt $60$ HZ with $R=1k\Omega$ and $C=1 \mu F$ capacitor.
First solve for $Xc=1/2(3.1416)60(.000001)$, the capacitive reactance of $C$ which $2.65k\Omega$.  Then solve for $Z=\sqrt{R^2 + Xc^2}$, vector sum of $R$ and $C$ which is $2.83k\Omega$.
Then solve for the current $I=V/Z=110 V/2.83k\Omega$ which is $38.8 mA$.  The true power $P_{\text{true}}=I^2R$, which is $38.8mA^2 \cdot 1k\Omega = 1.51 W$.  Hope this helps.
