\begin{problem*}{21.31}
Calculate the power delivered to each resistor in the circuit shown in
Figure P21.31.
\begin{center}
\begin{empfile}[3]
\begin{emp}(0,0)
  pair N[];
  numeric ddy, dx;
  ddy := 6pt;
  dx := 3cm;
  % draw the left branch components
  battery.L(origin, 90, "\mbox{18.0 V}", "$V$");
  N[0] := B.L.n-(0,ddy);
  N[1] := B.L.p+(0,ddy);
  % draw the horizontal components
  N[2] := N[0]+(dx,0);
  N[3] := N[1]+(dx,0);
  centreof.B(N[0], N[2], res);
  resistor.B(c.B, normal, phi.B, "R_4", "4.00\ohm");
  centreof.T(N[1], N[3], res);
  resistor.T(c.T, normal, phi.T, "R_1", "2.00\ohm");
  % draw the middle branch components
  centreof.M(N[2], N[3], res);
  resistor.M(c.M, normal, phi.M, "R_2", "3.00\ohm");
  % draw the right branch components
  N[4] := 2N[2]-N[0];
  N[5] := 2N[3]-N[1];
  centreof.R(N[4], N[5], res);
  resistor.R(c.R, normal, phi.R, "R_3", "1.00\ohm");
  % draw the currents
  %centreof.T(R.T.r, B.T.n, cur);
  %current.T(c.T, phi.T, "", "");
  % draw the wires
  wire(B.L.p, R.T.l, udsq);
  wire(B.L.n, R.B.l, udsq);
  wire(R.T.r, N[3], nsq);
  wire(R.B.r, N[2], nsq);
  wire(R.M.r, N[3], nsq);
  wire(R.M.l, N[2], nsq);
  wire(R.R.r, N[3], udsq);
  wire(R.R.l, N[2], udsq);
  % draw the nodes
  draw N[2] withpen pencircle scaled 3pt;
  draw N[3] withpen pencircle scaled 3pt;
\end{emp}
\end{empfile}
\end{center}
\end{problem*} % problem 21.31

\begin{solution}
Label the voltage $V = 18.0\U{V}$ and the resistors (starting in the
upper left) $R_1 = 2.00\Omega$, $R_2 = 3.00\Omega$, $R_3 =
1.00\Omega$, and $R_4 = 4.00\Omega$.

To find the total current through the circuit, we compute its
equivalent resistance.  First for the two resistors in parallel
\begin{equation}
 R_p = \left(\frac{1}{R_2} + \frac{1}{R_3}\right){-1} = 0.750\Omega
 %1/3 + 1 = 4/3
\end{equation}
And then for the complete circuit
\begin{equation}
 R_c = R_1 + R_p + R_4 = 6.75\Omega
\end{equation}

Using Ohm's law to calculate the total current $I_c = I_1 = I_4$ we have
\begin{equation}
 I_c = \frac{V}{R_c} = 18/6.75 = 2.67\U{A}
\end{equation}
And the powers dissipated through $R_1$ and $R_4$ are
\begin{align}
 P_1 &= I_1 V_1 = I_c^2 R_1 = \ans{14.2\U{W}} \\
 P_4 &= I_c^2 R_4 = \ans{28.4\U{W}}
\end{align}

The voltage across $R_p$ is given by
\begin{equation}
 V_p = I_c R_p = 2\U{V}
\end{equation}
And the powers dissipated through $R_2$ and $R_3$ are
\begin{align}
 P_2 &= I_2 V_2 = \frac{V_p^2}{R_2} = \ans{1.33\U{W}} \\
 P_3 &= \frac{V_p^2}{R_3} = \ans{4\U{W}}
\end{align}
\end{solution}
