\begin{problem*}{21.29}
Consider the circuit shown in Figure P21.29.
Find \Part{a} the current in the $R_1 = 20\Omega$ resistor and
 \Part{b} the potential difference between points $a$ and $b$.
\begin{center}
\begin{empfile}[2]
\begin{emp}(0,0)
  pair N[];
  numeric dx, ddx, dy;
  dx := 4cm;
  ddx := 1cm;
  dy := 1.5 cm;
  % draw the top branch components
  resistor.T(origin, normal, 0, "R_3", "10.0\ohm");
  battery.T(R.T.r, 0, "V", "25.0 V");
  N[0] := R.T.l;
  N[1] := B.T.p;
  % draw the top-middle branch components
  centerto.t(N[0], N[1], -dy, res);
  resistor.t(t, normal, 0, "R_4", "10.0\ohm");
  N[2] := (xpart N[0], ypart R.t.l);
  N[3] := (xpart N[1], ypart R.t.r);
  % draw the bottom-middle branch components
  centerto.b(N[0], N[1], -2dy, res);
  resistor.b(b, normal, 0, "R_5", "5.00\ohm");
  N[4] := (xpart N[0], ypart R.b.l);
  N[5] := (xpart N[1], ypart R.b.r);
  % draw the bottom branch components
  N[6] := N[2] - (ddx,0);
  N[7] := N[3] + (ddx,0);
  N[8] := N[6] - (0,2dy);
  N[9] := N[7] - (0,2dy);
  centreof.L(N[8], N[6], res);
  resistor.L(c.L, normal, phi.L, "\mbox{5.00\ohm}", "$R_2$");
  centreof.R(N[9], N[7], res);
  resistor.R(c.R, normal, phi.R, "R_1", "20.0\ohm");
  % draw the currents
  %centreof.T(R.T.r, B.T.n, cur);
  %current.T(c.T, phi.T, "", "");
  % draw the wires
  wire(R.t.l, N[0], rlsq);
  wire(R.t.r, N[1], rlsq);
  wire(R.b.l, N[2], rlsq);
  wire(R.b.r, N[3], rlsq);
  wire(N[2], R.L.r, rlsq);
  wire(N[3], R.R.r, rlsq);
  wire(N[8], R.L.l, nsq);
  wire(N[9], R.R.l, nsq);
  wire(N[8], N[9], nsq);
  % draw the nodes
  draw N[2] withpen pencircle scaled 3pt;
  draw N[3] withpen pencircle scaled 3pt;
  draw (N[2]+N[6])/2 withpen pencircle scaled 3pt;
  draw (N[3]+N[7])/2 withpen pencircle scaled 3pt;
  % label the connection points
  puttext.top("$a$", (N[2]+N[6])/2);
  puttext.top("$b$", (N[3]+N[7])/2);
\end{emp}
\end{empfile}
\end{center}
\end{problem*}

\begin{solution}
Label the voltage $V = 25.0\U{V}$ and the resistances (clockwise from
$b$) $R_1 = 20.0\Omega$, $R_2 = 5.00\Omega$, $R_3 = 10.0\Omega$, $R_4
= 10.0\Omega$, and $R_5 = 5.00\Omega$.

Computing some equivalent resistance of $R_1$ and $R_2$ in series we
have
\begin{equation}
 R_s = R_1 + R_2 = 25.0\Omega
\end{equation}
Computing the equivalent resistance of $R_s$, $R_4$, and $R_5$ in
parallel we have
\begin{equation}
 R_p = \left(\frac{1}{R_4} + \frac{1}{R_5} + \frac{1}{R_s}\right)^{-1}
     = 2.94\Omega
\end{equation}
And the equivalent resistance of the entire setup is
\begin{equation}
 R_e = R_p + R_3 = 12.94\Omega
\end{equation}

The total current is then (from Ohm's law)
\begin{equation}
 I_e = \frac{V}{R_e} = 1.93\U{A}
\end{equation}
And the voltage from $a$ to $b$ is
\begin{equation}
 V_{ab} = I_e R_p = \ans{5.68\U{V}}
\end{equation}
Which is what they were looking for in \Part{b}.

The current through the branch with $R_1$ and $R_2$ is then
\begin{equation}
 I_s = \frac{V_{ab}}{R_s} = \ans{227\U{mA}}
\end{equation}
Which is what they were looking for in \Part{a}.
\end{solution}
