\begin{problem*}{19.58} % internal resistance
A $45\U{V}$ battery of negligable internal resistance is connected to
a $38\U{k\Ohm}$ and a $27\U{k\Ohm}$ resistor in series.  What reading
will a voltmeter, of internal resistance $95\U{k\Ohm}$, give when used
to measure the voltage across each resistor?  What is the percent
inaccuracy due to meter resistance for each case?
\end{problem*}

\begin{solution}
Case 1: \\
The original situation looks like
\begin{center}
\begin{asy}
import Circ;
real u = 0.5cm;
TwoTerminal B = source((0,0), ang=90, type=DC, "$45\U{V}$", "$V$");
pair a = B.end+(0,u);
pair b = B.beg-(0,u);
TwoTerminal Ra = resistor(a, "$38\U{k\Ohm}$", "$R_1$");
TwoTerminal Rb = resistor(Ra.end, "$27\U{k\Ohm}$", "$R_2$");
TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), ang=-90, "", "$I$");
wire(Rb.end, I.beg, nsq);
wire(I.end, b, udsq);
wire(b, B.beg, nsq);
wire(a, B.end, nsq);
\end{asy}
\end{center}
Using Kirchhoff's loop rule
\begin{align*}
  V - IR_1 - IR_2 &= 0 \\
  I &= \frac{V}{R_1+R_2}
\end{align*}
so
\begin{align*}
  V_1 &= IR_1 = \frac{VR_1}{R_1+R_2} \approx 26.3\U{V} \\
  V_2 &= IR_2 = \frac{VR_2}{R_1+R_2} \approx 18.7\U{V}
\end{align*}

Case 2: \\
With the voltmeter across $R_1$ we have
\begin{center}
\begin{asy}
import Circ;
real u = 0.5cm;
TwoTerminal B = source((0,0), ang=90, type=DC, "$45\U{V}$", "$V$");
pair a = B.end+(0,u);
pair b = B.beg-(0,u);
TwoTerminal Ra = resistor(a, "$38\U{k\Ohm}$", "$R_1$");
TwoTerminal Ia = current(Ra.end, "", "$I_1$");
TwoTerminal Rv = resistor(a+(0,4u), "$95\U{k\Ohm}$", "$R_v$");
TwoTerminal Iv = current(Rv.end, "", "$I_v$");
TwoTerminal Rb = resistor(Ia.end, "$27\U{k\Ohm}$", "$R_2$");
TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), ang=-90, "", "$I_T$");
wire(Rb.end, I.beg, nsq);
wire(I.end, b, udsq);
wire(b, B.beg, nsq);
wire(a, B.end, nsq);
wire(a, Rv.beg, nsq);
wire(Iv.end, Ia.end, rlsq);
\end{asy}
\end{center}
Using our formula for resistors in parallel, we can bundle $R_v$ and $R_1$ into a single resistor $R_1'$, where
$$ R_1' = \p({\frac{1}{R_1} + \frac{1}{R_v}})^{-1} = 27.14285714\ldots\U{k\Ohm} $$

Once we've done that, we have the same situation as in Case 1, but with
\begin{align*}
  R_1 &\rightarrow R_1' \\
  I   &\rightarrow I_T
\end{align*}
so
\begin{align*}
  V_1' &= \frac{VR_1'}{R_1'+R_2} =
          \frac{ V \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} }
               { \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} + R_2} \approx \ans{22.6\U{V}} \\
  \frac{V_1'}{V_1} &= \frac{R_1'(R_1+R_2)}{R_1(R_1'+R_2)}
    = \frac{ \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1}\cdot(R_1+R_2) }
           { R_1\cdot\p[{\p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} + R_2}] }
    = \frac{ \p({\frac{R_v+R_1}{R_1 R_v}})^{-1}\cdot(R_1+R_2) }
           { R_1\cdot\p[{\p({\frac{R_v+R_1}{R_1 R_v}})^{-1} + R_2}] }
    = \frac{ \frac{R_1 R_v}{R_v+R_1}\cdot(R_1+R_2) }
           { R_1\cdot\p({\frac{R_1 R_v}{R_v+R_1} + \frac{R_2(R_v+R_1)}{R_v+R_1}}) } \\
    &=\frac{ R_1 \frac{R_v (R_1+R_2)}{R_v+R_1} }
           { R_1\cdot\p({\frac{R_1 R_v + R_2(R_v+R_1)}{R_v+R_1}}) }
    = \frac{ R_v R_1 + R_v R_2 }
           { R_1 R_v + R_2 R_v + R_1 R_2 }
    = \frac{ R_v (R_1 + R_2) }
           { R_v (R_1 + R_2) + R_1 R_2 } \\
    &= 0.8575\ldots \\
\end{align*}
Obviously, we could plug in known numbers and solve for
$\frac{V_1'}{V_1}$ after the first equality above, but crunching
through some simplifying algebra reveals the pretty spectacular final
form, from which you can trivially see that the fractional error in
Case 3 will be the same as that for Case 2.

Finally the percent error is given by
$$ \text{Error}_1 = 1-\frac{V_1'}{V_1} = 0.1425\ldots \approx \ans{14\%} $$

Case 3: \\
With the voltmeter across $R_2$ we have the same situation as Case 1, but with
\begin{align*}
  X_1 &\leftrightarrow X_2
\end{align*}
for any symbol $X$ (i.e. $R_1 \leftrightarrow R_2$, \ldots).
However, the equation for error in $V_1'$ is not effected by this exchange, so
$$ \text{Error}_2 = \text{Error}_1 \approx \ans{14\%} $$
The voltage $V_2'$ measured is given by
$$ V_2' = \frac{VR_2'}{R_2'+R_1} =
          \frac{ V \p({\frac{1}{R_2}+\frac{1}{R_v}})^{-1} }
               { \p({\frac{1}{R_2}+\frac{1}{R_v}})^{-1} + R_1} \approx \ans{16.0\U{V}} $$
\end{solution}
