1 \begin{problem*}{19.58} % internal resistance
2 A $45\U{V}$ battery of negligable internal resistance is connected to
3 a $38\U{k\Ohm}$ and a $27\U{k\Ohm}$ resistor in series. What reading
4 will a voltmeter, of internal resistance $95\U{k\Ohm}$, give when used
5 to measure the voltage across each resistor? What is the percent
6 inaccuracy due to meter resistance for each case?
11 The original situation looks like
16 TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
19 TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
20 TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
21 TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I$");
22 wire(Rb.end, I.beg, nsq);
28 Using Kirchoff's loop rule
30 V - IR_1 - IR_2 &= 0 \\
31 I &= \frac{V}{R_1+R_2}
35 V_1 &= IR_1 = \frac{VR_1}{R_1+R_2} \approx 26.3\U{V} \\
36 V_2 &= IR_2 = \frac{VR_2}{R_1+R_2} \approx 18.7\U{V}
40 With the voltmeter across $R_1$ we have
45 TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
48 TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
49 TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$");
50 TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$");
51 TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$");
52 TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
53 TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I_T$");
54 wire(Rb.end, I.beg, nsq);
59 wire(Iv.end, Ia.end, rlsq);
62 Using our formula for resistors in parallel, we can bundle $R_v$ and $R_1$ into a single resistor $R_1'$, where
63 $$ R_1' = \p({\frac{1}{R_1} + \frac{1}{R_v}})^{-1} = 27.14285714\ldots\U{k\Ohm} $$
65 Once we've done that, we have the same situation as in Case 1, but with
67 R_1 &\rightarrow R_1' \\
72 V_1' &= \frac{VR_1'}{R_1'+R_2} =
73 \frac{ V \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} }
74 { \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} + R_2} \approx \ans{22.6\U{V}} \\
75 \frac{V_1'}{V_1} &= \frac{R_1'(R_1+R_2)}{R_1(R_1'+R_2)}
76 = \frac{ \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1}\cdot(R_1+R_2) }
77 { R_1\cdot\p[{\p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} + R_2}] }
78 = \frac{ \p({\frac{R_v+R_1}{R_1 R_v}})^{-1}\cdot(R_1+R_2) }
79 { R_1\cdot\p[{\p({\frac{R_v+R_1}{R_1 R_v}})^{-1} + R_2}] }
80 = \frac{ \frac{R_1 R_v}{R_v+R_1}\cdot(R_1+R_2) }
81 { R_1\cdot\p({\frac{R_1 R_v}{R_v+R_1} + \frac{R_2(R_v+R_1)}{R_v+R_1}}) } \\
82 &=\frac{ R_1 \frac{R_v (R_1+R_2)}{R_v+R_1} }
83 { R_1\cdot\p({\frac{R_1 R_v + R_2(R_v+R_1)}{R_v+R_1}}) }
84 = \frac{ R_v R_1 + R_v R_2 }
85 { R_1 R_v + R_2 R_v + R_1 R_2 }
86 = \frac{ R_v (R_1 + R_2) }
87 { R_v (R_1 + R_2) + R_1 R_2 } \\
90 Obviously, we could plug in known numbers and solve for
91 $\frac{V_1'}{V_1}$ after the first equality above, but crunching
92 through some simplifying algebra reveals the pretty spectacular final
93 form, from which you can trivially see that the fractional error in
94 Case 3 will be the same as that for Case 2.
96 Finally the percent error is given by
97 $$ \text{Error}_1 = 1-\frac{V_1'}{V_1} = 0.1425\ldots \approx \ans{14\%} $$
100 With the voltmeter across $R_2$ we have the same situation as Case 1, but with
102 X_1 &\leftrightarrow X_2
104 for any symbol $X$ (i.e. $R_1 \leftrightarrow R_2$, \ldots).
105 However, the equation for error in $V_1'$ is not effected by this exchange, so
106 $$ \text{Error}_2 = \text{Error}_1 \approx \ans{14\%} $$
107 The voltage $V_2'$ measured is given by
108 $$ V_2' = \frac{VR_2'}{R_2'+R_1} =
109 \frac{ V \p({\frac{1}{R_2}+\frac{1}{R_v}})^{-1} }
110 { \p({\frac{1}{R_2}+\frac{1}{R_v}})^{-1} + R_1} \approx \ans{16.0\U{V}} $$