-\begin{problem*}{31}
- Calculate the currents in each resistor of Fig.~19-49.
+\begin{problem*}{19.31} % resistor networks
+Calculate the currents in each resistor of Fig.~19-49.
\end{problem*}
\begin{nosolution}
\begin{center}
\begin{asy}
- import Circ;
- real u = 3cm;
- TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
- TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
- TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
- pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
- TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
- TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
- TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
- TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
- wire(Ba.end, Raa.end, rlsq);
- wire(Rab.beg, Jbot, nsq);
- wire(Jbot, Rb.end, nsq);
- wire(Jbot, Rcb.end, rlsq);
+import Circ;
+real u = 3cm;
+TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
+TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
+TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
+pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
+TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
+TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
+TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
+TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
+wire(Ba.end, Raa.end, rlsq);
+wire(Rab.beg, Jbot, nsq);
+wire(Jbot, Rb.end, nsq);
+wire(Jbot, Rcb.end, rlsq);
\end{asy}
\end{center}
\end{nosolution}
\begin{solution}
\begin{center}
\begin{asy}
- import Circ;
- TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
- TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
- TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
- pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
- TwoTerminal Ic = current((Jbot+Rcb.end)/2, 0, "", "$I_3$");
- TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
- TwoTerminal Ib = current(Rb.end, -90, "", "$I_2$");
- TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
- TwoTerminal Ia = current(Ba.end, 180, "$I_1$", "");
- TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
- TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
- wire(Ia.end, Raa.end, rlsq);
- wire(Jbot, Ib.end, nsq);
- wire(Jbot, Ic.beg, nsq);
- wire(Ib.end, Rb.end, nsq);
- wire(Ic.end, Rcb.end, rlsq);
- dot("a", Jbot, S);
+import Circ;
+TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
+TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
+TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
+pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
+TwoTerminal Ic = current((Jbot+Rcb.end)/2, 0, "", "$I_3$");
+TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
+TwoTerminal Ib = current(Rb.end, -90, "", "$I_2$");
+TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
+TwoTerminal Ia = current(Ba.end, 180, "$I_1$", "");
+TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
+TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
+wire(Ia.end, Raa.end, rlsq);
+wire(Jbot, Ib.end, nsq);
+wire(Jbot, Ic.beg, nsq);
+wire(Ib.end, Rb.end, nsq);
+wire(Ic.end, Rcb.end, rlsq);
+dot("a", Jbot, S);
\end{asy}
\end{center}
Label the resistors from left to right: $R_1 = 12\U{\Ohm}$, $R_2 =
plug those currents into the junction rule and solve for $I_2$
\begin{align*}
\frac{V_1 + R_3 I_2}{R_{12}} + I_2 - \frac{V_2 - R_3 I_2}{R_{45}} &= 0 \\
- \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2 - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\
- \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2 &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\
- I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}} \\
+ \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2
+ - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\
+ \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2
+ &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\
+ I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}}
+ + 1 + \frac{R_3}{R_{45}}} \\
I_2 &= \ans{-28\U{mA}}
\end{align*}
Where the $-$ sign means the true current is in the opposite direction
I_3 &= \frac{V_2 - R_3 I_2}{R_{45}} = \ans{264\U{mA}}
\end{align*}
-Double-checking our algebra, we see $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$ where difference of $1\U{mA}$ is due to rounding errors from forcing our answers to milli-Volt precision.
+Double-checking our algebra, we see
+ $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$
+where difference of $1\U{mA}$ is due to rounding errors from forcing
+our answers to milli-Volt precision.
\end{solution}
-\begin{problem*}{58}
- 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?
+\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}
The original situation looks like
\begin{center}
\begin{asy}
- import Circ;
- real u = 0.5cm;
- TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
- pair a = B.end+(0,u);
- pair b = B.beg-(0,u);
- TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
- TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
- TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I$");
- wire(Rb.end, I.beg, nsq);
- wire(I.end, b, udsq);
- wire(b, B.beg, nsq);
- wire(a, B.end, nsq);
+import Circ;
+real u = 0.5cm;
+TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
+pair a = B.end+(0,u);
+pair b = B.beg-(0,u);
+TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
+TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
+TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -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 Kirchoff's loop rule
With the voltmeter across $R_1$ we have
\begin{center}
\begin{asy}
- import Circ;
- real u = 0.5cm;
- TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
- pair a = B.end+(0,u);
- pair b = B.beg-(0,u);
- TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
- TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$");
- TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$");
- TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$");
- TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
- TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -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);
+import Circ;
+real u = 0.5cm;
+TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
+pair a = B.end+(0,u);
+pair b = B.beg-(0,u);
+TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
+TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$");
+TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$");
+TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$");
+TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
+TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -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
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