Question 18.3: Equivalent Resistance Goal Solve a problem involving both se...
Equivalent Resistance
Goal Solve a problem involving both series and parallel resistors.
Problem Four resistors are connected as shown in Figure 18.11a. (a) Find the equivalent resistance between points a and c. (b) What is the current in each resistor if a 42-\mathrm{V} battery is connected between a and c ?
Strategy Reduce the circuit in steps, as shown in Figures 18.11 \mathrm{~b} and 18.11 \mathrm{c}, using the sum rule for resistors in series and the reciprocal-sum rule for resistors in parallel. Finding the currents is a matter of applying Ohm’s law while working backwards through the diagrams.
(a) Find the equivalent resistance of the circuit.
The 8.0- \Omega and 4.0-\Omega resistors are in series, so use the sum rule to find the equivalent resistance between a and b :
R_{\text {eq }}=R_{1}+R_{2}=8.0 \Omega+4.0 \Omega=12 \Omega
The 6.0- \Omega and 3.0-\Omega resistors are in parallel, so use the reciprocal-sum rule to find the equivalent resistance between b and c (don’t forget to invert!):
\begin{aligned} \frac{1}{R_{\mathrm{eq}}} & =\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{6.0 \Omega}+\frac{1}{3.0 \Omega}=\frac{1}{2.0 \Omega} \\ R_{\mathrm{eq}} & =2.0 \Omega \end{aligned}
In the new diagram, 18.11b, there are now two resistors in series. Combine them with the sum rule to find the equivalent resistance of the circuit:
R_{\mathrm{eq}}=R_{1}+R_{2}=12 \Omega+2.0 \Omega=14 \Omega
(b) Find the current in each resistor if a 42-\mathrm{V} battery is connected between points a and c.
Find the current in the equivalent resistor in Figure 8.11 \mathrm{c}, which is the total current. Resistors in series all carry the same current, so this is the current in the 12-\Omega resistor in Figure 8.11 \mathrm{~b}, and also in the 8.0-\Omega and 4.0-\Omega resistors in Figure 8.11a.
I=\frac{\Delta V_{a c}}{R_{\mathrm{eq}}}=\frac{42 \mathrm{~V}}{14 \Omega}=3.0 \mathrm{~A}
Apply the junction rule to point b :
\text { (1) } I=I_{1}+I_{2}
The 6.0-\Omega and 3.0-\Omega resistors are in parallel, so the voltage drops across them are the same:
\begin{aligned} \Delta V_{6 \Omega} & =\Delta V_{3 \Omega} \quad \rightarrow \quad(6.0 \Omega) I_{1}=(3.0 \Omega) I_{2} \quad \rightarrow \\ 2.0 I_{1} & =I_{2} \end{aligned}
Substitute this result into Equation (1), with I=3.0 \mathrm{~A} :
\begin{aligned} 3.0 \mathrm{~A} & =I_{1}+2 I_{1}=3 I_{1} \rightarrow I_{1}=1.0 \mathrm{~A} \\ I_{2} & =2.0 \mathrm{~A} \end{aligned}
Remarks As a final check, note that \Delta V_{b c}=(6.0 \Omega) I_{1}=(3.0 \Omega) I_{2}=6.0 \mathrm{~V} and \Delta V_{a b}=(12 \Omega) I_{1}=36 \mathrm{~V}; therefore, \Delta V_{a c}=\Delta V_{a b}+\Delta V_{b c}=42 \mathrm{~V}, as expected.