EQUIVALENT RESISTANCE GOAL Solve a problem involving both series and parallel resistors. PROBLEM Four resistors are connected as shown in Figure 18.12a. (a) Find the equivalent resistance between points a and c. (b) What is the current in each resistor if a 42−V battery is connected between a and c

Question

EQUIVALENT RESISTANCE

GOAL Solve a problem involving both series and parallel resistors.

PROBLEM Four resistors are connected as shown in Figure 18.12a. (a) Find the equivalent resistance between points [latex]a[/latex] and [latex]c[/latex]. (b) What is the current in each resistor if a [latex]42-\mathrm{V}[/latex] battery is connected between [latex]a[/latex] and [latex]c[/latex] ?

STRATEGY Reduce the circuit in steps, as shown in Figures 18.12b and 18.12c, 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.

Accepted Answer

(a) Find the equivalent resistance of the circuit.

The [latex]8.0-\Omega[/latex] and [latex]4.0-\Omega[/latex] resistors are in series, so use the sum rule to find the equivalent resistance between [latex]a[/latex] and [latex]b[/latex] :

[latex]R_{\mathrm{eq}}=R_{1}+R_{2}=8.0 \Omega+4.0 \Omega=12.0 \Omega[/latex]

The [latex]6.0-\Omega[/latex] and [latex]3.0-\Omega[/latex] resistors are in parallel, so use the reciprocal-sum rule to find the equivalent resistance between [latex]b[/latex] and [latex]c[/latex] (don't forget to invert!):

[latex]\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}[/latex]

In the new diagram, [latex]18.12 \mathrm{~b}[/latex], there are now two resistors in series. Combine them with the sum rule to find the equivalent resistance of the circuit:

[latex]R_{\mathrm{eq}}=R_{1}+R_{2}=12.0 \Omega+2.0 \Omega=14.0 \Omega[/latex]

(b) Find the current in each resistor if a [latex]42-\mathrm{V}[/latex] battery is connected between points [latex]a[/latex] and [latex]c[/latex].

Find the current in the equivalent resistor in Figure 18.12c, which is the total current. Resistors in series all carry the same current, so this value is the current in the [latex]12-\Omega[/latex] resistor in Figure [latex]18.12 \mathrm{~b}[/latex] and also in the [latex]8.0-\Omega[/latex] and [latex]4.0-\Omega[/latex] resistors in Figure 18.12a.

[latex]I=\frac{\Delta V_{a c}}{R_{\mathrm{eq}}}=\frac{42 \mathrm{~V}}{14 \Omega}=3.0 \mathrm{~A}[/latex]

Calculate the voltage drop [latex]\Delta V_{ ext {para }}[/latex] across the parallel circuit, which has an equivalent resistance of [latex]2.0 \Omega[/latex] :

[latex]\Delta V_{ ext {para }}=I R=(3.0 \mathrm{~A})(2.0 \Omega)=6.0 \mathrm{~V}[/latex]

Apply Ohm's law again to find the currents in each resistor of the parallel circuit:

[latex]\begin{aligned}&I_{1}=\frac{\Delta V_{ ext {para }}}{R_{6.0 \Omega}}=\frac{6.0 \mathrm{~V}}{6.0 \Omega}=1.0 \mathrm{~A} \&I_{2}=\frac{\Delta V_{ ext {para }}}{R_{3.0 \Omega}}=\frac{6.0 \mathrm{~V}}{3.0 \Omega}=2.0 \mathrm{~A}\end{aligned}[/latex]

REMARKS As a final check, note that [latex]\Delta V_{b c}=(6.0 \Omega) I_{1}=(3.0 \Omega) I_{2}=6.0 \mathrm{~V}[/latex] and [latex]\Delta V_{a b}=(12 \Omega) I_{1}=36 \mathrm{~V}[/latex]; therefore, [latex]\Delta V_{a c}=\Delta V_{a b}+\Delta V_{b c}=42 \mathrm{~V}[/latex], as expected.

Question 18.3: EQUIVALENT RESISTANCE GOAL Solve a problem involving both se...

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