Find the total current through R3 in Figure 8-26.

Question

Find the total current through [latex]R_{3}[/latex] in Figure 8-26.

Accepted Answer

Step 1: Find the current through [latex]R_{3}[/latex] due to source [latex]V_{S1}[/latex] by replacing source [latex]V_{S2}[/latex] with a short, as shown in Figure 8-27.

Looking from [latex]V_{S1}[/latex],

[latex]R_{T(S1)}= R_{1}+ \frac{R_{2}R_{3}}{R_{2}+ R_{3}}= 1.0 \ k\Omega + \frac{(1.0 \ k\Omega)(2.2 \ k\Omega)}{3.2 \ k\Omega} = 1.69 \ k\Omega[/latex]

[latex]I_{T(S1)}= \frac{V_{S1}}{R_{T(S1)}} = \frac{20 \ V}{1.69 \ k\Omega } = 11.8 \ mA[/latex]

Now apply the current-divider formula to get the current through [latex]R_{3}[/latex], due to source [latex]V_{S1}[/latex].

[latex]I_{3(S1)}= \left(\frac{R_{2}}{R_{2}+ R_{3}} \right) I_{T(S1)} = \left(\frac{1.0 \ k\Omega }{3.2 \ k\Omega } \right) 11.8 \ mA = 3.69 \ mA[/latex]

Notice that this current is upward through [latex]R_{3}[/latex].

Step 2: Find [latex]I_{3}[/latex] due to source [latex]V_{S2}[/latex] by replacing source [latex]V_{S1}[/latex] with a short, as shown in Figure 8-28.

Looking from [latex]V_{S2}[/latex].

[latex]R_{T(S2)}= R_{2}+ \frac{R_{1}R_{3}}{R_{1}+ R_{3}}= 1.0 \ k\Omega + \frac{(1.0 \ k\Omega)(2.2 \ k\Omega)}{3.2 \ k\Omega} = 1.69 \ k\Omega[/latex]

[latex]I_{T(S2)}= \frac{V_{S2}}{R_{T(S2)}} = \frac{15 \ V}{1.69 \ k\Omega } = 8.88 \ mA[/latex]

Now apply the current-divider formula to find the current through [latex]R_{3}[/latex] , due to source [latex]V_{S2}[/latex].

[latex]I_{3(S2)}= \left(\frac{R_{1}}{R_{1}+ R_{3}} \right) I_{T(S2)} = \left(\frac{1.0 \ k\Omega }{3.2 \ k\Omega } \right) 88.8 \ mA = 2.78 \ mA[/latex]

Notice that this current is downward through [latex]R_{3}[/latex].

Step 3: Calculate the total current through [latex]R_{3}[/latex],.
[latex]I_{3(tot)}[/latex] = [latex]I_{3(S1)}[/latex]- [latex]I_{3(S2)}[/latex] = 3.69 mA - 2.78 mA = 0.91 mA = 910 μA
This current is upward through [latex]R_{3}[/latex].

Question 8.9: Find the total current through R3 in Figure 8-26.

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