Find the Thevenin equivalent circuit between A and B of the circuit in Figure 8-33.

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Accepted Answer

First, remove [latex]R_{L}[/latex]. Then [latex]V_{TH}[/latex] equals the voltage across [latex]R_{2}[/latex] + [latex]R_{3}[/latex] as shown in Figure 8-34(a), because [latex]V_{4}[/latex] = 0 V since there is no current through it.

[latex]V_{TH}= \left(\frac{R_{2} + R_{3}}{R_{1}+ R_{2} + R_{3}} \right) V_{S} = \left(\frac{690 \ \Omega }{1690 \ \Omega } \right) 10 \ V = 4.08 \ V[/latex]

To find [latex]R_{TH}[/latex], first replace the source with a short to simulate a zero internal resistance. Then [latex] R_{1}[/latex] appears in parallel with [latex]R_{2}[/latex] + [latex]R_{3}[/latex], and [latex]R_{4}[/latex] is in series with the series-parallel combination of [latex]R_{1}[/latex], [latex]R_{2}[/latex], and [latex]R_{3}[/latex], as indicated in Figure 8-34(b).

[latex]R_{TH}= R_{4} + \frac{R_{1}(R_{2}+ R_{3})}{R_{1}+ R_{2}+ R_{3}}= 1000 \ \Omega + \frac{(1000 \ \Omega )(690 \ \Omega )}{1690 \ \Omega }= 1410 \ \Omega [/latex]

The resulting Thevenin equivalent circuit is shown in Figure 8-34(c),

Question 8.10: Find the Thevenin equivalent circuit between A and B of the ...

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