Refer to Figure 19-16. Determine the Thevenin voltage for the circuit within the beige box as viewed from terminals A and B.

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Accepted Answer

Thevenin's voltage for the circuit between terminals A and B is the voltage that appears across A and B with [latex]R_{L}[/latex] removed from the circuit. There is no voltage drop across [latex]R_{2}[/latex] because the open between terminals A and B prevents current through it. Thus, [latex] V_{AB}[/latex] is the same as [latex]V_{c2}[/latex] and can be found by the voltage-divider formula.

[latex]V_{AB}= V_{C2}= \left(\frac{X_{C2}\angle -90°}{R_{1}- jX_{C1}- jX_{C2}} \right)V_{s}= \left(\frac{1.5\angle -90°k\Omega }{1.0 \ k\Omega - j3 \ k\Omega } \right)10 \angle 0°V[/latex]

[latex] \ \ \ \ \ \ \ =\left(\frac{1.5\angle -90°k\Omega }{3.16\angle -71.6°k\Omega} \right)10\angle 0°V= 4.75\angle -18.4°V[/latex]

[latex] V_{th}= V_{AB}= 4.75\angle -18.4°V[/latex]

Question 19.5: Refer to Figure 19-16. Determine the Thevenin voltage for th...

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