Refer to Figure 19-20. Determine Zth, for the circuit within the beige box as viewed from terminals A and 8. This is the same circuit used in Example 19-5.

Question

Refer to Figure 19-20. Determine [latex]Z_{th}[/latex], for the circuit within the beige box as viewed from terminals A and B. This is the same circuit used in Example 19-5.

Accepted Answer

First, replace the voltage source with its internal impedance (zero in this case), as shown in Figure 19-21.

Looking from terminals A and B. [latex]C_{2}[/latex] appears in parallel with the series combination of [latex] R_{1}[/latex] and[latex]C_{1}[/latex]. This entire combination is in series with[latex]R_{2}[/latex], The calculation for [latex]Z_{th}[/latex], is as follows:

[latex]Z_{th}= R_{2}\angle 0°+ \frac{(X_{C2}\angle -90°) (R_{1}- jX_{C1})}{R_{1}-jX_{C1}- jX_{C2}} [/latex]

[latex] \ \ \ \ \ \ =560\angle 0°\Omega + \frac{(1.5\angle -90°k\Omega )(1.0 \ k\Omega - j1.5 \ k\Omega )}{1.0 \ k\Omega -j3 \ k\Omega } [/latex]

[latex] \ \ \ \ \ \ =560\angle 0°\Omega + \frac{(1.5\angle -90°k\Omega )(1.8\angle -56.3°k\Omega )}{3.16\angle -71.6°k\Omega } [/latex]

[latex]\ \ \ \ \ \ = 560\angle 0°\Omega + 854\angle -74.7°\Omega=560 \ \Omega + 225 \ \Omega - j824 \ \Omega [/latex]

[latex]\ \ \ \ \ \ =785 \ \Omega - j824 \ \Omega = 1138 \angle -46.4°\Omega [/latex]

Question 19.8: Refer to Figure 19-20. Determine Zth, for the circuit within...

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