Find Z_{th} for the part of the circuit in Figure 19-18 that is within the beige box as viewed from terminals A and B. This is the same circuit used in Example 19-4.
Question 19.7: Find Zth for the part of the circuit in Figure 19-18 that is...
First, replace V_{s} with its internal impedance (zero in this case), as shown in Figure 19-19. Looking in between terminals A and B. R_{1} and X_{L}are in parallel. Thus.
Z_{th}=\frac{(R_{1}\angle 0°)(X_{L}\angle 90°)}{R_{1} + jX_{L}} = \frac{(100\angle 0°\Omega) (50\angle 90°\Omega)}{100 \ \Omega + j50 \ \Omega }\ \ \ \ \ \ =\frac{(100\angle 0°\Omega) (50\angle 90°\Omega) }{112\angle 26.6°\Omega} =44.6\angle 63.4°\Omega
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