Find Z_{n} for the circuit in Figure 19-33 (Example 19-13) viewed from the open across terminals A and B
Find Z_{n} for the circuit in Figure 19-33 (Example 19-13) viewed from the open across terminals A and B
First, replace V_{s} with its internal impedance (zero), as indicated in Figure 19-35.
Looking in between terminals A and B. C_{2} is in scries with the parallel combination of R and C_{1}. Thus.
Z_{n}= X_{C2}+ \frac{RX_{C1}}{R+ X_{C1}}= 100 \angle -90°\Omega + \frac{(56\angle 0°\Omega )(50\angle -90°\Omega )}{56 \ \Omega – j50 \ \Omega }\ \ \ \ \ =100 \angle -90°\Omega + 37.3 \angle -48.2\Omega
\ \ \ \ \ =-j100 \ \Omega + 24.8 \ \Omega – j27.8 \ \Omega = 24.8 \ \Omega – j128 \ \Omega
The Norton equivalent impedance is a 24.8 Ω resistance in series with a 128 Ω capacitive reactance.