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Impedance and Admittance Objective of Lecture Demonstrate how to apply Thévenin and Norton transformations to simplify circuits that contain one or more ac sources, resistors, capacitors, and/or inductors. Source Transformation A voltage source plus one impedance in series is said to be equivalent to a current source plus one impedance in parallel when the current into the load and the voltage across the load are the same. Equivalent Circuits Thévenin Vth = In Zn Norton In = Vth/Zth Example 1 First, convert the current source to a cosine function and then to a phasor. I1 = 5mA sin(400t+50o) = 5mA cos(400t+50o-90o)= 5mA cos(400t-40o) I1 = 5mA -40o Example 1 (con’t) Determine the impedance of all of the components when w = 400 rad/s. In rectangular coordinates Z C1 = j / wC1 = j /400rad / s 1F = j 2.5k Z R1 = R1 = 3k Z L1 = jwL1 = j (400rad / s )(0.3H ) = j120 Z L2 = jwL2 = j (400rad / s )( 2 H ) = j800 Z R2 = R2 = 5k Z C2 = j / wC2 = j /400rad / s 0.7 F = j 3.57 k Example 1 (con’t) Convert to phasor notation Z C1 = 2.5k 90o Z R1 = 3k0o Z L1 = 12090o Z L2 = 80090o Z R2 = 5k0o Z C2 = 3.57k 90o Example 1 (con’t) Vt h1 = I1Z C1 = 5mA 40o 2.5k 90o Vt h1 = (5mA)2.5k 40o 90o Vt h1 = 7.5V 130o Example 1 (con’t) Find the equivalent impedance for ZC1 and ZR1 in series. This is best done by using rectangular coordinates for the impedances. Z eq = Z R Z C = 3k j 2.5k 1 Z eq1 = 1 1 3k2 2.5k2 tan 1 2.5k Z eq1 = 3.91k 39.8o 3k Example 1 (con’t) Perform a Norton transformation. In1 = Vt h1 /Z eq1 3.91k 39.8 = 7.5V / 3.91k 130 39.8 In1 = 7.5V 130o In1 In1 = 1.92mA 90.2o Z n1 = Z eq1 o o o Example 1 (con’t) Since it is easier to combine admittances in parallel than impedances, convert Zn1 to Yn1 and ZL1 to YL1. As Yeq2 is equal to YL1 + Yn1, the admittances should be written in rectangular coordinates, added together, and then the result should be converted to phasor notation. Example 1 (con’t) Yn1 = 1 Z n1 = 0.256m 139.8o Yn1 = 0.256m 1 cos 39.8o j sin 39.8o Yn1 = 0.198 j 0.164m 1 YL1 = 1 Z L1 = 8.33m 1 900 Yn1 = j8.33m 1 Yeq 2 = 0.198 j 0.164 m 1 j8.33m 1 Yeq 2 = 0.198 j8.17 m 1 Yeq2 = 0.1982 8.17 2 m 1 tan 1 8.17 0.198 Yeq2 = 0.817 m 1 88.60 Example 1 (con’t) Next, a Thévenin transformation will allow Yeq2 to be combined with ZL2. Example 1 (con’t) Vt h2 = In1 / Yt h2 Vt h2 Vt h2 1.92mA 90.2o = 8.17m 1 88.6o = 0.235V 1.6o Example 1 (con’t) Z t h2 = 1 / Yt h2 = 12288.6 o Z th2 = 122 cos 88.6 o j sin 88.6 o Z th2 = 2.98 j122 Z L2 = j800 Z eq3 = Z th2 Z L2 Z eq3 = 2.98 j122 j800 Z eq3 = 2.98 j 922 Z eq3 = 2.982 9222 tan 1 922 2.98 Z eq3 = 92289.8o Example 1 (con’t) Perform a Norton transformation after which Zeq3 can be combined with ZR2. Example 1 (con’t) In2 = Vt h2 Z eq3 0.235V 1.60 In2 = 0 92289.8 In2 = 0.255mA 91.40 Example 1 (con’t) Yeq4 = 1 / Z eq3 1 / Z R 2 Yeq4 = 1.08m 1 89.8o 0.2m 10o Yeq 4 = 1.08m 1 cos( 89.8o ) j sin( 89.8o ) 0.2m 1 Yeq 4 = 0.204 j1.08m 1 Yeq4 = 1.10m 1 79.4o Z eq4 = 90679.4o Example 1 (con’t) Use the equation for current division to find the current flowing through ZC2 and Zeq4. IC 2 = YC 2 YC 2 Yeq 4 In2 YC 2 = 1 / Z C 2 = 0.280m90o YC2 = j 0.280m Yeq 4 = 0.204 j1.08m IC 2 0.280m90o = 0.255mA 91.40 j 0.280m 0.204 j1.08m IC 2 0.280m90o = 0.255mA 91.40 0.204 j 0.8m IC 2 IC 2 0.280m90o 0 = 0 . 255 mA 91 . 4 0.826m 75.7 o = 86.0 A74.3o Example 1 (con’t) Then, use Ohm’s Law to find the voltage across ZC2 and then the current through Zeq4. VC 2 = IC 2 Z C 2 = 86.0A74.3o 3.57k 90o VC 2 = 0.309V 15.7 o VC 2 = Veq4 Ieq4 = Veq4 Z eq4 0.309V 15.7 o = 90679.4o Ieq4 = 0.341mA 95.1o Example 1 (con’t) Note that the phase angles of In2, Ieq4, and IC2 are all different because of the imaginary components of Zeq4 and ZC2. The current through ZC2 leads the voltage, which is as expected for a capacitor. The voltage through Zeq4 leads the current. Since the phase angle of Zeq4 is positive, it has an inductive part to its impedance. Thus, it should be expected that the voltage would lead the current. Electronic Response Explain why the circuit on the right is the result of a Norton transformation of the circuit on the left. Also, calculate the natural frequency wo of the RLC network. Summary Circuits containing resistors, inductors, and/or capacitors can simplified by applying the Thévenin and Norton Theorems. Transformations can easily be performed using currents, voltages, impedances, and admittances written in phasor notation. Calculation of equivalent impedances and admittances requires the conversion of phasors into rectangular coordinates. Use of the current and voltage division equations also requires the conversion of phasors into rectangular coordinates.