Question

j312 -j52 HE 2290° V 412 ZL Find the Thevenin equivalent of the circuit with respect to Z. Express the answer in terms of the Thevenin voltage (Vr ), short-circuit current (isc), and Thevenin impedance (Z). Round answer to 2 decimal places, add trailing O's as needed. TI .V | А Zr-

Answer 1

To find the Thevenin's equivalent of circuit with respect to ZL, we first find the Thevenin voltage Vt.

For this we replace ZL by an open circuit and the modified circuit will be:

-j5 ohm j3 ohm HH + +2 2V 4 ohm Vt 90° -

Being an open circuit, no current flows through inductor(j3). So, voltage drop across it is zero.

So, Thevenin voltage will be the voltage across 4 ohm resistor. To find this, we use voltage divider rule as:

Vt = voltage across 4 ohm resistor =   $2\angle 90*\left [ \frac{4}{4-j5} \right ]$

= 1.25$\angle$141.34 V

Next step is to find Thevenin impedance Zt. For this, we replace ZL by an open circuit and voltage source 2$\angle$90 by a short circuit. Modified circuit will be:

As per circuit, Zt = [(-j5)||4] + j3

= [-j20/(4 - j5)] + j3

= (-j20 + j12 +15)/(4 - j5)

= (15 - j8)/(4 - j5)

= 2.65$\angle$23.26 ohm

and short circuit current Isc = Vt/Zt

= (1.25$\angle$141.34)/(2.65$\angle$23.26)

= 0.47$\angle$118.08 A

So, Vt = 1.25$\angle$141.34 V

Zt = 2.65$\angle$23.26 ohm

Isc = 0.47$\angle$118.08 A