Solved Problems: Thevenin Equivalent Circuit, Source Transformation, and Capacitor Voltage

Solved problems:
Step 1:
To determine the thevenin equivalent circuit it is necessary to consider the Vth and Rth
In order to find Rth we have to short circuit the voltage source and open the current source
By doing so,
Rth = [2 + 4j || 6] + -2j
= [(2 + 4j × 6) / (2 + 4j + 6)] -2j
Rth =2.4 – 0.2j Ω = 2.41∠-5°
Next step is to find the thevenin voltage Vth we can use the method of either mesh analysis or
nodal analysis,
Here we have considered the nodal analysis,
[(60 ∠0° -Vth) / (2 +4j)] + 5 ∠90° = Vth / 6
[6 (60 ∠0° -Vth) / (2 +4j)] + 5 ∠90° = Vth
1.34 ∠-70° (60 ∠0° -Vth) + 30 ∠90° = Vth
2∠-40° Vth = 58.83∠-50°
Vth = 29.41∠-10° V
The load is considered to be the resistor and the capacitor
RL = 4 – 3j = 5∠-40°
Ans: RL = 5∠-40°
Considering the Vth and Rth

We get Ix = Vth / Rth + RL
Hence, Ix = 29.41∠-10° / 7.17∠-28°
Ans: Ix = 4.10∠18°
c) To determine the maximum power the thevenin’s impedance should be considered as the load
impedence
i.e., Rth = RL (Rth*)
For this condition,
Ix = Vth / Rth + Rth*
Ix = 6.10∠-5°
Maximum power transfer
P = Ix2 RL
Ans: Pmax = 89.68 W
2) Source transformation:
By the method of source transformation the voltage source becomes current source and vice-
versa
Therefore, I = 60 ∠0° / (2 + 4j)[ we have considered the series resistance of 2Ω and 4jΩ]
I = 13.41 ∠-70°
And V = 5 ∠90° × (4-3j)
V = 25∠49°
With the nodal analysis/ mesh analysis
We determine I1 = 13.41 ∠-70° (mesh 1)
2+ 4j (I1 – I2) + 6 I2 + 2j (I2 - I3) = 0 (mesh 2)
(4-2j) I2 - 2j I3 = -4.472 ∠70°
-2j (I2 - I3) + (4.3j) I3 – V =0 (mesh 3)
-2j I2 + 6.3j I3 = 25∠49°
From the above equations we could determine
I3 = 3.54∠-8°