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

Solve problems related to Thevenin equivalent circuit, source transformation, and capacitor voltage using nodal analysis and mesh analysis. Find the maximum power transfer, voltage across the capacitor, and more.

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°

I2 = 7.94∠-119°
I = I1 + I2 + I3
I = 21.26∠-77°
Ix = I. Opposite resistance / (sum of resistance)
= 21.26∠-77°* (4-3j) / (12 – 1j)
Ans: Ix = 51.23∠-42°
To find: Voltage across the capacitor
Source 1:
Considering the voltage source V1= 12 cos 3t V = 12 ∠0°, then the another voltage source
should be short circuited and the current source should be opened
ω = 3 rad/sec
R = 6 Ω, L = 2H =jωL => L = 6j
C=1/12 F = 1/ jωL => -4j
Solution:
By imagining with the above condition, it is found that the capacitor and the inductor are parallel
to each other
L || C = ( L× C / L + C)
= (6j × -4j)/ (6j + -4j) = -12j
Zeq1 = 6 -12j
Current I1 = V1/ Zeq1 = 12 ∠0°/ (6 -12j)
I1 = 0.894 ∠70°
Voltage across capacitor V01 = I1 × C
V01 = 3.58
∠-30°

Source 2:
Now the voltage source V2 = 10 V is considered
Solution:
R || C = ( R× C / R + C) = (6 × -4j)/ (6 + -4j) = -1.846 – 2.769 j
Considering the series network with inductance
Zeq2 = -1.846 – 3.23 j
Current I2 = V2/ Zeq2 = 10 / (-1.846 – 3.23 j) = -1.3337 + 2.334j
Voltage across capacitor V02 = I2 × C
V02 = 10.75
∠33°
Source 3:
Now considering the current source I = 4 sin 2t, we should make the voltage source as open
circuit
Imagining this situation we have three elements in the series
V03 = I × (R + L + C) = 4 ∠0° × (6 +4j -6j) = 0.6 + 0.2j
Hence V03 = 0.632
∠20°
Adding the above voltages we could determine
V0 = V01 + V02 + V03
Ans: V0 = 13.694
∠18°
The above circuit looks like an integrator circuit since it has the capacitor forms as a feedback
circuit
Vout = -Vs / (jωRC)

Vout = -3∠0° / 1000 j RC (ω = 1000)
REFERENCES:
1) S. Huseinbegovic, B. Perunicic-Draenovic, N. Hadimejlic, Discrete-time sliding mode
direct power control for grid connected inverter with comparative study, IEEE 23rd
International Symposium on Industrial Electronics (ISIE) (2014).
2) Nam Sung Kim A. D, Ann Arbor S. Hu Mary Jane Irwin, “Leakage Current: Moore‟s
Law Meets Static Power,” Published by the IEEE Computer Society, 2003.