Thevenin Theorem's is an important topic from gate point of view. Experts at GATE- prepladder shares 6 simple steps to solve questions on Thevenin's Theorem
The article highlights six steps through which questions on Thevenin Theorem can be solved.
Electric Circuit can be analysed through Thevenin’s Theorem through the following steps:
6 simple steps to solve questions on Thevenin Theorem
First of all, open the load resistor
Secondly, calculate/measure the open circuit voltage. Denote this by Thevenin Voltage (VTH)
Next, you must Open Current Sources and Short Voltage Sources
Next, you must calculate the Open Circuit Resistance. Denote this by Thevenin Resistance (Rt)
Next, you must redraw the circuit with Voc as the Voltage source and Rt as the resistance. Then, you must connect the load resistor which you had removed in Step 1. The resulting circuit would be the Equivalent Thevenin Circuit.
Finally, by applying the Ohm’s law, you can find the total current flowing through the load resistor by using the formula,
i = Voc/ (Rt + RL)
Applying this steps at GATE question
Question : In the circuit shown below, if the source voltage Vs = 100∟53.13o volts then the thevenin's equivalent voltage in volts as seen by the load resistance RL is
We need to find out the Thevenin’s voltage across the two terminals as seen by the RL. Name the two terminals as A and B.
Since the Thevenin’s theorem states that any complex network can be equivalently represented by applying Thevenin’s voltage in series with Thevenin’s Resistance.
Let VTH denote Thevenin’s voltage which is the open circuit voltage across A and B and Rt denotes Thevenin’s Resistance.
If we use A and B terminals, the entire complex network will be represented by A and B terminals, then the value of VTH will be the answer.
VTH – Open circuit voltage across the interested terminals
Rt denotes the Equivalent Resistance across terminals A and B
- Firstly, Remove RL . This is the only way to calculate VTH. Now, no current will flow in the circuit as terminals A and B are open, therefore, I2 = 0
Voltage Drop across Resistance j6 Ω = I2 * XL
Voltage Drop across Resistance 5 Ὠ = I2 * RL
Voltage Drop across the resistance j6 Ω and 5Ὠ will also be 0 since I2 = 0
VTH = 10 VL1 + Vj6 +V5Ω
Since Vj6 and V5Ω are 0,
Therefore, VTH = 10* VL1 ..................................(1)
- VL1 is the voltage drop across j4 Ω
I1 is the current flowing across the circuit
j4 Ω = XL is the Inductive Reactance
VL1 = I1 * j4 where
I1 = (Vs – j40*I2)/ (3+ j4)
I1 = Voltage Difference/ Denter Resistance
Since A and B are open terminals, Therefore, I2 = 0
Therefore, I1 = Vs/(3+ j4)
Therefore, VL1 = I1 *j4
VL1 = (j4 * Vs) / (3+ j4) ...........................................(2)
Solving equations (1) and (2)
VTH = 10 * VL1
VTH = 10 * (j4 / 3+ j4) * Vs
Now putting (3 + j4) using polar form,
Magnitude of (3+ J4) = 5
Angle = tan -1 (Imaginary part/ Real Part) = tan-1 (4/3)
VTH = Vs * j40/ 5[tan-1(4/3)]
VTH = 100(∟13º) * j40 / 5(∟53.13º)
j4 can be written as j4 + 0
Using the polar form of (j4 + 0),
Magnitude of (0 + j4) = 4
Angle = tan-1(4/0) = tan-1 (∞) = 90º
So, j40 = 40 ∟90º
VTH = j40 * 100 / 5 = j40 * 20
Therefore, VTH = 40 * 20 [90º + 53.13º +(- 53.13º)]
Therefore, VTH= 800 ∟90º
Best Wishes !!