Thievens Theorem

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

#step1

        First of all, open the load resistor

#step2

       Secondly, calculate/measure the open circuit voltage. Denote this by Thevenin Voltage (VTH)

#step3

        Next, you must Open Current Sources and Short Voltage Sources

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#step4

        Next, you must calculate the Open Circuit Resistance. Denote this by Thevenin Resistance (Rt)

#step5

       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.

#step6

        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)

 

 must read : How to prepare Root Locus for GATE ECE 2017 || Solving Root Locus Simplified

 

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

 

Answer:  C

Explanation :

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

 

Step 1

  1. 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

Therefore,

 VTH = 10 VL1 + Vj6 +V5

 

Since Vj6 and V5 are 0,

 

Therefore, VTH = 10* VL1   ..................................(1)

Step 2

  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)

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Step 3

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)

Step 4 

VTH = Vs * j40/ 5[tan-1(4/3)]

Step 5

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

 

Step 6

Therefore,

VTH = j40 * 100 / 5 = j40 * 20

 

Therefore, VTH = 40 * 20 [90º + 53.13º +(- 53.13º)]

Therefore, VTH= 800 ∟90º

 

must read : Expected cut-off of GATE 2017(ECE) with topic wise analysis of previous papers

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