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 (V_{TH})

**#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 R_{t }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 V_{s} = 100∟53.13^{o} volts then the thevenin's equivalent voltage in volts as seen by the load resistance R_{L} is

Answer: C

Explanation :

We need to find out the Thevenin’s voltage across the two terminals as seen by the R_{L}. 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 V_{TH} 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 V_{TH} will be the answer.

V_{TH }– Open circuit voltage across the interested terminals

Rt denotes the Equivalent Resistance across terminals A and B

**Step 1**

- Firstly, Remove R
_{L}. This is the only way to calculate V_{TH}. Now, no current will flow in the circuit as terminals A and B are open, therefore, I_{2}= 0

Voltage Drop across Resistance j6 Ω = I_{2} * X_{L}

Voltage Drop across Resistance 5 Ὠ = I_{2} * R_{L}

Voltage Drop across the resistance j6 Ω and 5Ὠ will also be 0 since I_{2} = 0

Therefore,

V_{TH} = 10 V_{L1} + V_{j6} +V_{5}_{Ω}

Since V_{j6} and V_{5}_{Ω} are 0,

Therefore, V_{TH} = 10* V_{L1} ..................................(1)

**Step 2 **

- V
_{L1 }is the voltage drop across j4 Ω

I1 is the current flowing across the circuit

j4 Ω = X_{L} is the Inductive Reactance

V_{L1} = I_{1} * j4 where

I_{1} = (V_{s} – j40*I_{2})/ (3+ j4)

I1 = Voltage Difference/ Denter Resistance

Since A and B are open terminals, Therefore, I_{2} = 0

Therefore, I_{1} = V_{s}/(3+ j4)

Therefore, V_{L1} = I_{1} *j4

V_{L1} = (j4 * Vs) / (3+ j4) ...........................................(2)

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**Step 3**

Solving equations (1) and (2)

V_{TH} = 10 * V_{L1}

V_{TH} = 10 * (j4 / 3+ j4) * V_{s}

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 **

V_{TH} = V_{s} * j40/ 5[tan^{-1}(4/3)]

**Step 5 **

V_{TH} = 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,

V_{TH} = j40 * 100 / 5 = j40 * 20

Therefore, V_{TH }= 40 * 20 [90º + 53.13º +(- 53.13º)]

Therefore, V_{TH}= 800 ∟90º

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

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