May 16, 2022
/ 8 /# Thevenin's Theorem Explained With Solved Examples

### But why do we need this theorem

### Steps to solve circuit problem using thevenin theorem

### Example problem on thevenin’s theorem

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#### Step 2:

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## More For You ☄

**Thevenin theorem** is a theorem used to model complex circuits into a simple circuit with an equivalent voltage source and resistance.

As the theorem states that;

‘any two terminals of a network can be replaced by an equivalent voltage source and an equivalent series resistance.

The voltage source is the voltage across the two terminals with load, if any, removed.

The series resistance is the resistance of the network measured between two terminals with load removed and constant voltage source being replaced by its internal resistance (or if it is not given with zero resistance, i.e., short circuit) and constant current source replaced by infinite resistance, i.e., open circuit.’

Because it is not like mesh analysis or nodal analysis which we use to analyze circuit but rather it is a technique that makes engineers unique (to my opinion). Here’s is what I mean, if two engineers are tasked to design a circuit each with an output of (5V and 1A).

The first engineer let’s say *MR X* decided to design his own circuit with 2 resistors and a 20V battery as shown below while *MR Y* decided to design his own circuit with just one resistor and a 9V battery as shown below also.

As you can see, both circuit gives the required output but what both circuits contain are different and this is what makes the thevenin theorem helpful as it lets us to approximately guess what another engineer has designed even if one does not know what is inside the box containing the circuit just as our home charger.

There are different variations but at the end of the day, the output of those charger are quite same.

If there is a load resistor R

_{L}then you need to remove it.Find the thevenin voltage V

_{th}across the terminal of interest. The V_{th}can be found by finding the open circuit voltage across the terminalFind the short circuit current I

_{sc}between terminal a → bWith the short circuit current known, the thevenin resistance can be found by dividing V

_{th}by I_{sc}To solve for V

_{th}, you can use whatever technique you are most comfortable with or that suit the problem at hand maybe mesh analysis, nodal analysis or source transform.After finding the V

_{th}and R_{th}then you need to sketch your circuit as shown below.Also if there is R

_{L}then, you need to replace the open terminal a → b with R_{L}as shown below and lastly find the current that flow through R_{L}as the current that flows through it is not I_{sc}.

Lets solve some examples to understand the concept better.

**Ex.1** – find the current through the 2Ω resistor using thevenin theorem?

Remove the 2Ω resistor and leave the terminal open.

Using mesh analysis, lets analyse the circuit. Since there is only one loop as the terminal a → b has made the second loop to no longer be a loop.

The equation gotten is :

40-5i-20-10i = 0

20-15i = 0

15i=20

I=20/15 = 1.33A.

Now to find the open terminal voltage V_{th}, we need to add the voltage across the 10Ω resistor and the 10V battery.

Thus, the voltage across the 10Ω is V_{10Ω} = 10 * 1.33 = 13.3v

Hence, V_{th} = 13.3 + 10 = 23.3v

To find I_{sc} we need to short circuit the a → b terminal.

Now since short circuiting of the circuit has cause the loop to be two as oppose to one that we previously did.

So equation for **loop I**:

40- 51_{1}-20-10(1_{1}-1_{2}) =0;

40-51_{1}-20-101_{1}+101_{2} =0; *by expanding the bracket*

20-151_{1}+101_{2} = 0;

151_{1}-101_{2}=20 ………(1)

**Loop II**

-10(1_{2}-1_{1})+10 =0;

-101_{2}+101_{1}+10 =0;

101_{1}-101_{2}=-10……………..(2)

By solving the simultaneous equation, we get

1_{1} = 6A;

1_{2} = 7A;

But I_{sc} = 1_{2} = 7A

hence, R_{th} = V_{th}/I_{sc} = 23.3/7 = 3.33Ω

now replacing the 2Ω resistor we removed earlier then the circuit becomes this;

but we need to calculate for the current that flows through the load resistor (2Ω),

I_{L} = V_{th} / (R_{th} + R_{L}) = 23.3/(3.33 +2) =23.3/(5.33) = 4.37A