Thevenin’s Theorem Simplified

Thevenin’s Theorem Simplified

Thevenin’s Theorem

Thevenin’s Theorem is simply a method used in simplifying a complex circuits or networks,easier and faster. In 1893, a French Engineer called M.L Thevenin , made this theorem.

Thevenin’s Theorem may be stated as “It is possible to simplify any complex circuit/linear circuit with Current and Voltage sources with an equivalent circuit containing a single Voltage Source connected in series to a load.


Simple Steps To Solve Electric Circuit By Thevenin’s Theorem

1. Open the load resistor.

2. Calculate the Open Circuit Voltage. This is the Thevenin Voltage (V TH).

3. Short Voltage Sources and calculate the Open Circuit Resistance. This is the Thevenin Resistance (R TH ).

4. Now, Redraw the circuit with the calculated open circuit Voltage (V TH ) in Step (2) as voltage Source and the calculated open circuit resistance (R TH ) in step (3) connected in series to the load resistor that was removed in step (1). This is the Equivalent Thevenin Circuit of that Linear Electric Network or Complex circuit which had to be simplified and solved by Thevenin’s Theorem you have done.

5. Finally, find the Total current flowing through Load resistor by using the Ohm’s Law I_{{L}}=\frac{V_{TH}}{R_{TH}+R_{L}}


Find the load current and load voltage across the load resistor in the above circuit  by using Thevenin’s Theorem.



Step 1

Open the load circuit



Step 2

Calculate the open voltage which will be the thevenin theorem.

For the 10Ω resistor,since it is an open-circuit,thus no current flows through it. The voltage across it zero

Since the 20Ω and 50Ω resistors are connected in series,the same current flows through each of the resistors.

Using Voltage Divider Rule

V_{{50}}=\frac{V_{T}R_{50}}{R_{{20}}+R_{{50}}    V_{{50}}=\frac{6*50}{20+50}=3.6V



Step 3

Short Voltage Sources and calculate the Open Circuit Resistance. This is the Thevenin Resistance (R TH ).

Thevenin's theorem


It is very clear that the 10Ω resistor is in series with a parallel connection of 20Ω and 50Ω resistor. I.e 10Ω+ (20Ω ll 50Ω)…(ll means parallel with)

R_{{TH}}=10+\frac{100}{7} ; R_{{TH}}=24.3\Omega

Step 4

Connect the VTH and RTH in series with the removed load resistor.

Thevenin Theorem

Step 5

Using  I_{{L}}=\frac{V_{TH}}{R_{TH}+R_{L}} to find the current in the load resistor.


I_{{L}}=\frac{3.6}{24.3+5} ; I_{{L}}=0.1mA

Load voltage; from Ohm’s law 

V_{{L}}=I_{{L}}R_{{L}}  ;  V_{{L}}=0.1\times5=0.5V

Loading Facebook Comments ...

Leave a Reply

Your email address will not be published. Required fields are marked *