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thevenin and norton theorem solved problems -Electricaldiary

 What is Thevenin's Theorem?

Thevenin's theorem is a circuit theory that states any complex circuit or network can be reduced to a simple network that has a voltage source (Called Thevenin's Voltage) connected in series with total circuit equivalent resistance (Called Thevenin's Resistance) along with Load resistance.

By following the given step we can apply Thevenin's theorem 

  • First Remove the Load Resistance from the Circuit.
  • Calculate Open terminal Voltage Across Load terminal and called it Vth.
  • Remove all Current Sources and Open the Terminal of the Current Source.
  • Short the Voltage Source by its Internal Resistance
  • Calculate The Circuit Total resistance across the Load terminal and called it Rth.
  • Now redraw the circuit that consists of calculated Voltage(Vth) in series with Calculated Resistance(Rth).
  • Connect the Load Across in series with the circuit.
Now we will take an Example and Solve it by using Thevenin's Theorem

Example:-
Solve the circuit given below and find the Current Flowing through the given circuit below

thevenin theorem example

Step - Remove the load and short the Voltage source by its internal resistance
after removing load and short the battery by its internal resistance it will look like
from the above circuit it is clear that
Resistor  R and R2 are parallel and these two are in series  R3with the Resistance of the circuit across terminal AB

 Re=  R1|| R2 + R3

 Re= (100/2) + 100 = 50 +100 = 150Ω
Re= 150Ω this is called Thevenin's equivalent Resistance (Rth)


Step - 2 Calculate the Open terminal Voltage across AB
Let V is voltage Across Terminal AB then 
By Voltage division Rule

V_{th} = \frac{VR_{2}}{R_{1} + R_{2}} 

V_{th} = \frac{100\times100}{100 + 100} = \frac{100\times100}{200} = 50

Now Draw Thevenin's Equivalent Circuit like this
thevenin theorem example
Let I is load current flowing through the load then 

I = \frac{V_{th}}{R_{th}+R_{load}}

I = \frac{50}{150+100} = \frac{50}{250} = 0.2A

Current Flowing through Load I = 0.2A
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