Thevenin's Theorem

                  Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of “linear” is identical to that found in the Superposition Theorem, where all the underlying equations must be linear (no exponents or roots). If we're dealing with passive components (such as resistors, and later, inductors and capacitors), this is true. However, there are some components (especially certain gas-discharge and semiconductor components) which are nonlinear: that is, their opposition to current changes with voltage and/or current. As such, we would call circuits containing these types of components, nonlinear circuits.

                Thevenin's Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load.


 Steps to follow for Thevenin's Theorem:

• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for series circuits.

Example:

             Find V_TH, R_THand the load current flowing through and load voltage across the load resistor in fig (1) by using Thevenin’s Theorem.

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1.Open the 5kΩ load resistor

2.Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V_TH). Fig (3)

We have already removed the load resistor from figure 1, so the circuit became an open circuit as shown in fig 2. Now we have to calculate the Thevenin’s Voltage. Since 3mA Current flows in both 12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor as it is open.

So 12V (3mA  x 4kΩ) will appear across the 4kΩ resistor. We also know that current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in parallel with 4k resistor. So the same voltage (i.e. 12V) will appear across the 8kΩ resistor as 4kΩ resistor. Therefore 12V will appear across the AB terminals. So,

V_TH = 12V 

3. Open Current Sources and Short Voltage Sources. Fig (4)

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4.Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (R_TH)

We have Reduced the 48V DC source to zero is equivalent to replace it with a short in step (3), as shown in figure ()  We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and  12k Ω resistor. i.e.:

8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)

R_TH = 8kΩ +  [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]

R_TH = 8kΩ + 3kΩ

R_TH = 11kΩ

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5. Connect the R_THin series with Voltage Source V_TH and re-connect the load resistor. This is shown in fig (6) i.e. Thevenin circuit with load resistor.

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6. Now apply the last step i.e. calculate the total load current & load voltage as shown in fig 6.

I_L = V_TH/ (R_TH + R_L)

= 12V / (11kΩ + 5kΩ) → = 12/16kΩ

I_L= 0.75mA

And

V_L = I_Lx R_L

V_L = 0.75mA x 5kΩ

V_L= 3.75V

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