Thevenin Theorem This theorem is very conceptual. If we think deeply about an electrical circuit, we can visualize the statements made in Thevenin theorem. Suppose we have to calculate the current through any particular branch in a circuit. This branch is connected with rest of the circuits at its two terminal. Due to active sources in the circuit, there is one electric potential difference between the points where the said branch is connected. The current through the said branch is caused by this electric potential difference that appears across the terminals. So rest of the circuit can be considered as a single voltage source, that's voltage is nothing but the open circuit voltage between the terminals where the said branch is connected and the internal resistance of the source is nothing but the equivalent resistance of the circuit looking back into the terminals where, the branch is connected.

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.

Step 1.

Open the 5kΩ load resistor (Fig 2).

Step 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 

Step 3.

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

Step 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 (3)  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

Step 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. This the Thevenin’s equivalent circuit 

Step 6.

Now apply the last step i.e Ohm’s law . 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

With Matildo