THEVININ'S THEOREM

Statement: A linear network consisting of a number of voltage sources and resistances can be
replaced by an equivalent network having a single voltage source called Thevenin’s voltage (VTh )
and a single resistance called Thevenin’s resistance (RTh ) .

STEPS INVOLVED

Any linear circuit connected to two terminals can be simplified down to just a Thevenin voltage and a Thevenin impedance. This simplified version will have the same effect on any external circuit as the original (more complex) circuit you started with. That's why it is called an "equivalent".

1. Find the Thevenin Voltage, Vth, also known as the open circuit voltage. Open the circuit at the terminals of interest, removing any circuitry that is not part of the network to be simplified. Compute the voltage across the open terminals using your favorite technique (node-voltage, superposition, or whatever).
2. Find the Thevenin Resistance (or Impedance) by deactivating all independent sources. Voltage sources are shorted, current sources are opened. Then look into the terminals and find the total resistance (using series and parallel combinations. 

          LET US SLOVE THE PROBLEMS

QUES1

 

Find the Thévenin equivalent with respect to the 7k ohm resistor.

Remove the 7k ohm, since it is not part of the circuit we wish to simplify. Keep the terminals open since we are finding the Thevenin.

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Find Vth, the voltage across the terminals (in this case it is the voltage over the 3k ohm). Combine the 1k and 2k in parallel.
1k || 2k = (1k*2k) / (1k+2k) = 2M/3k = 2/3k = 667 ohms

Now use a voltage divider to compute Vth across the 3k ohm.
Vth = [3k/(667+3k)] * 5V = 4.1V

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Find the Thevenin Resistance by deactivating all sources and computing the total resistance across the terminals. The voltage sources is shorted, as shown:

Now let's redraw the circuit, bringing the 1k and 2k into a vertical position (but still keeping them connected the same way electrically).

They are all in parallel, so:
Rth = 1k || 2k || 3k = 1 / (1/1k + 1/2k + 1/3k) = 545 ohms
Note, as a check, the equivalent resistance for parallel resistors is always smaller than the smallest resistor in the combination. For example, 545 is smaller than 1k.
The final Thevenin equivalent is then:



QUES2


Find the Thévenin equivalent with respect to the capacitor in the circuit shown. Then replace the capacitor with a resistor chosen for maximum power transfer. What is the value of the resistor? What is the power absorbed by this resistor?

Solution

The Thevenin equivalent has two parts, Vth and Rth. We will do the easier one first -- Rth. To find the Thevenin resistance, deactivate all sources (short voltages and open currents). Since we are finding the Thevenin with respect to the capacitor, we also take the cap out of the circuit and consider the resistance seen from the terminals where the cap was.

The 3K and 1K are in series, but then that combination is shorted out by the wire that replaced the voltage source. Another way to think about it is that we have 1K+3K =4K, and then 4K || 0 = 0. That is, a 0 ohm resistor in parallel with anything else is still 0. Thus we have:

Redrawing it slightly (but maintaining the same connections):

We now see the 4K is in parallel to the 5K, so
Rth = 4K || 5K = 4K*5K/(4K+5K) = 2.22K ohms
Now we must find Vth. For this, we must find the open circuit voltage at the terminals:

Note that there is 2V across the 4K and 5K in series. It does not matter that the 2V is also across the 3K and 1K in series. We will use a voltage divider for the 4K and 5K in series with a know total voltage of 2V:
Vth = 5K/(4K+5K)*2V = 1.11V
So the final Thevenin equivalent is: