Thevenin’s Theorem states that “Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor.” Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis.
Thevenins equivalent circuit
As far as the load resistor RL is concerned, any “one-port” network consisting of resistive circuit elements and energy sources can be replaced by one single equivalent resistance Rs and equivalent voltage Vs, where Rs is the source resistance value looking back into the circuit and Vs is the open circuit voltage at the terminals.
Basic steps to determining Thevenin equivalent are
1. Remove the load resistor RL or component concerned.
2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
3. Find VS by the usual circuit analysis methods.
4. Find the current flowing through the load resistor RL.
EXAMPLE1. Thevenin’s Theorem-
Firstly, we have to remove the centre 40Ω resistor and short out all the emf´s connected to the circuit, or open circuit any current sources. The value of resistor Rs is found by calculating the total resistance at the terminals A and B with all the emf´s removed, and the value of the voltage required Vs is the total voltage across terminals A and B with an open circuit and no load resistor Rs connected. Then, we get the following circuit.
Find Equivalent resistance:
10 ohm resistance is in parallel with 20 ohm resistance.
Rt = (R1* R2) / (R1+ R2) = 20 * 10 / 20 + 10 = 6.67 ohm
Find the Equivalent Voltage (Vs)
We now need to reconnect the two voltages back into the circuit, and as VS = VAB the current flowing around the loop is calculated as:
I = 20-10 / 20+ 10 = 0.33 Amps
so the voltage drop across the 20Ω resistor can be calculated as:
Vab = 20 – (20Ω x 0.33amps) = 13.33 volts.
Then the Thevenins Equivalent circuit is shown below with the 40Ω resistor connected.
and from this the current flowing in the circuit is given as:
I = 13.33 / (6.67 + 40) = 0.286