THEVENIN’S THEOREM
Thevenin’s Theorem states that:
“Any linear circuit containing several voltages and resistances can be replaced by just
one single voltage in series with a single resistance connected across the load”.
“Any linear network having a number of voltage sources and resistances can be
replaced by a simple equivalent circuit consisting of a single voltage source (VTH) in
series with a resistance (RTH), where VTH is the open-circuit voltage at the terminals of
the load and RTH is the equivalent resistance measured across the terminals while
independent sources are turned off”
It is possible to simplify any electrical circuit, no matter how complex, to an equivalent
two-terminal circuit with just a single constant voltage source in series with a resistance
(or impedance) connected to a load.
Thevenin’s Theorem is especially useful in the circuit analysis of power or battery
systems and other interconnected resistive circuits where it will have an effect on the
adjoining part of the circuit.
Thevenin's Theorem enables analyses of power circuits having a load that changes
value during the analysis process.
Fig. Thevenin’s equivalent circuit
As far as the load resistor RL is concerned, any complex “one-port” network consisting
of multiple resistive circuit elements and energy sources can be replaced by one single
equivalent resistance Rs and one single equivalent voltage Vs. Rs is the source
resistance value looking back into the circuit and Vs is the open circuit voltage at the
terminals.
To analyse the circuit above, we have to remove the centre 40Ω load resistor connected
across the terminals A-B, and remove any internal resistance associated with the
voltage source(s). This is done by shorting out all the voltage sources connected to the
circuit, that is v = 0, or open circuit any connected current sources making i = 0. The
reason for this is that we want to have an ideal voltage source or an ideal current source
for the circuit analysis.
The value of the equivalent resistance, Rs is found by calculating the total resistance
looking back from the terminals A and B with all the voltage sources shorted. We then
get the following circuit.
i) Find the Equivalent Resistance (Rs)
10 Ω resistor in parallel with the 20 Ω resistor.
𝑅1 × 𝑅2 20 × 10
𝑅𝑇 = ( )=( ) = 6.67Ω
𝑅1 + 𝑅2 20 + 10
The voltage Vs is defined as the total voltage across the terminals A and B when there
is an open circuit between them. That is without the load resistor RL connected.
ii) Find the Equivalent Voltage (Vs)
Reconnect the two voltages back into the circuit, and as VS = VAB the current flowing
around the loop is calculated as:
𝑉 20𝑉 − 10𝑉
𝐼= =( ) = 0.33𝑎𝑚𝑝𝑠
𝑅 20𝛺 + 10𝛺
This current of 0.33 amperes (330mA) is common to both resistors so the voltage drop
across the 20Ω resistor or the 10Ω resistor can be calculated as:
VAB = 20 – (20Ω x 0.33 amps) = 13.33 volts.
Or
VAB = 10 + (10Ω x 0.33 amps) = 13.33 volts
Then the Thevenin’s Equivalent circuit would consist or a series resistance of 6.67Ω and
a voltage source of 13.33V. With the 40Ω resistor connected back into the circuit we
get:
….From this, the current flowing around the circuit is given as:
𝑉 13.33𝑉
𝐼= =( ) = 0.286 𝑎𝑚𝑝𝑠
𝑅 6.67𝛺 + 40𝛺
Recap….Using Kirchoff’s circuit law, we get the same answer as shown below:
Find the current flowing in the 40Ω Resistor, R3
The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.
Using Kirchhoffs Current Law, the equations are given as:
At node A: I1 + I2 = I3
At node B: I3 = I1 + I2
Using Kirchhoffs Voltage Law, KVL the equations are given as:
Loop 1 is given as: 10 = R1 I1 + R3 I3 = 10I1 + 40I3
Loop 2 is given as: 20 = R2 I2 + R3 I3 = 20I2 + 40I3
Loop 3 is given as: 10 – 20 = 10I1 – 20I2
As I3 is the sum of I1 + I2, the equation can be rewritten as;
Eq. No 1: 10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2
Eq. No 2: 20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2
We now have two Simultaneous Equations that can be reduced to give the values
of I1 and I2.
Substitution of I1 in terms of I2 gives us the value of I1 as: - 0.143 Amps
Substitution of I2 in terms of I1 gives us the value of I2 as: + 0.429 Amps
As: I3 = I1 + I2
The current flowing in resistor R3 is given as: - 0.143 + 0.429 = 0.286 Amps
The voltage across the resistor R3 is given as: 0.286 x 40 = 11.44 volts
The negative sign for I1 means that the direction of current flow initially chosen was
wrong, but never the less still valid. In fact, the 20v battery is charging the 10v battery.
Thevenin’s theorem can be used as another type of circuit analysis method and is
particularly useful in the analysis of complicated circuits consisting of one or more
voltage or current source and resistors that are arranged in the usual parallel and series
connections.
Thevenin’s Theorem Summary
Thevenin’s theorem is another type of circuit analysis tool that can be used to reduce
any complicated electrical network into a simple circuit consisting of a single voltage
source, Vs in series with a single resistor, Rs.
When looking back from terminals A and B, this single circuit behaves in exactly the
same way electrically as the complex circuit it replaces. That is the I - v relationships at
terminals A-B are identical.
The basic procedure for solving a circuit using Thevenin’s Theorem is as follows:
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.
Example 2:
Find the Thevenin’s equivalent circuit for the circuit below;
Solution
When calculating the Thevenin’s equivalent resistance, all voltage sources must be
turned off, meaning it acts like a short circuit and all current sources act like an open
circuit, as shown in the figure below:
Let us calculate the Thevenin’s resistance for the above circuit:
Thevenin’s resistance, Rth = 4 || 12 +1 = 4 x 12 / 16 + 1 = 4 ohms
Finding Thevenin’s voltage:
4i1 + 12(i1 – i2) = 32V, i2 = -2A
Solving the above equations, we get i1 = 0.5A
Therefore Vth = 12(i1 – i2) = 12(0.5 + 2) = 30V
The equivalent Thevenin’s circuit is shown in the figure below:
Question to do….
Calculate the Thevenin's voltage and Thevenin's resistance of the circuit below
