Thevenin’s theorem and its application are used to solve dc circuits and the advantage of Thevenin’s theorem over conventional circuit reduction techniques in situations where load changes. A network could be converted into a very simple equivalent circuit that represents either in the form of a practical voltage source known as Thevenin’s voltage source.
The reduction of computational complexity that involves solving the current through a branch for different values of
load resistance. In many applications, a network may contain a variable component or element while other elements in the circuit are kept constant. If the solution for current or voltage or power in any component of the network is desired, in such cases the whole circuit needs to be analyzed each time with the change in component value.
Any linear active circuit containing independent voltages and resistances can be replaced by just one single voltage in series with a single resistance connected across the load
Thevenin’s Theorem: Thevenin’s theorem states that any two output terminals of an active linear network containing independent sources (that includes voltage and current sources) can be replaced by a simple voltage source of magnitude VTH in series with a single resistor RTH . Where RTH is the equivalent resistance of the network when looking from the output terminals with all sources (voltage and current) removed and replaced by their internal resistances and the magnitude of is equal to the open-circuit voltage across the VTH terminal.
Steps for applying Thevenin’s Theorem
To find a current L I through the load resistance RL in fig using Thevenin’s theorem, the following steps are followed:
Disconnect the load resistance RL from the circuit, as indicated in fig
Calculate the open-circuit voltage VTH as shown in fig. at the load terminals between A& B after disconnecting the load resistance RL. In general, we can apply any of the techniques i.e. mesh-current, node-voltage and superposition method that learned in earlier
Redraw the circuit as shown in fig with each practical source replaced by its internal resistance as shown in fig Note that voltage sources should be short-circuited (just remove them and replace them with plain wire) and current sources should be open-circuited (just removed).
Look backward into the resulting circuit from the load terminals between A& B, as suggested by the eye in fig Calculate the resistance that would exist between the load terminals in order to calculate the resistance across the load terminals. The resistance RTH is described in the statement of Thevenin’s theorem.
calculating this resistance may be a difficult task but one can try to use the standard circuit reduction technique or
Δ-Y and Y-Δ Conversions transformation techniques.
Place RTH in series with VTH to form the Thevenin’s equivalent circuit as shown in fig
Reconnect the original load to the Thevenin voltage circuit as shown in fig the load’s voltage, current and power may be calculated by a simple arithmetic operation only
Voltage across Load
Power absorbe by Load
Application of Thevenin’s theorem
Example: For the circuit shown in fig.8.3(a), find the current through resistor R2 = RL=1 ohm ( Ib-b branch) using Thevenin’s theorem & hence calculate the voltage across the current source( Vcg)
Step-1: Disconnect the load resistance RL and redraw the circuit as shown in fig
Step-2: Apply any method (say node-voltage method) to calculate VTH
At node C:
Step-3: Redraw the circuit fig indicating the direction of currents in different branches. One can find the Thevenin’s voltage using KVL around the closed path ‘gabg’
VTh = Vag – Vbg = 3-2 =1 Volt
Step-4: Thevenin’s resistance of the fixed part of the circuit can be computed by looking at the load terminals ‘a’- ‘b’
RTH = (R1+R 3) ||R4 = (3+4 )||2 = 1.55 ohm
Step-5: Place RTH in series with VTH to form the Thevenin’s equivalent circuit
Step-6: The circuit shown in fig is redrawn to indicate different branch currents. Referring to fig, one can calculate the voltage Vbg and voltage across the current source Vcg using the following equations.
Vbg = Vag– Vab =3 – 1 *0.39 = 2.61 volt
Ibg = 2.61/2 = 1.305 Amp
Icb = 1.305-0.39 = 0.915 Amp
Vcg = 4*0.915+2*1.305 = 6.27 Volt
- One great advantage of Thevenin’s theorem over the normal circuit reduction technique or any other technique is that once the Thevenin equivalent circuit has been formed, it can be reused in calculating load current, load voltage and load power for different loads using the equations
- With the help of Thevenin’s theorem, one can find the choice of load resistance that results in the maximum power transfer to the load
- On the other hand, the effort necessary to solve this problem-using node or mesh analysis methods can be quite complex and tedious from the computational point of view.