**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 V_{TH} in series with a single resistor R_{TH} . Where R_{TH} 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 V_{TH} terminal.

## Steps for applying Thevenin’s Theorem

To find a current L I through the load resistance R_{L} in fig using Thevenin’s theorem, the following steps are followed:

**Step-1**:

Disconnect the load resistance R_{L} from the circuit, as indicated in fig

**Step-2**:

Calculate the open-circuit voltage V_{TH} as shown in fig. at the load terminals between A& B after disconnecting the load resistance R_{L}. In general, we can apply any of the techniques i.e. mesh-current, node-voltage and superposition method that learned in earlier

**Step-3:**

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).

**Step-4:**

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 R_{TH} 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.

**Step-5:**

Place R_{TH} in series with V_{TH} to form the **Thevenin’s equivalent circuit** as shown in fig

**Step-6:**

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

Load Current

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 R_{2} = R_{L}=1 ohm ( I_{b-b} branch) using Thevenin’s theorem & hence calculate the voltage across the current source( V_{cg})

**Solution:****Step-1**: Disconnect the load resistance R_{L} and redraw the circuit as shown in fig

**Step-2**: Apply any method (say node-voltage method) to calculate V_{TH}

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’

V_{Th} = V_{ag }– V_{bg} = 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’

R_{TH }= (R_{1}+R _{3}) ||R_{4} = (3+4 )||2 = 1.55 ohm

**Step-5:** Place R_{TH} in series with V_{TH} 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 V_{bg }and voltage across the current source V_{cg} using the following equations.

V_{bg }= V_{ag}– V_{ab} =3 – 1 *0.39 = 2.61 volt_{ }

I_{bg }= 2.61/2 = 1.305 Amp

I_{cb} = 1.305-0.39 = 0.915 Amp

V_{cg} = 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.