**Thevenin’s Theorem Definition**

Suppose we are given an arbitrary circuit containing any or all of the following elements: resistors, voltage sources, current sources (the source can be dependent as well as independent). Let us identify a pair of nodes, say node a and b, such that the circuit can be partitioned into two parts as shown in figure 1.

Figure.1: Circuit partitioned into two parts

Furthermore, suppose that circuit A contains no dependent source that is dependent on a variable in circuit B and vice versa. Then, we can model circuit A by an appropriate independent voltage source, call it V_{oc} that is connected in series with an appropriate resistance, call it R_{TH}. This series combination of a voltage source and a resistance is called the Thevenin’s equivalent of circuit A. in other words, circuit A in figure 1 and the circuit in the shaded box in figure 2 have the same effect on circuit B. This result is known as **Thevenin’s theorem** and is one of the most useful and significant concepts in circuit theory.

Figure.2: Thevenin’s Equivalent Circuit

**Circuit B** (which is often called a load) may consist of many circuit elements, a single element (a load resistor), or no element.

- You May Also Read: Norton’s Theorem with Solved Examples

**Thevenin’s Equivalent Open Circuit Voltage**

To obtain the voltage V_{oc}– called the open circuit voltage- remove circuit B from circuit A, and determine the voltage between node a and b ( where the + is at node a). This voltage, as shown in figure 3(a), is V_{oc}.

Figure.3 (a): determination of Open Circuit Voltage

**Thevenin’s Equivalent Resistance **

There are several ways to find the Thevenin’s equivalent resistant, which are given below:

**1.** To obtain the resistance R_{TH}– called the Thevenin’s equivalent resistance of circuit A:

**i)** Remove circuit B from circuit A.

**ii)** Set all independent sources in circuit A to zero. (A zero voltage source is equivalent to a short circuit, and zero current source is equivalent to an open circuit).

**iii)** Determine the resistance between nodes a and b- this is R_{TH}– as shown in figure 3(b).

Figure.3 (b): Determination of Thevenin’s Equivalent Resistance R_{TH}

**2.** Short-circuit the terminals a and b then find the short-circuit current I_{sc}. The Thevenin equivalent resistance is given by

\[{{\text{R}}_{\text{th}}}\text{= }{{\text{V}}_{\text{oc}}}\text{/}{{\text{I}}_{\text{sc}}}\text{= }{{\text{V}}_{\text{th}}}\text{/}{{\text{I}}_{\text{sc}}}\]

**3.** When the source network has a ladder structure and contains no controlled (dependent) sources, R_{TH} is easily found by series-parallel reduction of the dead network.

When the source network contains** controlled sources**, the Thevenin’s resistance can be found using the method represented by figure 4.

Figure.4: determination of Thevenin’s Resistance

Here, the dead source network has been connected to an external test source. This test source may be any independent voltage or current source that establishes v_{o} at the terminals. Since the dead network contains only resistors and controlled sources, and since R_{TH} equals the equivalent resistance of the dead network, the equivalent resistance theorem tells us that

\[{{R}_{TH}}={{v}_{o}}/{{i}_{o}}\]

**Thevenin’s Theorem Solved Example with Voltage Source**

Find Thevenin equivalent circuit of the following network.

By applying nodal voltage method,

$\frac{{{V}_{1}}-25}{5}+\frac{{{V}_{1}}}{20}-3=0$

${{V}_{1}}=32~V$

${{V}_{th}}=32V$

** **To find short circuit current, we short-circuited terminals a and b as shown in the following figure:

Again using nodal voltage method,

$\frac{{{V}_{2}}-25}{5}+\frac{{{V}_{2}}}{20}-3+\frac{{{V}_{2}}}{4}=0$

${{V}_{2}}=16~V$

${{I}_{sc}}=\frac{{{V}_{2}}}{4}=\frac{16}{4}=4A$

So,

${{R}_{th}}=\frac{{{V}_{th}}}{{{I}_{sc}}}=\frac{32}{4}$

${{R}_{th}}=8~\Omega $

**Now, **we have following equivalent circuit:

**Thevenin’s Theorem Solved Problem with Dependent Sources**

Consider a circuit shown in the following figure. We will determine V_{th }and R_{th} across terminals a and b of the circuit.

Applying KCL at node 1:

\[\begin{matrix} \frac{{{V}_{1}}-5}{2000}+\frac{{{V}_{1}}}{6000}+\frac{{{V}_{1}}-{{V}_{2}}}{1000}=0 & \cdots & (1) \\\end{matrix}\]

Multiply (1) by 6000 yields

\[3{{V}_{1}}\text{- }15\text{ }+\text{ }{{V}_{1}}+\text{ }6{{V}_{1}}\text{ -}6{{V}_{2}}=\text{ }0\]

\[10{{V}_{1}}=\text{ }6{{V}_{2}}+\text{ }15\]

\[{{V}_{1}}=\text{ }0.6{{V}_{2}}+\text{ }1.5\]

Applying KCL at node 2:

\[\begin{matrix} \frac{{{V}_{2}}-{{V}_{1}}}{1000}+0.0005{{V}_{1}}+\frac{{{V}_{2}}}{10000}=0 & \cdots & (2) \\\end{matrix}\]

Multiply (2) by 10000 yields

\[10{{V}_{2}}\text{ -}10{{V}_{1}}+\text{ }5{{V}_{1}}+\text{ }{{V}_{2}}=\text{ }0\]

\[11{{V}_{2}}\text{ -}5{{V}_{1}}=\text{ }0\]

\[11{{V}_{2}}\text{- }5(0.6{{V}_{2}}+\text{ }1.5)\text{ }=\text{ }0\]

\[8{{V}_{2}}=\text{ }7.5\]

Finally we have

\[{{V}_{th}}=\text{ }{{V}_{oc}}=\text{ }{{V}_{2}}=\text{ }7.5/8\text{ }=\text{ }0.9375\text{ }V\]

Now, we have dependent source, so method 1 cannot be used for finding Equivalent resistance Rth. Either Method 2 or Method 3 can be utilized here. We will prefer to employ second method. Terminals a and b are to be shorted as shown in the following figure and V_{2} = 0.

Applying KCL at node 1:

\[\begin{matrix} \frac{{{V}_{1}}-5}{2000}+\frac{{{V}_{1}}}{6000}+\frac{{{V}_{1}}}{1000}=0 & \cdots & (3) \\\end{matrix}\]

Multiplying (3) by 6000 yields:

\[3{{V}_{1}}\text{- }15\text{ }+\text{ }{{V}_{1}}+\text{ }6{{V}_{1}}=\text{ }0\]

\[10{{V}_{1}}=\text{ }15\]

\[{{V}_{1}}=\text{ }1.5\text{ }V\to \text{ }v\text{ }=\text{ }{{V}_{1}}=\text{ }1.5\text{ }V\]

The current through R_{3} (⟶) is given by

\[{{I}_{{{R}_{3}}}}=\text{ }{{V}_{1}}/{{R}_{3}}=~1.5\text{ }V/1k\Omega =\text{ }1.5mA\]

The current through VCCS (↓) is given by

\[{{I}_{VCCS}}=\text{ }0.0005{{V}_{1}}=~0.0005*1.5\text{ }A\text{ }=\text{ }0.75mA\]

Now, short-circuit current would be

\[{{I}_{sc}}=\text{ }{{I}_{{{R}_{3}}}}\text{ -}{{I}_{VCCS}}=\text{ }1.5mA\text{ -}0.75mA=\text{ }0.75mA\]

And Thevenin’s equivalent resistance is

\[{{R}_{th}}={{V}_{th}}/{{I}_{sc}}=\text{ }0.9675\text{ }V/0.75mA=\text{ }1.25\text{ }k\Omega \]

**Summary of Thevenin’s Theorem**

Following steps are being used to determine Thevenin’s equivalent circuit.

- The portion of circuit considered as load is removed
- The open circuit voltage V
_{oc}is calculated at terminals. - To calculate R
_{th}

- Short circuit all independent voltage sources.
- Open circuit all independent current sources.

- Draw equivalent circuit, connect the load and determine load current.

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