# Thevenins Theorem   Whatsapp  It often occurs in practice that a particular element in a circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outlet terminal may be connected to different appliances constituting a variable load. Each time the variable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenin's theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit.
According to Thevenin's theorem, the linear circuit in [Fig. 1(a)] can be replaced by that in [Fig. 1(b)]. (The load in [Fig. 1] may be a single resistor or another circuit.) The circuit to the left of the terminals $a$-$b$ in [Fig. 1(b)] is known as the Thevenin equivalent circuit.  Fig. 1: Replacing a linear two-terminal circuit by its Thevenin equivalent: (a) original circuit, (b) the Thevenin equivalent circuit.
Thevenin's theorem developed in 1883 by M. Leon Thevenin (1857 — 1926), a French telegraph engineer, states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source $V_{Th}$ in series with a resistor $R_{Th}$, where $V_{Th}$ is the open-circuit voltage at the terminals and $R_{Th}$ is the input or equivalent resistance at the terminals when the independent sources are turned off.
Our major concern right now is how to find the Thevenin equivalent voltage $V_{Th}$ and resistance $R_{Th}$. To do so, suppose the two circuits in [Fig. 1] are equivalent. Two circuits are said to be equivalent if they have the same voltage-current relation at their terminals. Let us find out what will make the two circuits in [Fig. 1] equivalent.
If the terminals $a$-$b$ are made open-circuited (by removing the load), no current flows, so that the open-circuit voltage across the terminals $a$-$b$ in [Fig. 1(a)] must be equal to the voltage source $V_{Th}$ in [Fig. 1(b)], since the two circuits are equivalent. Thus $V_{Th}$ is the open-circuit voltage across the terminals as shown in [Fig. 3(a)]; that is,
$$V_{Th} = v_{oc} \tag{1}$$  Fig. 3: Finding VTh and RTh.
Again, with the load disconnected and terminals $a$-$b$ open-circuited, we turn off all independent sources. The input resistance (or equivalent resistance) of the dead circuit at the terminals $a$-$b$ in [Fig. 1(a)] must be equal to $R_{Th}$ in [Fig. 1(b)] because the two circuits are equivalent.
Thus, $R_{Th}$ is the input resistance at the terminals when the independent sources are turned off, as shown in [Fig. 3(b)]; that is,
$$R_{Th}= Rin \tag{2}$$
To apply this idea in finding the Thevenin resistance $R_{Th}$, we need to consider two cases.
##### CASE 1
:
If the network has no dependent sources, we turn off all independent sources. RTh is the input resistance of the network looking between terminals a and b, as shown in [Fig. 3(b)].
##### CASE 2
:
If the network has dependent sources, we turn off all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage source $V_o$ at terminals $a$ and $b$ and determine the resulting current $i_o$. Then $R_{Th} = v_o/i_o$, as shown in [Fig. 4(a)]. Alternatively, we may insert a current source io at terminals a-b as shown in [Fig. 4(b)] and find the terminal voltage vo. Again $R_{Th} = v_o/i_o$.  Fig. 4: Finding RTh when circuit has dependent sources.
Either of the two approaches will give the same result. In either approach we may assume any value of $v_o$ and $i_o$. For example, we may use $v_o = 1 V$ or $i_o = 1 A$, or even use unspecified values of $v_o$ or $i_o$.
It often occurs that $R_{Th}$ takes a negative value. In this case, the negative resistance ($v = -iR$) implies that the circuit is supplying power.
This is possible in a circuit with dependent sources; Example 3 will illustrate this.
Thevenin's theorem is very important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a single independent voltage source and a single resistor. This replacement technique is a powerful tool in circuit design.
As mentioned earlier, a linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit.  Fig. 5: A circuit with a load: (a) original circuit, (b) Thevenin equivalent.
Consider a linear circuit terminated by a load $R_L$, as shown in [Fig. 5(a)]. The current $I_L$ through the load and the voltage $V_L$ across the load are easily determined once the Thevenin equivalent of the circuit at the load's terminals is obtained, as shown in [Fig. 5(b)]. From [Fig. 5(b)], we obtain $$I_L = {V_{Th} \over R_{Th} + R_L} \tag{3}$$ $$V_L = R_LI_L = {R_L \over R_{Th} + R_L} V_{Th} \tag{4}$$ Note from [Fig. 5(b)] that the Thevenin equivalent is a simple voltage divider, yielding $V_L$ by mere inspection.
Example 1: Find the Thevenin equivalent circuit of the circuit shown in Fig. 6, to the left of the terminals a-b. Then find the current through $R_L = 6$, $16$, and $36 Ω$. Fig. 6: For Example 1.
Solution: We find $R_{Th}$ by turning off the 32V voltage source (replacing it with a short circuit) and the 2A current source (replacing it with an open circuit). The circuit becomes what is shown in [Fig. 7(a)]. Thus,
$$R_{Th} = 4 || 12 + 1 = {4 \times 12 \over 16} + 1 = 4 Ω$$  Fig. 7: For Example 1: (a) finding $R_{Th}$, (b) finding $V_{Th}$.
To find $V_{Th}$, consider the circuit in $\text{Fig. 7(b)}$. Applying mesh analysis to the two loops, we obtain
$$-32 + 4i_1 + 12(i_1 - i_2) = 0$$
$$i_2 = -2 A$$
Solving for $i_1$, we get $i_1 = 0.5 A$. Thus,
$$V_{Th} = 12(i_1 - i_2) = 12(0.5 + 2.0) = 30 V$$
Alternatively, it is even easier to use nodal analysis. We ignore the $1Ω$ resistor since no current flows through it. At the top node, KCL gives
$${32 - V_{Th} \over 4} + 2 = {V_{Th} \over 12}$$
or
$$96 - 3 V_{Th} + 24 = V_{Th}\Rightarrow V_{Th} = 30 V$$
as obtained before. Fig. 8: The Thevenin equivalent circuit for Example 1.
The Thevenin equivalent circuit is shown in [Fig. 8]. The current through $R_L$ is
$$I_L = {V_{Th} \over R_{Th} + R_L}$$
$$= {30 \over 4 + R_L}$$
When $R_L = 6$,
$$I_L= 3 A$$
When $R_L = 16$,
$$I_L = {30 \over 20} = 1.5 A$$
When $R_L = 36$,
$$I_L = {30 \over 40} = 0.75 A$$
Example 2: Find the Thevenin equivalent of the circuit in [Fig. 9]. Fig. 9: For Example 2.
Solution: This circuit contains a dependent source, unlike the circuit in the previous example. To find $R_{Th}$, we set the independent source equal to zero but leave the dependent source alone. Because of the presence of the dependent source, however, we excite the network with a voltage source $v_o$ connected to the terminals as indicated in [Fig. 10(a)]. We may set $v_o = 1 V$ to ease calculation, since the circuit is linear. Our goal is to find the current io through the terminals, and then obtain $R_{Th} = 1/i_o$. (Alternatively, we may insert a $1A$ current source, find the corresponding voltage $v_o$, and obtain $R_{Th} = v_o/1$.)  Fig. 10: Finding $R_{Th}$ and $V_{Th}$ for Example 2.
Applying mesh analysis to loop 1 in the circuit in [Fig. 10(a)] results in
$$-2v_x + 2(i_1 - i_2) =0$$
or
$$v_x = i_1 - i_2$$
But
$$-4i_2 = v_x = i_1 - i_2$$
hence,
$$i_1 = -3i_2 \tag{1.1}$$
For loops 2 and 3, applying KVL produces
$$4i_2 + 2(i_2-i_1) + 6(i_2 - i_3) = 0 \tag{1.2}$$
$$6(i_3 - i_2) + 2i_3 + 1 = 0 \tag{1.3}$$
Solving these equations gives
$$i_3 = {-1 \over 6A}$$
But $i_o = -i_3 = 1/6 A$. Hence,
$$R_{Th} = {1 V \over i_o} = 6 Ω$$
To get $V_{Th}$, we find $v_{oc}$ in the circuit of [Fig. 10(b)]. Applying mesh analysis, we get
$$i_1 = 5 \tag{1.4}$$
$$-2v_x + 2(i_3 - i_2) = 0 \Rightarrow v_x = i_3 - i_2$$
$$4(i_2 - i_1) + 2(i_2 - i_3) + 6i_2 = 0$$
or
$$12i_2 - 4i_1 - 2i_3 = 0 \tag{1.5}$$
But $4(i_1- i2) = v_x$ . Solving these equations leads to $i_2 = 10/3$. Hence,
$$V_{Th} = v_{oc} = 6i_2 = 20 V$$
The Thevenin equivalent is as shown in [Fig. 11]. Fig. 11: The Thevenin equivalent of the circuit in Fig. 10.
Example 3: Determine the Thevenin equivalent of the circuit in [Fig. 12(a)].  Fig. 12:
Solution:
Since the circuit in [Fig. 12(a)] has no independent sources, $V_{Th} = 0 V$. To find $R_{Th}$, it is best to apply a current source $i_o$ at the terminals as shown in [Fig. 12(b)]. Applying nodal analysis gives
$$i_o + i_x = 2i_x + {v_o \over 4} \tag{1}$$
But
$$i_x = 0 - {v_o \over 2}= -{v_o \over 2} \tag{2}$$
Substituting Eq. 2 into Eq. 1 yields
$$i_o = i_x + {v_o \over 4} = - {v_o \over 2}+ {v_o \over 4} = -{v_o \over 4}$$
or
$$v_o = -4i_o$$
Thus,
$$R_{Th} = {v_o \over i_o} = -4 Ω$$
The negative value of the resistance tells us that, according to the passive sign convention, the circuit in [Fig. 12(a)] is supplying power. Of course, the resistors in [Fig. 12(a)] cannot supply power (they absorb power); it is the dependent source that supplies the power. This is an example of how a dependent source and resistors could be used to simulate negative resistance.

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