#### Leon Charles Thevenin

##### Leon Charles Thevenin

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#### Leon Charles Thevenin

##### Leon Charles Thevenin

Leon Charles Thevenin was a French telegraph engineer who worked on Ohm's law and extended it to the analysis of complicated electrical networks. He is remembered today almost entirely for one small piece of work. His theorem, published in 1883, was based on his study of Kirchhoff's Laws and is found in every basic textbook on electrical circuits. It has made his name familiar to every student of electrical circuits and to every electrical and electronics engineer.

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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,
$$V_{Th} = v_{oc} \quad \text{. . . (Eq.1)}$$

**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,
$$R_{Th}= Rin \quad \text{. . . (Eq.2)}$$

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
$$I_L = {V_{Th} \over R_{Th} + R_L} \quad \text{. . . (Eq.3)}$$
$$V_L = R_LI_L = {R_L \over R_{Th} + R_L} V_{Th} \quad \text{. . . (Eq.4)}$$
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 RTh, (b) finding VTh.

To find VTh, consider the circuit in 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 RTh and VTh 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,
$$i_1 = -3i_2 \tag{1}$$
For loops 2 and 3, applying KVL produces
$$4i_2 + 2(i_2-i_1) + 6(i_2 - i_3) = 0 \tag{2}$$
$$6(i_3 - i_2) + 2i_3 + 1 = 0 \tag{3}$$
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
$$i_1 = 5 \tag{4}$$
$$-2v_x + 2(i_3 - i_2) = 0 \Rightarrow v_x = i_3 - i_2$$
$$4(i_2 - i_1) + 2(i_2 - i_3) + 6i_2 = 0$$
or
$$12i_2 - 4i_1 - 2i_3 = 0 \tag{5}$$
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
$$i_o = i_x + {v_o \over 4} = - {v_o \over 2}+ {v_o \over 4} = -{v_o \over 4}$$00
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.