Thevenin's theorem states that any
linear
circuit containing several voltage
sources and resistors can be simplified to a *Thevenin-equivalent circuit*
with a single voltage source and resistance connected in series with a load.
Specifically, the three components connected in series are (see Figure 7.3.1(b)):

_{L};

Thevenin voltage V_{th}, found by removing R_{L} from the
original circuit and calculating the potential difference from one load
connection point to the other (e.g. from a to b in Figure 7.3.1(a), either
across R_{1}, V_{1}, and R_{3} or across R_{2}
and R_{3});

Thevenin resistance R_{th}, found by removing R_{L} from the
original circuit and calculating the total equivalent resistance between the
two load connection points (e.g. between a and b in Figure 7.3.1(a), thus as
the equivalent resistance of the parallel combination of R_{1} and
R_{2}, connected in series with R_{3}).

**Figure 7.3.1** (a) An example of a DC resistive circuit with load resistor
R_{L} identified, and (b) its Thevenin equivalent. In fact, (b) shows
the general form of all Thevenin-equivalent circuits.

Thevenin's theorem is particularly useful when the load resistance in a circuit
is subject to change. When the load's resistance changes, so does the current
it draws and the power transferred to it by the rest of the circuit. In fact,
currents everywhere in a circuit will be subject to change whenever a single
resistance changes, and the entire circuit would need to be re-analysed to find
the new current through and power transferred to a load. Repeating circuit
analysis to find the new current through a load every time its resistance changes
would be very time-consuming. In contrast, according to Thevenin's theorem once
V_{th} and R_{th} are determined for the rest of the circuit,
the current through the load is always simply calculated as

${I}_{L}=\frac{{V}_{\mathrm{th}}}{{R}_{\mathrm{th}}+{R}_{\mathrm{Lh}}}$ (7.3.1)

from which the voltage drop across, and power transferred to the load are, respectively,

V_{L} = R_{L}I_{L}
$=\frac{{R}_{L}{V}_{\mathrm{th}}}{{R}_{\mathrm{th}}+{R}_{L}}$ (7.3.2)

${P}_{L}={{I}_{L}}^{2}{R}_{L}=\frac{{{V}_{L}}^{2}}{{R}_{L}}=\frac{{R}_{L}{{V}_{\mathrm{th}}}^{2}}{{({R}_{\mathrm{th}}+{R}_{L})}^{2}}$ (7.3.3)

Equations 7.3.1-7.3.3 are easily applied, and the problem of repeated circuit analysis
each time a load's resistance changes is mainly reduced to the one-time problem of finding
the Thevenin voltage V_{th} and resistance R_{th} with respect to R_{L}.
Example 7.3.1 shows the procedure for doing this for the circuit in Figure 7.3.1(a).

**Example 7.3.1**

_{th}and R

_{th}for the circuit in Figure 7.3.1(a).

Strategy

1. Find V_{th}: note that with the circuit open between a and b there is no current
through, and therefore no voltage drop across R_{3}. Therefore, the potential
difference between a and b must occur in the loop containing V_{1}, R_{1},
and R_{2}. We are free to choose either parallel branch of that loop, as the
potential difference across R_{2} must equal the potential difference across
V_{1} and R_{1} by the loop rule. Therefore, we will first determine the
current in this loop and apply Ohm's law to find V_{th} - V_{R2}.

Find R_{th}: Proceeding from *a* to *b* we encounter a junction
where the circuit branches in two directions, towards R_{1}and R_{2}.
V_{1} is an ideal voltage source with no resistance, and can therefore be ignored
when calculating equivalent resistance. We then encounter another junction where the
two branches reconnect, so R_{1} and R_{2} are connected in parallel.
Proceeding on, we encounter R_{3} in series with the parallel connection of
R_{1} and R_{2}, and eventually reach *b*. We will add these
resistances using the rules for adding series and parallel resistors.

Solution

The current through the loop with V_{1}, R_{1} and R_{2} all
connected in series is

By Ohm's law, the voltage across R_{2} is therefore

By our above reasoning, we therefore have

${V}_{\mathrm{th}}={V}_{{R}_{2}}=\frac{{R}_{2}{V}_{1}}{{R}_{1}+{R}_{2}}$To find R_{th}, first write

Then, by our above reasoning, ${R}_{\mathrm{th}}={R}_{12}+{R}_{3}=\frac{{R}_{2}{V}_{1}}{{R}_{1}+{R}_{2}}+{R}_{3}$

**Significance**

The potential difference from *a* to *b* was calculated as a drop in
potential across R_{2} as current flows from the positive to the negative
terminal of the voltage source V_{1}. Along the parallel branch (that is,
parallel from the perspective of the load connection points *a* and *b*),
potential rises at V_{1}, then drops across R_{1}, travelling in
the clockwise direction. By the loop rule, there must be an overall potential rise
in the clockwise direction along this branch that equals negative the potential
drop in the clockwise direction across R_{2}. Thus, between *a* and
*b* along the left branch, travelling in the counter-clockwise direction there
is also a drop in potential, equal to

as required.

It is important to note that perspective matters when treating components as being
connected in series or parallel. Here, when determining the current through R_{2}
in the open circuit, we noted that current flows through a single circuit loop with
V_{1}, R_{1}, and R_{2} all connected in series, and determined
the current through R_{2} as the potential drop across the series combination
of resistors, divided by the equivalent resistance. However, when calculating R_{th}
we found that from the perspective of the connection points R_{1} and R_{2}
are connected along parallel branches of the circuit.

The procedure used here to calculate V_{th} and R_{th} is the same
as that which we apply to more complex circuits. When doing so, it is important to
correctly account for voltage rises and drops across between the two load connection
points, although to this end we do have freedom of choice in which branch to follow
and can always choose the simplest path.

**CHECK YOUR UNDERSTANDING 7.3**

The circuit is the same as the one from Example 7.3.1, but with R_{1} replaced
by a short. Determine V_{th} and R_{th} in this case.

**Maximum Power Transfer Theorem**

Thevenin's theorem finds a useful application in the **maximum power transfer theorem**,
which states that maximum power will be transferred to a load when its resistance is equal
to the Thevenin resistance of the network supplying the power. This interesting and highly
useful fact is easily proven by taking the derivative of Equation 7.3.3 with respect to
R_{L}, setting the result equal to 0, and solving for the value of R_{L}
that maximises the function.

$\u27f91-\frac{2{R}_{L}}{({R}_{\mathrm{th}}+{R}_{L})}=0$

$\u27f9{R}_{L}={R}_{\mathrm{th}}$

**EXAMPLE 7.3.2**

Applying Maximum Power Transfer Theorem

What is the maximum amount of power that can be dissipated in R_{L}?

Strategy

The maximum amount of power that can be dissipated in R_{L} is, by the maximum power
transfer theorem, the power dissipated when R_{L} = R_{th}, for the Thevenin
equivalent circuit calculated with respect to R_{L}. To find this, we first determine
V_{th} and R_{th}, as follows.

With R_{L} replaced by an open circuit, there are two loops: one, passing through
V_{1}, R_{1}, R_{2}, and V_{2}; the other, passing through
V_{2}, R_{2}, and R_{3}. We will calculate the current through R_{3}
using Mesh Analysis techniques developed earlier, then determine V_{th} = V_{R3}
using Ohm's law. Note that we do not actually need to calculate any other currents, since V_{th},
the potential difference between a and b, must equal V_{R3} regardless which branch
is taken.

To find R_{th}, note that with respect to connection points *a* and *b*,
R_{1}, R_{2},, and R_{3} are all connected in parallel.

Finally, when R_{L} = R_{th}, the current in the load is I_{L} =
$\frac{{V}_{\mathrm{th}}}{2{R}_{\mathrm{th}}}$
(see Equation 7.3.1), and the power dissipated in R_{L} is P_{L,,max} =
I_{L}^{2}R_{L} = V_{th}^{2} / 4R_{th} (Equation 7.3.3).

Solution

Using the strategies developed in Mesh Analysis, we can write the matrix equations for this network as

$\left[\begin{array}{c}-{V}_{1}+{V}_{2}\\ -{V}_{2}\end{array}\right]=\left[\begin{array}{cc}-({R}_{1}+{R}_{2})& {R}_{2}\\ {R}_{2}& -({R}_{2}+{R}_{3})\end{array}\right]$ $\left[\begin{array}{c}{I}_{1}\\ {I}_{2}\end{array}\right]$where I_{1} and I_{2} are the clockwise mesh currents in the left and right loops, respectively.

To find I_{2} (the actual current in R_{3}), we apply Cramer's rule:

The Thevenin-equivalent voltage is therefore

${V}_{\mathrm{th}}={R}_{3}{I}_{2}=\frac{R3(V2R1+V1R2)}{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{2}{R}_{3}}$The Thevenin-equivalent resistance is

${P}_{\mathrm{L,max}}=\frac{{{V}_{\mathrm{th}}}^{2}}{4{R}_{\mathrm{th}}}=\frac{{R}_{1}{R}_{2}{R}_{3}3{(V2R1+V1R2)}^{2}}{{({R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{2}{R}_{3})}^{3}}$Significance

It is important to be clear that P_{L,max} is the power dissipated in R_{L} only when
R_{L} = R_{th}. The general expression for power dissipated in R_{L}is given by Equation 7.3.3.

Authored by: Daryl Janzen. Provided by: Department of Physics and Engineering Physics, University of Saskatchewan. License: CC BY: Attribution

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