**Thevenin’s Equivalent
Circuit Example: The Bridge Circuit**

The Thevenin theorem states that any real
source may be represented as an ideal potential source in series
with a resistor. In many cases, one may use the Thevenin circuit
to solve electronics problems that might otherwise be tedious at
best. Take, for example a "bridge" circuit shown in
Figure 1, wherein an element is placed such as to bridge two
parallel arms of a resistor network. We may need to know the
current through and potential across the bridging resistor, *R*_{5}.

Figure 1 Resistor bridge circuit with 25 volt source potential.

We start by removing (in theory) *R*_{5}
to determine the Thevenin equivalent to the resistor network
composed of *R*_{1} through *R*_{4}.
The resulting circuit may be used to determine the potential and
current through *R*_{5} upon inserting this element
into the Thevenin equivalent circuit. To make the example more
concrete, consider the following for resistance; *R*_{1}=
1.0 kW
, *R*_{2}=900 W , *R*_{3}=800 W, and *R*_{4}=1.1
kW .
The value of *R*_{5} does not have to specified to
derive a formula.

Figure 2 Bridge circuit with *R*_{5}
removed.

The potential is more easily calculated since
the equivalent circuit reduces to two parallel voltage dividers.
Using Ohm’s law, one find the current through the series
arms, then the potential drop across the second resistor. For
example, the potential at point *A* is the current through
this arm times the resistance of *R*_{3}. Thus

In this case, 25 V x 800 W /(1000 W + 800 W ) = 11.11 V. The
potential at point *B* is calculated similarly, except we
use *R*_{2} and *R*_{4}. In this case, *E*_{B }= 25 V x 1100 W /(900
W +1100
W ) =
13.57 V. The differential potential is the potential of
Thevenin’s ideal potential source, *E*_{EQ}=*E*_{B}*-E*_{A}=13.75
- 11.11 = 2.64 V.

The Thevenin equivalent resistance found by
shorting-out the ideal potential source (in theory) and
determining resistance between points *A* and *B*.

Figure 3 Equivalent circuit with voltage source short-circuited.

In this case the resistance is the sum of the two parallel circuits

The resistance of the Thevenin equivalent circuit can be found to be 939 W .

The resulting Thevenin equivalent circuit shown in Figure 4. This circuit is functionally equivalent to that in Figure 2.

Figure 4 The Thevenin equivalent
circuit of the bridge circuit with *R*_{5} restored.

It is now relatively easy to calculate the
current through the circuit. The total resistance for the series
resistors is *R*_{TOT}*=R*_{EQ}*+R*_{5}.
Thus the current through both the equivalent resistor and *R*_{5}
is

*I*=*E*_{EQ} /(*R*_{EQ}*+R*_{5}).

The potential between points *A *and *B*
changes when *R*_{5} is restored. This new potential
is calculated using the voltage divider formula, or can be found
by considering the voltage drop across *R*_{5} due
to Ohm’s law. In either case

*E*_{AB}=*E*_{EQ}*
R*_{5}/(*R*_{EQ}*+R*_{5}).