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, R5.

Figure 1 Resistor bridge circuit with 25 volt source potential.

We start by removing (in theory) R5 to determine the Thevenin equivalent to the resistor network composed of R1 through R4. The resulting circuit may be used to determine the potential and current through R5 upon inserting this element into the Thevenin equivalent circuit. To make the example more concrete, consider the following for resistance; R1= 1.0 kW , R2=900 W , R3=800 W, and R4=1.1 kW . The value of R5 does not have to specified to derive a formula.

Figure 2 Bridge circuit with R5 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 R3. 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 R2 and R4. In this case, EB = 25 V x 1100 W /(900 W +1100 W ) = 13.57 V. The differential potential is the potential of Thevenin’s ideal potential source, EEQ=EB-EA=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 R5 restored.

It is now relatively easy to calculate the current through the circuit. The total resistance for the series resistors is RTOT=REQ+R5. Thus the current through both the equivalent resistor and R5 is

I=EEQ /(REQ+R5).

The potential between points A and B changes when R5 is restored. This new potential is calculated using the voltage divider formula, or can be found by considering the voltage drop across R5 due to Ohm’s law. In either case

EAB=EEQ R5/(REQ+R5).

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