# Circuit Analysis with Thevenin's theorem by Daryl Janzen

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)):

Thevenin voltage Vth, found by removing RL 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 R1, V1, and R3 or across R2 and R3);

Thevenin resistance Rth, found by removing RL 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 R1 and R2, connected in series with R3).

Figure 7.3.1 (a) An example of a DC resistive circuit with load resistor RL 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 Vth and Rth 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,

VL = RLIL $=\frac{{R}_{L}{V}_{\mathrm{th}}}{{R}_{\mathrm{th}}+{R}_{L}}$ (7.3.2)

${P}_{L}={{I}_{L}}^{2}{R}_{L}=\frac{{{V}_{L}}^{2}}{{R}_{L}}=\frac{{R}_{L}{{V}_{\mathrm{th}}}^{2}}{{\left({R}_{\mathrm{th}}+{R}_{L}\right)}^{2}}$ (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 Vth and resistance Rth with respect to RL. Example 7.3.1 shows the procedure for doing this for the circuit in Figure 7.3.1(a).

Example 7.3.1

Applying Thevenin's Theorem Find Vth and Rth for the circuit in Figure 7.3.1(a).

Strategy

1. Find Vth: note that with the circuit open between a and b there is no current through, and therefore no voltage drop across R3. Therefore, the potential difference between a and b must occur in the loop containing V1, R1, and R2. We are free to choose either parallel branch of that loop, as the potential difference across R2 must equal the potential difference across V1 and R1 by the loop rule. Therefore, we will first determine the current in this loop and apply Ohm's law to find Vth - VR2.

Find Rth: Proceeding from a to b we encounter a junction where the circuit branches in two directions, towards R1and R2. V1 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 R1 and R2 are connected in parallel. Proceeding on, we encounter R3 in series with the parallel connection of R1 and R2, 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 V1, R1 and R2 all connected in series is

$I=\frac{{V}_{1}}{{R}_{1}+{R}_{2}}$

By Ohm's law, the voltage across R2 is therefore

${V}_{{R}_{2}}={R}_{2}I=\frac{{R}_{2}{V}_{1}}{{R}_{1}+{R}_{2}}$

By our above reasoning, we therefore have

${V}_{\mathrm{th}}={V}_{{R}_{2}}=\frac{{R}_{2}{V}_{1}}{{R}_{1}+{R}_{2}}$

To find Rth, first write

${R}_{12}={\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\right)}^{-1}=\frac{{R}_{2}{V}_{1}}{{R}_{1}+{R}_{2}}$

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 R2 as current flows from the positive to the negative terminal of the voltage source V1. Along the parallel branch (that is, parallel from the perspective of the load connection points a and b), potential rises at V1, then drops across R1, 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 R2. Thus, between a and b along the left branch, travelling in the counter-clockwise direction there is also a drop in potential, equal to

$-\left({V}_{1}-{R}_{1}I\right)={\mathrm{-V}}_{1}+\frac{{R}_{2}{V}_{1}}{{R}_{1}+{R}_{2}}=\frac{{R}_{2}{V}_{1}}{{R}_{1}+{R}_{2}}={R}_{\mathrm{th}}$

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 R2 in the open circuit, we noted that current flows through a single circuit loop with V1, R1, and R2 all connected in series, and determined the current through R2 as the potential drop across the series combination of resistors, divided by the equivalent resistance. However, when calculating Rth we found that from the perspective of the connection points R1 and R2 are connected along parallel branches of the circuit.

The procedure used here to calculate Vth and Rth 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.

The circuit is the same as the one from Example 7.3.1, but with R1 replaced by a short. Determine Vth and Rth 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 RL, setting the result equal to 0, and solving for the value of RL that maximises the function.

$\frac{d{P}_{L}}{d{R}_{L}}=\frac{{{V}_{\mathrm{th}}}^{2}}{{\left({R}_{\mathrm{th}}+{R}_{L}\right)}^{2}}-\frac{2{R}_{L}{{V}_{\mathrm{th}}}^{2}}{{\left({R}_{\mathrm{th}}+{R}_{L}\right)}^{3}}=0$

$⟹1-\frac{2{R}_{L}}{\left({R}_{\mathrm{th}}+{R}_{L}\right)}=0$

$⟹{R}_{L}={R}_{\mathrm{th}}$

EXAMPLE 7.3.2

Applying Maximum Power Transfer Theorem

What is the maximum amount of power that can be dissipated in RL?

Strategy

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

With RL replaced by an open circuit, there are two loops: one, passing through V1, R1, R2, and V2; the other, passing through V2, R2, and R3. We will calculate the current through R3 using Mesh Analysis techniques developed earlier, then determine Vth = VR3 using Ohm's law. Note that we do not actually need to calculate any other currents, since Vth, the potential difference between a and b, must equal VR3 regardless which branch is taken.

To find Rth, note that with respect to connection points a and b, R1, R2,, and R3 are all connected in parallel.

Finally, when RL = Rth, the current in the load is IL = $\frac{{V}_{\mathrm{th}}}{2{R}_{\mathrm{th}}}$ (see Equation 7.3.1), and the power dissipated in RL is PL,,max = IL2RL = Vth2 / 4Rth (Equation 7.3.3).

Solution

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

$\left[\begin{array}{c}-{V}_{1}+{V}_{2}\\ -{V}_{2}\end{array}\right]=\left[\begin{array}{cc}-\left({R}_{1}+{R}_{2}\right)& {R}_{2}\\ {R}_{2}& -\left({R}_{2}+{R}_{3}\right)\end{array}\right]$ $\left[\begin{array}{c}{I}_{1}\\ {I}_{2}\end{array}\right]$

where I1 and I2 are the clockwise mesh currents in the left and right loops, respectively.

To find I2 (the actual current in R3), we apply Cramer's rule:

${I}_{2}=\frac{\left[\begin{array}{cc}-\left({R}_{1}+{R}_{2}\right)& -{V}_{1}+{V}_{2}\\ {R}_{2}& -{V}_{2}\end{array}\right]}{\left[\begin{array}{cc}-\left({R}_{1}+{R}_{2}\right)& {R}_{2}\\ {R}_{2}& -\left({R}_{2}+{R}_{3}\right)\end{array}\right]}=$ $\frac{{V}_{2}{R}_{1}+{V}_{2}{R}_{2}+{V}_{1}{R}_{2}+{V}_{2}{R}_{2}}{{R}_{1}{R}_{2}+{{R}_{2}}^{2}+{R}_{1}{R}_{3}+{R}_{2}{R}_{3}-{{R}_{2}}^{2}}=\frac{{V}_{2}{R}_{1}+{V}_{1}{R}_{2}}{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{2}{R}_{3}}$

The Thevenin-equivalent voltage is therefore

${V}_{th}={R}_{3}{I}_{2}=\frac{R3\left(V2R1+V1R2\right)}{{R}_{1}{R}_{2}+{R}_{1}{R}_{3}+{R}_{2}{R}_{3}}$

The Thevenin-equivalent resistance is

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

Significance

It is important to be clear that PL,max is the power dissipated in RL only when RL = Rth. The general expression for power dissipated in RLis 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