Circuit Analysis with Thevenin's theorem

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)):

Load resistor, R_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 R₁, V₁, and R₃ or across R₂ and R₃);

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 R₁ and R₂, connected in series with R₃).

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

$I_{L} = \frac{V_{th}}{R_{th} + R_{Lh}}$ (7.3.1)

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

V_L = R_LI_L $= \frac{R_{L} V_{th}}{R_{th} + R_{L}}$ (7.3.2)

$P_{L} = {I_{L}}^{2} R_{L} = \frac{{V_{L}}^{2}}{R_{L}} = \frac{R_{L} {V_{th}}^{2}}{{(R_{th} + R_{L})}^{2}}$ (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

Applying Thevenin's Theorem Find V_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 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 V_th - V_R₂.

Find R_th: 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_{th} = V_{R_{2}} = \frac{R_{2} V_{1}}{R_{1} + R_{2}}

To find R_th, first write

R_{12} = {(\frac{1}{R_{1}} + \frac{1}{R_{2}})}^{-1} = \frac{R_{2} V_{1}}{R_{1} + R_{2}}

Then, by our above reasoning, $R_{th} = R_{12} + R_{3} = \frac{R_{2} V_{1}}{R_{1} + R_{2}} + R_{3}$

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

- (V_{1} - R_{1} I) = {-V}_{1} + \frac{R_{2} V_{1}}{R_{1} + R_{2}} = \frac{R_{2} V_{1}}{R_{1} + R_{2}} = R_{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 R_th 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 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 R₁ 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.

\frac{d P_{L}}{d R_{L}} = \frac{{V_{th}}^{2}}{{(R_{th} + R_{L})}^{2}} - \frac{2 R_{L} {V_{th}}^{2}}{{(R_{th} + R_{L})}^{3}} = 0

⟹ 1 - \frac{2 R_{L}}{(R_{th} + R_{L})} = 0

⟹ R_{L} = R_{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 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 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, R₁, R₂,, and R₃ are all connected in parallel.

Finally, when R_L = R_th, the current in the load is I_L = $\frac{V_{th}}{2 R_{th}}$ (see Equation 7.3.1), and the power dissipated in R_L is P_L,,max = I_L²R_L = V_th² / 4R_th (Equation 7.3.3).

Solution

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

[\begin{matrix} - V_{1} + V_{2} \\ - V_{2} \end{matrix}] = [\begin{matrix} - (R_{1} + R_{2}) & R_{2} \\ R_{2} & - (R_{2} + R_{3}) \end{matrix}]

[\begin{matrix} I_{1} \\ I_{2} \end{matrix}]

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{[\begin{matrix} - (R_{1} + R_{2}) & - V_{1} + V_{2} \\ R_{2} & - V_{2} \end{matrix}]}{[\begin{matrix} - (R_{1} + R_{2}) & R_{2} \\ R_{2} & - (R_{2} + R_{3}) \end{matrix}]} =

\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{R 3 (V 2 R 1 + V 1 R 2)}{R_{1} R_{2} + R_{1} R_{3} + R_{2} R_{3}}

The Thevenin-equivalent resistance is

P_{L,max} = \frac{{V_{th}}^{2}}{4 R_{th}} = \frac{R_{1} R_{2} R_{3} 3 {(V 2 R 1 + V 1 R 2)}^{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_Lis 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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