7.3 Thévenin’s Theorem
LEARNING OBJECTIVES
- Find the Thévenin equivalent circuit for any linear circuit
- Calculate the maximum power that can be transferred to a load at any point in a circuit, and the value of the load resistance required to draw maximum power
Thévenin’s theorem states that any linear circuit containing several voltage sources and resistors can be simplified to a Thévenin-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, ;
- Thévenin voltage , found by removing from the original circuit and calculating the potential difference from one load connection point to the other (e.g. from to in Figure 7.3.1(a), either across and or across and );
- Thévenin resistance , found by removing from the original circuit and calculating the total equivalent resistance between the two load connection points (e.g. between and in Figure 7.3.1(a), thus as the equivalent resistance of the parallel combination of and , connected in series with ).
(Figure 7.3.1)
Thévenin’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 Thévenin’s theorem once and are determined for the rest of the circuit, the current through the load is always simply calculated as
(7.3.1)
from which the voltage drop across, and power transferred to the load are, respectively,
(7.3.2)
(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 Thévenin voltage and resistance with respect to . Example 7.3.1 shows the procedure for doing this for the circuit in Figure 7.3.1(a).
EXAMPLE 7.3.1
Applying Thévenin’s Theorem
Find and for the circuit in Figure 7.3.1(a).
Strategy
- Find : note that with the circuit open between and there is no current through, and therefore no voltage drop across . Therefore, the potential difference between and must occur in the loop containing and We are free to choose either parallel branch of that loop, as the potential difference across must equal the potential difference across and by the loop rule. Therefore, we will first determine the current in this loop and apply Ohm’s law to find .
- Find : Proceeding from to we encounter a junction where the circuit branches in two directions, towards and . is an ideal voltage source with no resistance, and