Thevenins Theorem

It often occurs in practice that a particular element in a circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outlet terminal may be connected to different appliances constituting a variable load. Each time the variable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenin's theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit. According to Thevenin's theorem, the linear circuit in [Fig. 1(a)] can be replaced by that in [Fig. 1(b)]. (The load in [Fig. 1] may be a single resistor or another circuit.) The circuit to the left of the terminals $a$-$b$ in [Fig. 1(b)] is known as the Thevenin equivalent circuit. Our major concern right now is how to find the Thevenin equivalent voltage $V_{Th}$ and resistance $R_{Th}$. To do so, suppose the two circuits in [Fig. 1] are equivalent. Two circuits are said to be equivalent if they have the same voltage-current relation at their terminals. Let us find out what will make the two circuits in [Fig. 1] equivalent. If the terminals $a$-$b$ are made open-circuited (by removing the load), no current flows, so that the open-circuit voltage across the terminals $a$-$b$ in [Fig. 1(a)] must be equal to the voltage source $V_{Th}$ in [Fig. 1(b)], since the two circuits are equivalent. Thus $V_{Th}$ is the open-circuit voltage across the terminals as shown in [Fig. 3(a)]; that is, Again, with the load disconnected and terminals $a$-$b$ open-circuited, we turn off all independent sources. The input resistance (or equivalent resistance) of the dead circuit at the terminals $a$-$b$ in [Fig. 1(a)] must be equal to $R_{Th}$ in [Fig. 1(b)] because the two circuits are equivalent. Thus, $R_{Th}$ is the input resistance at the terminals when the independent sources are turned off, as shown in [Fig. 3(b)]; that is, To apply this idea in finding the Thevenin resistance $R_{Th}$, we need to consider two cases.

CASE 1

: If the network has no dependent sources, we turn off all independent sources. RTh is the input resistance of the network looking between terminals a and b, as shown in [Fig. 3(b)].

CASE 2

: If the network has dependent sources, we turn off all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage source $V_o$ at terminals $a$ and $b$ and determine the resulting current $i_o$. Then $R_{Th} = v_o/i_o$, as shown in [Fig. 4(a)]. Alternatively, we may insert a current source io at terminals a-b as shown in [Fig. 4(b)] and find the terminal voltage vo. Again $R_{Th} = v_o/i_o$. Either of the two approaches will give the same result. In either approach we may assume any value of $v_o$ and $i_o$. For example, we may use $v_o = 1 V$ or $i_o = 1 A$, or even use unspecified values of $v_o$ or $i_o$. It often occurs that $R_{Th}$ takes a negative value. In this case, the negative resistance ($v = -iR$) implies that the circuit is supplying power.
This is possible in a circuit with dependent sources; Example 3 will illustrate this. Thevenin's theorem is very important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a single independent voltage source and a single resistor. This replacement technique is a powerful tool in circuit design.
As mentioned earlier, a linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit. Consider a linear circuit terminated by a load $R_L$, as shown in [Fig. 5(a)]. The current $I_L$ through the load and the voltage $V_L$ across the load are easily determined once the Thevenin equivalent of the circuit at the load's terminals is obtained, as shown in [Fig. 5(b)]. From [Fig. 5(b)], we obtain Note from [Fig. 5(b)] that the Thevenin equivalent is a simple voltage divider, yielding $V_L$ by mere inspection.