Thevenin’s Theorem Basic Formula Electric Circuits

Figure 1

If the terminals a-b are made open-circuited (by removing the load), no current flows, so that the open-circuit voltage across terminal a-b in Figure.(1a) must be equal to the voltage source V_Th in Figure.(1b), since the two circuits are equivalent.

Thus V_Th is the open-circuit voltage across the terminals as shown in Figure.(2a); that is,

Figure 2

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 terminals a-b in Figure.(1a) must be equal to R_Th in Figure.(2b) because the two circuits are equivalent.

Thus, R_Th is the input resistance at terminals when the independent sources are turned off, as shown  in Figure.(2b); that is

Case 1 If the network has no dependent sources, we turn off all independent sources. R_Th is the input resistance of the network looking between terminals a and b, as shown in Figure.(2b)

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 Figure.(3a).

Figure 3

Alternatively, we may insert a current source i_o at terminals a-b as shown in Figure.(3b) and find the terminal voltage v_o.

Again R_Th = v_o/i_o. Either of the two approaches will give the same results. In either approach, we may assume any value of v_o and i_o.

For example, we may use v_o = 1V or i_o = 1A, 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.

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. 

Figure 4

Consider a linear circuit terminated by a load R_L, as shown in Figure.(4a).

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 Figure.(4b). From Figure.(4b), we get

(3a)

(3b)

Note from Figure.(4b) that the Thevenin equivalent is a simple voltage divider, yielding V_L by mere inspection.

To understand better, let us review the examples below :

1.Find the Thevenin equivalent circuit shown in Figure.(5), to the left of the terminals a-b. Then find the current through R_L = 6, 16, and 36 Ω

Figure 5

Solution :

We find R_Th by turning off the 32 V voltage source (replacing it with a short circuit) and the 2 A current source (replacing it with an open circuit). The circuit becomes what is shown in Figure.(6a). Thus,

Figure 6

To find V_Th, consider the circuit in Figure.(6b). Applying mesh analysis to the two loops, we obtain

Solving for i₁, we get i₁ = 0.5 A. Thus,

Alternatively, it is even easier to use nodal analysis. We ignore the 1 Ω resistor since no current flows through it. At the top node, KVL gives

or

as obtained before. We could also use source transformation to find V_Th.

The Thevenin equivalent circuit is shown in Figure.(7). 

Figure 7

The current through R_L is

When R_L = 6,

When R_L = 16,

When R_L = 36,

2.Find the Thevenin equivalent of the circuit in Figure.(8) at terminals a-b.

Figure 8

Solution :

This circuit contains a dependent source, unlike the circuit in the previous example. To find R_Th, we set the independent source equal to zero but leave the dependent source alone.

Because of the presence of the dependent source, however, we excite the network with a voltage source v_o connected to the terminals as indicated in Figure.(9a).

We may set v_o = 1 V to ease calculation since the circuit is linear. Our goal is to find the current i_o through the terminals and then obtain R_Th = 1/i_o.

Alternatively, we may insert a 1 A current source, find the corresponding voltage v_o, and obtain R_Th = v_o/1.

Figure 9

Applying mesh analysis to loop 1 in the circuit of Figure.(9a) results in

But -4i₂ = v_x = i₁ – i₂; hence

(2.1)

For loops 2 and 3, applying KVL produces

(2.2)

(2.3)

Solving these equations gives

But i_o = –i₃ = 1/6 A. Hence,

To get V_Th, we find v_oc in the circuit of Figure.(9b). Applying mesh analysis, we get

(2.4)

(2.5)

or

(2.6)

But 4(i₁-i₂) = v_x. Solving these equations leads to i₂ = 10/3. Hence

The Thevenin equivalent is as shown in Figure.(10)

Figure 10