For the network of Fig. 9.54,
a. Find the Thévenin equivalent circuit for the portion of the network in the shaded area.
b. Reconstruct the network of Fig. 9.54 with the Thévenin equivalent network in place.
c. Using the resulting network of part (b) find the voltage $V_a$ .

Question

For the network of Fig. 9.54,
a. Find the Thévenin equivalent circuit for the portion of the network in the shaded area.
b. Reconstruct the network of Fig. 9.54 with the Thévenin equivalent network in place.
c. Using the resulting network of part (b) find the voltage [latex]V_a[/latex].

Accepted Answer

a. Steps 1 and 2: See Fig. 9.55.

Step 3: See Fig. 9.56.

[latex]R_{Th} = 12 Ω \parallel (4 Ω + 6 Ω) = 12 Ω \parallel 10 Ω = 5.45 [/latex]

Step 4: Applying the superposition theorem, we will first find the effect of the voltage source on the Thévenin voltage using the network of Fig. 9.57. Applying the voltage divider rule:

[latex]R^{\prime }_{Th} = \frac{12 Ω (18 V)}{6 Ω + 4 Ω + 12 Ω}=\frac{216}{22}V= 9.82 V[/latex]

The contribution due to the current source is determined using the network of Fig. 9.58(a) redrawn as shown in Fig. 9.58(b). Applying the current divider rule:

[latex]I^{\prime}=\frac{4 Ω (2 A)}{4 Ω + 18 Ω} =\frac{8}{22}A=0.364 A [/latex]

and [latex]E^{\prime\prime}_{Th}=-I^{\prime }(12 Ω) =-(0.364 A)(12 Ω) = -4.37 V[/latex]

so that [latex]E_{Th}=E^{\prime}_{Th}+E^{\prime\prime}_{Th}=9.82 V - 4.37 V = 5.45 V[/latex]

b. The reconstructed network is shown in Fig. 9.59.

c. Using the voltage divider rule:

[latex]V_a =\frac{8 Ω (5.45 V + 16 V)}{5.45 Ω + 8 Ω}=\frac{8 (21.45)}{13.45}V= \frac{171.6}{3.45}V=2.76 V [/latex]

Instead of using the superposition theorem, the current source could first have been converted to a voltage source and the series elements combined to determine the Thévenin voltage. In any event both approaches would have yielded the same results.

Question 9.11: For the network of Fig. 9.54, a. Find the Thévenin equivalen...

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