Question 9.10: Find the Thévenin equivalent circuit with respect to termina...

We first determine the Thévenin equivalent voltage. This voltage is the open-circuit voltage appearing at terminals a,b. We choose the reference for the Thévenin voltage as positive at terminal a. We can make two source transformations relative to the 120 V,12Ω, and  60Ω circuit elements to simplify this portion of the circuit. At the same time, these transformations must preserve the identity of the controlling voltage V_{x }because of the dependent voltage source. We determine the two source transformations by first replacing the series combination of the 120 V source and 12Ω resistor with a 10 A current source in parallel with 12Ω. Next, we replace the parallel combination of the 12 and 60Ω resistors with a single 10Ω resistor. Finally, we replace the 10 A source in parallel with 10Ω with a 100 V source in series with 10Ω. Figure 9.31 shows the resulting circuit. We added the current I to Fig. 9.31 to aid further discussion. Note that once we know the current I, we can compute the Thévenin voltage. We find I by summing the voltages around the closed path in the circuit shown in Fig. 9.31. Hence

100 = 10\pmb{I} – j40\pmb{I} + 120\pmb{I} + 10\pmb{V}_{x}=(130 – j40)\pmb{I} + 10\pmb{V}_{x}.

We relate the controlling voltage \pmb{V}_{x} to the current I by noting from Fig. 9.31 that

\pmb{V}_{x} = 100 – 10\pmb{I} .

Then,

\pmb{I} =\frac{-900}{30 – j40 }= 18 \angle -126.87° A  .

we now calculate \pmb{V}_{x} :

\pmb{V}_{x} = 100 – 180 \angle -126.87° = 208 + j144 V.

Finally, we note from Fig. 9.31 that

=784 – j288 = 835.22 \angle -20.17° V.

To obtain the Thévenin impedance, we may use any of the techniques previously used to find the Thévenin resistance. We illustrate the test-source method in this example. Recall that in using this method, we deactivate all independent sources from the circuit and then apply either a test voltage source or a test current source to the terminals of interest. The ratio of the voltage to the current at the source is the Thévenin impedance. Figure 9.32 shows the result of applying this technique to the circuit shown in Fig. 9.30. Note that we chose a test voltage source\pmb{V}_{T}. Also note that we deactivated the independent voltage source with an appropriate short-circuit and preserved the identity of \pmb{V}_{x} .

The branch currents\pmb{I}_{a} and\pmb{I}_{b} have been added to the circuit to simplify the calculation of \pmb{I}_{T} By straightforward applications of Kirchhoff’s circuit laws, you should be able to verify the following relationships:

\pmb{I}_{a} =\frac{\pmb{V}_{T}}{10 – j40^{,}}\pmb , \quad {V}_{x} = 10\pmb{I}_{a},

=\frac{-\pmb{V}_{T}(9 + j4)}{120\left(1 – j4\right)},

=\frac{\pmb{V}_{T}\left(3- j4\right)}{12\left(10 – j40\right)},

Z_{Th}=\frac{\pmb{V}_{T}}{\pmb{I}_{T}}= 91.2 – j 38.4 Ω.

Figure 9.33 depicts the Thévenin equivalent circuit.