Question 5.10: Thévenin and Norton Equivalents Find the Thévenin and Norton...
Thévenin and Norton Equivalents
Find the Thévenin and Norton equivalent circuits for the circuit shown in Figure 5.32(a).
We must find two of the three quantities: Voc, Isc, or Zt . Often, it pays to look for the two that can be found with the least amount of work. In this case, we elect to start by zeroing the sources to find Zt . After that part of the problem is finished, we will find the short-circuit current.
If we zero the sources, we obtain the circuit shown in Figure 5.32(b). The Thévenin impedance is the impedance seen looking back into terminals a–b. This is the parallel combination of the resistance and the impedance of the capacitance. Thus, we have
Z_t=\frac{1}{1/100+1/\left(-j100\right) } =\frac{1}{0.01+j0.01} =\frac{1}{0.01414\angle 45^\circ } =70.71\angle -45^\circ =50-j50~\Omega
Now, we apply a short circuit to terminals a–b and find the current, which is shown in Figure 5.32(c). With a short circuit, the voltage across the capacitance is zero. Therefore, IC = 0. Furthermore, the source voltage Vs appears across the resistance, so we have
\mathrm{I}_R=\frac{\mathrm{V}_s}{100}=\frac{100}{100}=1\angle0^\circ \mathrm{~A}
Then applying KCL, we can write
\mathrm{I_{sc}}=\mathrm{I}_R-\mathrm{I}_s=1-1\angle90^\circ =1-j=1.414-45^\circ \mathrm{~A}
Next, we can solve Equation 5.80 for the Thévenin voltage:
Z_t=\frac{\mathrm{V_{oc}}}{\mathrm{I_{sc}}}=\frac{\mathrm{V}_t}{\mathrm{I_{sc}}} (5.80)
\mathrm{V}_t=\mathrm{I_{sc}} Z_t=1.414\angle -45^\circ \times 70.71\angle -45^\circ =100\angle -90^\circ \mathrm{~V}
Finally, we can draw the Thévenin and Norton equivalent circuits, which are shown in Figure 5.33.
