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## Q. 5.8

It is instructive to examine the use of Thévenin’s and Norton’s theorems in the solution of the network in Fig. 5.4a, which is redrawn in Fig. 5.10a.  ## Verified Solution

If we break the network at the 6-kΩ load, the open-circuit voltage is found from Fig. 5.10b. The equations for the mesh currents are

-6 + 4k$I_{1} + 2k(I_{1} – I_{2})$ = 0

and

$I_{2} = 2 × 10^{-3}$

from which we easily obtain $I_{1}$ = 5/3 mA. Then, using KVL, $V_{oc}$ is

$V_{oc} = 4kI_{1} + 2kI_{2}$
= 4k $\huge(\frac{5}{3} × 10^{-3} \huge) + 2k(2 × 10^{-3})$
= $\frac{32}{3}$ V

$R_{Th}$ is derived from Fig. 5.10c and is

$R_{Th}$ = (2k // 4k) + 2k = $\frac{10}{3}$

Attaching the Thévenin equivalent to the load produces the network in Fig. 5.10d. Then using voltage division, we obtain

$V_{o} = \frac{32}{3} \huge( \frac{6k}{6k + \frac{10}{3}k} \huge)$
= $\frac{48}{7}$ V

In applying Norton’s theorem to this problem, we must find the short-circuit current shown in Fig. 5.10e. At this point the quick-thinking reader stops immediately! Three mesh equations applied to the original circuit will immediately lead to the solution, but the three mesh equations in the circuit in Fig. 5.10e will provide only part of the answer, specifically the short-circuit current. Sometimes the use of the theorems is more complicated than a straightforward attack using node or loop analysis. This would appear to be one of those situations. Interestingly, it is not. We can find $I_{sc}$ from the network in Fig. 5.10e without using the mesh equations. The technique is simple, but a little tricky, and so we ignore it at this time. Having said all these things, let us now finish what we have started. The mesh equations for the network in Fig. 5.10e are

-6 + 4k′($I_{1} – I_{sc}$) + 2k′($I_{1} – 2 × 10^{-3})$ = 0

2k′($I_{sc} – 2 × 10^{-3}$) + 4k′ ($I_{sc} – I_{1})$ = 0

where we have incorporated the fact that $I_{2} = 2 × 10^{-3}$ A. Solving these equations yields $I_{sc}$ = 16/5 mA. $R_{Th}$ has already been determined in the Thévenin analysis. Connecting the Norton equivalent to the load results in the circuit in Fig. 5.10f. Solving this circuit yields $V_{o}$ = 48/7 V.