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## Q. 5.7

Let us use Thévenin’s theorem to find $V_{o}$ in the network in Fig. 5.9a.

## Verified Solution

If we break the network to the left of the current source, the open-circuit voltage $V_{oc_{1}}$ is as shown in Fig. 5.9b. Since there is no current in the 2-kΩ resistor and therefore no voltage across it, $V_{oc_{1}}$ is equal to the voltage across the 6-kΩ resistor, which can be determined by voltage division as

$V_{oc_{1}} = 12 \huge( \frac{6k}{6k + 3k} \huge)$= 8 V

The Thévenin equivalent resistance, $R_{Th_{1}}$′, is found from Fig. 5.9c as

$R_{Th_{1}} = 2k + \frac{(3k)(6k)}{3k + 6k}$ = 4 kΩ

Connecting this Thévenin equivalent back to the original network produces the circuit shown in Fig. 5.9d. We can now apply Thévenin’s theorem again, and this time we break the network to the right of the current source as shown in Fig. 5.9e. In this case $V_{oc_{2}}$ is

$V_{oc_{2}} = (2 × 10^{-3}$)(4k) + 8 = 16 V

and $R_{Th_{2}}$ obtained from Fig. 5.9f is 4 kΩ. Connecting this Thévenin equivalent to the remainder of the network produces the circuit shown in Fig. 5.9g. Simple voltage division applied to this final network yields $V_{o}$ = 8 V. Norton’s theorem can be applied in a similar manner to solve this network; however, we save that solution as an exercise.