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## Q. 5.6

Let us use Thévenin’s and Norton’s theorems to find $V_{o}$ in the network in Example 5.3.

## Verified Solution

The circuit is redrawn in Fig. 5.8a. To determine the Thévenin equivalent, we break the network at the 6-kΩ load as shown in Fig. 5.8b. KVL indicates that the open-circuit voltage, $V_{oc}$′, is equal to 3 V plus $V^{\prime}_{1}$, the voltage which is the voltage across the current source. The 2 mA from the current source flows through the two resistors (where else could it possibly go!) and, therefore,
$V_{1} = (2 × 10^{-3})$(1k + 2k) = 6 V. Therefore, $V_{oc}$ = 9 V. By making both sources zero, we can find the Thévenin equivalent resistance, $R_{Th}$′, using the circuit in Fig. 5.8c. Obviously, $R_{Th}$ = 3 k. Now our Thévenin equivalent circuit, consisting of $V_{oc}$ and $R_{Th}$′, is connected back to the original terminals of the load, as shown in Fig. 5.8d. Using a simple voltage divider, we find that $V_{o}$ = 6 V.

To determine the Norton equivalent circuit at the terminals of the load, we must find the short-circuit current as shown in Fig. 5.8e. Note that the short circuit causes the 3-V source to be directly across (i.e., in parallel with) the resistors and the current source. Therefore, $I_{1}$ = 3/(1k + 2k) = 1 mA. Then, using KCL, $I_{sc}$ = 3 mA. We have already determined $R_{Th}$, and, therefore, connecting the Norton equivalent to the load results in the circuit in Fig. 5.8f. Hence, $V_{o}$ is equal to the source current multiplied by the parallel resistor combination, which is 6 V.