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## Q. 5.9

We wish to determine the Thévenin equivalent of the network in Fig. 5.11a at the terminals A-B. ## Verified Solution

Our approach to this problem will be to apply a 1-V source at the terminals as shown in Fig. 5.11b and then compute the current $I_{o}$ and $R_{Th} = 1/I_{o}.$
The equations for the network in Fig. 5.11b are as follows. KVL around the outer loop specifies that

$V_{1} + V_{x}$ = 1

The KCL equation at the node labeled $V_{1}$ is

$\frac{V_{1}}{1k} + \frac{V_{1} – 2V_{x}}{2k} + \frac{V_{1} – 1}{1k}$ = 0

Solving the equations for $V_{x}$ yields $V_{x}$ = 3/7 V. Knowing $V_{x}$′, we can compute the currents $I_{1}$′, $I_{2}$′, and $I_{3}$′. Their values are

$I_{1} = \frac{V_{x}}{1k} = \frac{3}{7}$ mA
$I_{2} = \frac{1 – 2V_{x}}{1k} = \frac{1}{7}$ mA
$I_{3} = \frac{1}{2k} =\frac{1}{2}$ mA

Therefore,

$I_{o} = I_{1} + I_{2} + I_{3}$
= $\frac{15}{14}$ mA

and

$R_{Th} = \frac{1}{I_{o}}$

= $\frac{14}{15}$