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Chapter 5

Q. 5.9

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

Step-by-Step

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}