Let us find both Vo and V1 in the circuit in Fig. 3.22.

Question

Let us find both [latex]V_{o}[/latex] and [latex]V_{1}[/latex] in the circuit in Fig. 3.22.

Accepted Answer

Although it appears that there are two unknown mesh currents, the current [latex]I_{1}[/latex] goes directly through the current source and, therefore, [latex]I_{1}[/latex] is constrained to be 2 mA. Hence, only the current [latex]I_{2}[/latex] is unknown. KVL for the rightmost mesh is

2k( [latex]I_{2} - I_{1}) - 2 + 6k I_{2}[/latex] = 0

And, of course,

[latex]I_{1} = 2 × 10^{-3}[/latex]

These equations can be written as

- 2k[latex]I_{1} + 8k I_{2}[/latex] = 2
[latex]I_{1}[/latex] = 2/k

Solving these equation for [latex]I_{2}[/latex] yields [latex]I_{2}[/latex] = 3/4kA and thus

[latex]V_{o} = 6k I_{2} =\frac{9}{2}[/latex]V

To obtain [latex]V_{1}[/latex] we apply KVL around any closed path. If we use the outer loop, the KVL equation is

-[latex]V_{1} + 4kI_{1} - 2 + 6kI_{2}[/latex] = 0

And therefore,

[latex]V_{1} =\frac{21}{2}[/latex] V

Note that since the current [latex]I_{1}[/latex] is known, the 4-kΩ resistor did not enter the equation in finding [latex]V_{o}[/latex]. However, it appears in every loop containing the current source and, thus, is used in finding [latex]V_{1}[/latex].

Question 3.14: Let us find both Vo and V1 in the circuit in Fig. 3.22.

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