Find V1 and V2 in the circuit in Fig. P3.5 using nodal analysis.
Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.
Solution:
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We will label two nodes to perform nodal analysis on this circuit as follows:
Let us first look at node V_1, the orange node.
We can write a KCL expression like so (note that k=10^3 and m=10^{-3}):
(remember that while I_1, I_2 and the 6mA leave the node, 4mA is coming into the node)
We can express these currents in terms of voltage and resistance using (I=V/R).
\dfrac{V_1}{6k}+\dfrac{V_1-V_2}{4k}+6m=4m\,\,\,\color{orange} {\text{(eq.1)}}
Let us now look at node V_2, the purple node and write a KCL expression.
Again, expressing these currents in terms of voltage and resistance, we get:
\dfrac{V_2-V_1}{4k}+\dfrac{V_2}{6k}+\dfrac{V_2}{3k}=6m\,\,\,\color{purple} {\text{(eq.2)}}
Solving equations 1 and 2 simultaneously gives us (see full steps):
V_2=8 v
Final Answer:
V_2=8 v