# Problem 1-8: Nodal Analysis - Power of Current Source

Solve the circuit using nodal analysis and find the power of $Is_1$.

Solution
a) Choose a reference node, label the voltages:

b) Apply KCL to nodes:
Node #1:
$Is_2 + \frac{V_1 }{R_2} = 0 \rightarrow V_1 = -4 v$

Node #2:
$-Is_1+\frac{V_2}{R_1} = 0 \rightarrow V_2 = 6v$

Node #3:
$Is_1 +\frac{V_3}{R_3} = 0 \rightarrow V_3 = -4 v$.
c) Find the unknown:
$P_{Is_1}=Is_1 \times (V_3 - V_2) = -20 W$ supplying.

## 3 thoughts on “Problem 1-8: Nodal Analysis - Power of Current Source”

1. Eli says:

Can we ignore R2-Is2 loop and say that R1 and R3 are in series so R13=5ohm and the power is RI^2=5*2^2=20W?

1. Yes, it is ok to ignore the $latex R_2$ – $latex Is_2$ loop because it has only one common point with the other loop and current cannot flow between. Your solution is also correct.

2. Reuben says:

I got 20W for the answer. How do you know if it is negative or positive? I know SRS, however I am not sure of how to apply it in this problem. Where do the plus and minus go?