Solve the circuit using nodal analysis and find the power of  Is_1 .
Problem 1-8 - Nodal Analysis and Power Calculation

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

Nodal Analysis - Power Calculation for Current Source

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.


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3 Comments

  1. 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. 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?

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