
#1
Sep711, 03:11 AM

P: 22

1. Find the power delivered by the 5V Voltage source in the network of Fig. This is my attept at solution: At node 1 V1...assuming current leaving V1: KCL equation: 1A+1A+[(V15)/3]+[(V1(V210))/2]=0 At node 2 V2...assuming current leaving V2: KCL equation: 2A1A+[V2(V1+10)]=0 But, I'm not certain about the sign convention when there is a voltage source with a resistor between V1 and V2.... I'm also confused that how to proceed when there is a current source and resistor between two nodes..... Where did I go wrong...? Please help me.... 



#2
Sep711, 06:07 AM

Mentor
P: 11,443

If you look at the node labelled 2, you'll note that three branches come together and two of them have specified currents. What does KCL tell you about the current in the remaining branch? Can you now repeat this feat at node 1?




#3
Sep711, 10:01 AM

P: 22

Yes, there are four branches coming together at node 1.Currents are specified for two of the branches. At node 1 KCL: 1+1+[(v1(5))/3]+[v1(v2+10)]/2=0 Did I write it correctly? Thanks... 



#4
Sep711, 10:36 AM

Mentor
P: 11,443

Problem with Node Voltage Methodflowing through the node, the last one is thus determined y the others. So what must i_{2} be? 



#5
Sep711, 10:43 AM

P: 22

I think i2=(v1+5)/2, am I correct?




#6
Sep711, 10:58 AM

Mentor
P: 11,443

Look again at node 1. You've determined what i_{1} must be, coming from node 2, right? So all branches leading to node 1 have known currents except for the branch with i_{2}. So what's i_{2}?




#7
Sep711, 11:08 AM

P: 22

As 2A and 1A are reaching node 2....3A must leave it..
So, i1=3A, am I correct..? and wow i2=3A...I got it now ....I think I'm correct...amn't I? Thank you very very much... 



#8
Sep711, 11:26 AM

Mentor
P: 11,443

Now you can figure out the power delivered by the 5V supply. 


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