Power generated by dependent source

[hogrampage]

Homework Statement

Determine the power generated by the dependent source in the circuit shown below:

Homework Equations

I_x=V/R
P_gen=V*I

The Attempt at a Solution

I made an equivalent circuit, but it may not be correct (resistors in parallel?):

I_x=-(12V/(12/7)Ω)=-7A
2I_x=2(-7a)=-14A

Passive sign convention is satisfied for the dependent source, so P_gen = -P_abs:
P_gen=-(12V)(-14A)=-168W

I am not sure I did this correctly, as this is my first time in a class like this. If I am going about this wrong, could someone point me in the right direction? I understand most of the homework in this class, but this one kind of confused me.

Thanks

 

[NascentOxygen]

I think you are straying way off course. The independent voltage source produces a current in that 3 ohm resistor. You can determine its value.

i_x has a direction opposite to what the voltage source produces, so you need to attach a negative sign to the magnitude of the current and call it i_x

continue ...

 

[hogrampage]

Yeah, my professor showed us how to do it in class today and it makes a lot more sense now. I had just learned the material and the book we have isn't very good at giving examples/explanations.

 

[crixus]

Yeah it can get tricky for a newbie but stick to basics...

First of all realize that the wire connecting the two parts of the circuit cannot provide for both forward and return path for current. So current in the middle wire has to be zero. Now analyse both the parts of the circuit individually following what nascent oxygen told in the previous post...

As far as a book is concerned you can refer to :
Engineering Circuit Analysis by Hayt
Best of luck

 

[DeeNos]

for your help!
Hello,

I can confirm that your approach is correct. You have correctly applied Ohm's law to determine the current through the dependent source, and then used the formula for power (P = V*I) to calculate the power generated by the source.

Your equivalent circuit is also correct, as the two resistors in parallel can be combined into a single equivalent resistance. This does not affect the current through the dependent source, as the voltage drop across the equivalent resistance remains the same, and thus the current through the dependent source remains unchanged.

Overall, your solution is correct and your understanding of the concepts is sound. Keep up the good work!