# just starting circuits....need a little help

#### sideline81

Joined Dec 1, 2011
4
I'm just starting to learn circuits so I may just be missing something small here, but I need some help.

I'm given a single-node pair circuit and need to find iA, iB, and iC. I think whats throwing me off is that there is a dependent current source in the middle of the circuit and I'm not sure how I should handle it.

Any help would be great.
Thanks

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#### syed_husain

Joined Aug 24, 2009
61
I'm just starting to learn circuits so I may just be missing something small here, but I need some help.

I'm given a single-node pair circuit and need to find iA, iB, and iC. I think whats throwing me off is that there is a dependent current source in the middle of the circuit and I'm not sure how I should handle it.

Any help would be great.
Thanks
lets assume, the bottom node is grounded. you can make ground anywhere you like but sometimes for certain problems choosing ground makes calculation a lot easier. for your problem you have only two options either top or bottom.

i will give you a hint. if you follow the current direction you can write according Kirchoff's Current Law (KCL) at the top of the node:

Rich (BB code):
$$i_{A}+i_{B}+i_{C}+2 = 5.6$$
and
Rich (BB code):
$$i_{B} = 0.1*V_{x}$$
the rest is up to you. if you have still problem post your working.

#### sideline81

Joined Dec 1, 2011
4
That's the setup I was using and came up with 13.5V

Here is how I got 13.5V

5.6 = iA + iB + iC + 2
5.6 = V/18 + .1V + V/9 +2
100.4 = V + 1.8V + 2V + 36
64.8 = 4.8V
V = 13.5

Then to find iA i calculated 13.5V/18ohms = .75A, but that isn't right.

not sure what I'm doing wrong

#### syed_husain

Joined Aug 24, 2009
61
That's the setup I was using and came up with 13.5V

Here is how I got 13.5V

5.6 = iA + iB + iC + 2
5.6 = V/18 + .1V + V/9 +2
100.4 = V + 1.8V + 2V + 36
64.8 = 4.8V
V = 13.5

Then to find iA i calculated 13.5V/18ohms = .75A, but that isn't right.

not sure what I'm doing wrong
i am also getting the same answer as yours. are you sure about the answer? can you make sure the sum of currents in your answer (not what you calculated) is 5.6?

#### sideline81

Joined Dec 1, 2011
4
I get 1A = .75A , iB = 1.35A , and iC = 1.5A
The problem is an online assignment and when I try those answers it says they are incorrect. When I add up all the currents, they do equal 5.6, which I would normally take to mean they are correct. I don't know what else to try

#### syed_husain

Joined Aug 24, 2009
61
I get 1A = .75A , iB = 1.35A , and iC = 1.5A
The problem is an online assignment and when I try those answers it says they are incorrect. When I add up all the currents, they do equal 5.6, which I would normally take to mean they are correct. I don't know what else to try
i think i got the solution. the equation

Rich (BB code):
$$i_{A} +i_{B}+i_{c}+2=5.6$$
from this we can write following(i took the ground connection at bottom):
Rich (BB code):
$$\frac {-V_{x}}{18} +0.1*V_{x} -\frac{V_{x}}{9} +2 = 5.6$$
in previous post i did not notice the polarity sign of Vx given.
now
Rich (BB code):
$$V_{x} = - 54 V$$
,
Rich (BB code):
$$i_{A} = 3 A,i_{B}= -5.4 A, i_{C} = 6 A$$
hope this will help

cheers