# KVL Question

Discussion in 'Homework Help' started by hacker804, Mar 3, 2015.

1. ### hacker804 Thread Starter New Member

Mar 3, 2015
6
0
Hi there.I am new to this forum and recently started a linear circuit analysis course.I have a question regarding Kirchhoff's Voltage Law.Consider the circuit below:

I need to find the current in this circuit.I can apply KCL and write the loop equation as:
-120+V30+2Va-Va=0
-120+V30+Va=0
Now I know that V30=30I and Va=15I
-120+30I+15I=0
45I=120
I=2.7A

However the answer in the textbook comes out to be 8A.What did I do wrong?
Can anyone kindly help?.Thanks

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Va is -15I

hacker804 likes this.
3. ### Marco foschi New Member

Mar 4, 2015
4
0
-120+30i+2(-15i)-15i=0
so based on my calculation I have i= - 8

Is it correct?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783

Go back to your very first equation. The very last term on the RHS is -Va. If Va is -15*I then -Va=+15*I.

5. ### james hudson New Member

Mar 4, 2015
1
0
i am stuck as
well on this.

6. ### hacker804 Thread Starter New Member

Mar 3, 2015
6
0
@t n k can you please tell how Va=-15I ?

7. ### hacker804 Thread Starter New Member

Mar 3, 2015
6
0
Can you please tell how Va=-15I ?

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
1,116
The reason is exactly the same as in the case of -120V. Do you understand why you use -120V

9. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Here's a novel idea -- label your diagram and start from basics. You've shown enough of an attempt for me to justify showing you how to do it explicitly because I think that will help you the most.

KVL says that the sum of the voltage drops around the circuit is equal to zero. The voltage drop from point 'a' to point 'b' is denoted by Vab which equals (Va-Vb). With that in mind, we have

Vag + Vba + Vcb + Vgc = 0

We also have the following:
Vag = 120V
Vab = I·30Ω = -Vba
Vbc = 2VA = -Vcb
VA = Vgc = -Vcg = -I·15Ω

Plugging these in we get

Vag - Vab - Vbc + VA = 0
Vag - Vab - 2VA + VA = 0
Vag - Vab - VA = 0
120V - I·30Ω + I·15Ω = 0
120V - I·15Ω = 0

I = 120V/15Ω = 8A

Now let's check our work.
Vab = (8A)(30Ω) = 240V
VA = -(8A)(15Ω) = -120V

If we set Vg = 0V, then
Va = 120V
Vb = Va - Vab = 120V - 240V = -120V
Vc = -VA = 120V

That means that the voltage of the dependent source is

Vbc = (-120V) - (120V) = -240V
Vbc = 2VA = 2(-120V) = -240V

These agree, so our answer is correct.

Doing a check is important. I actually made a mistake in one of my equations and originally got an answer of I=2.67A and only caught it because partway through the check it became apparent that I was going to get 0V across the dependent source, which would have required that I=0.

Notice three key things:

1) I tracked my units throughout my work.