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
No your sign is incorrect. 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.
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. 2) I asked if the answer made sense. 3) I checked my work to verify its validity.