Engineering Solving for Unknown Current in RC Circuit: Understanding Capacitor Behavior

[dwn]

Homework Statement

Image

The Attempt at a Solution

I know how to solve for a and c.

a: 300V / (300Ω) = 1 A
c: 120V / (200Ω) = 0.6 A

what I don't understand is how they solved for B (0.6A), considering the current is 3A provided by the capacitor.

I initially tried to use a current divider, but that didn't give me the correct result.
Since the capacitor acts as a voltage source, I considered trying to find the voltage of the capacitor, v = Ve^(-t/RC), but the value of C is not given.

 

[donpacino]

I cannot view the image. Try re-uploading it

 

[dwn]

Hopefully it will work now.

 

[donpacino]

Do you know what the 'u' means in the equations?

u is the identifier for unit step function.

if I have an equation X=u(t), that mean X=0 when t<0, and 1 when t>=0

so if iC=3u(-t). iC=3u(-t) when t<=0, and iC=0 when t>0

 

[dwn]

Yes, I was aware of that. That's how I solved the for a and c. For part b we are to solve for -0.5 second, which means that neither voltage sources are active, and the only source comes from the capacitor (3 amps). Correct?

 

[donpacino]

so by 'a' you mean you're solving for i1 at t=-1.5?
if that is the case, then your method of solving the problem is incorrect, and you happened to get the correct answer.

at t=-1.5:
Va=0
Vb=0
iC=3
you use current division to find i1
i1=iC*100/(100+200)=1A

same with C:
at t=1.5
vA=300
vB=-120
iC=0

write a kvl
vA-100*i1+vB-200*i1=0
i1=0.6 A

at t=-0.5:
vA=300u(t-1) since t-1=-1.5, vA=0
vB=120(t+1) since t+1=0.5, vB=-120
iC=3u(-t) since -t=0.5, iC=3

...

 

[dwn]

Oh wow, I think I was jumping to conclusions and just got lucky. I see the mistake now, thank you very much! Thought I was doing right thing since the answer just happened to work out that way. Wish the textbook wouldn't do that, set's one off in the wrong direction! haha

 

[donpacino]

you're welcome. Mistakes will always happen. The important thing is to rectify them!