• Support PF! Buy your school textbooks, materials and every day products Here!

Current Through a Capacitor in a RC Circuit

  • Engineering
  • Thread starter Drakkith
  • Start date
  • #1
Drakkith
Staff Emeritus
Science Advisor
20,702
4,384

Homework Statement


Find ##i(0^+) ## and ##i(t), t≥0^+##

pr_7-51.jpg

Homework Equations




The Attempt at a Solution


[/B]
I'm having trouble finding the current through this capacitor. I tried using a KVL loop for ##t=0^+##, but I'm doing it incorrectly or something.

Earlier in the problem I found: ##v_c(0^+) = -120 V##
Sign convention for resistors: Positive on the top for the 150k resistor, positive on the right side for the other 2.

KVL1: Starting at node ##b## and moving clockwise with current ##i_1## :
##-2.5ki_1+150k(i_1-i_2)-v_c=0##
##-2.5ki_1+150ki_1-150ki_2-(-120)=0##
##147.5ki_1-150ki_2=-120##

KVL2: Starting just to the left of the 50k resistor and moving clockwise:
##-50ki_2+200-150k(i_2-i_1)=0##
##-50ki_2-150ki_2+150ki_1=-200##
##150ki_1-200ki_2=-200##

Solving, I get:
##i_1=0.85 mA##
##i_2=1.6mA##

##i=-i_1=-0.85 mA##

However, this is appears to be incorrect.
 

Attachments

Answers and Replies

  • #2
berkeman
Mentor
56,439
6,358
Earlier in the problem I found: vc(0+)=−120V
Looks correct.
I tried using a KVL loop
For me, the KVL equations are generally less intuitive and move complicated. Not to make extra work for you, but can you write the KCL equation for the main node in the circuit for t>0 and see if that makes it easier to solve?
 
  • #3
Drakkith
Staff Emeritus
Science Advisor
20,702
4,384
For me, the KVL equations are generally less intuitive and move complicated. Not to make extra work for you, but can you write the KCL equation for the main node in the circuit for t>0 and see if that makes it easier to solve?
Sure. From the main node, ##i## goes left, ##i_1## goes down, and ##i_2## goes right.
##i+i_1+i_2=0##
##\frac{v-v_c}{2.5}+\frac{v}{150k}+\frac{v-200}{50k}=0##
Multiplying both sides by 150k:
##60v-60v_c+v+3v-600=0##
##64v-60(-120)=600##
##64v+7200=600##
##64v=-6600##
##v=-103.125 V##

##i=\frac{v-v_c}{2500}=6.75 mA##
Which is correct...

##i(t)## is then: ##i(t)=6.75e^{(1000t)} mA##

Why in the world did one method work but not the other?
 
  • #4
berkeman
Mentor
56,439
6,358
Which is correct...
Yahoo! :smile:
Why in the world did one method work but not the other?
Both methods should work, but at least for me, the KVL equations are much less intuitive and therefore much easier to make a mistake when using them. I'll leave it to other (smarter) members (EDIT -- like vela) to find the small typo in your work with the KVL equations... :smile:
 
Last edited:
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149
Looks like the signs aren't consistent in your KVL equations. If you're going clockwise around the left loop, for example, you should get
$$-2500i_1 + 150000(i_2-i_1) + v_c = 0.$$ In the second KVL equation, the sign on 200 should be negative since you're moving from higher potential to lower potential as you go clockwise around the loop.
 
  • #6
Drakkith
Staff Emeritus
Science Advisor
20,702
4,384
Looks like the signs aren't consistent in your KVL equations. If you're going clockwise around the left loop, for example, you should get
$$-2500i_1 + 150000(i_2-i_1) + v_c = 0.$$ In the second KVL equation, the sign on 200 should be negative since you're moving from higher potential to lower potential as you go clockwise around the loop.
Isn't that already consistent though? Moving from positive to negative voltage across a resistor (moving from higher to lower potential), the sign is +, whereas it's - when moving from negative to positive (as in the ##-2500i_1##).
 
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149
Isn't what already consistent?

Which way are you assuming the currents ##i_1## and ##i_2## flow? If it's clockwise around the loops, then according to your sign convention, the first term should be ##+2500i_1##.
 
  • #8
Drakkith
Staff Emeritus
Science Advisor
20,702
4,384
Isn't what already consistent?
The way I was doing it already.

Which way are you assuming the currents ##i_1## and ##i_2## flow? If it's clockwise around the loops, then according to your sign convention, the first term should be ##+2500i_1##.
How so? The negative terminal of the 2500 ohm resistor is on the left.
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149
Potential drops as the current flows through the resistor. If the current enters on the left, the left end of the resistor is at a higher potential than the right end.
 
  • #10
Drakkith
Staff Emeritus
Science Advisor
20,702
4,384
Looking back at my homework, I can't reconcile what I did there with what I've done here or what you're telling me vela. I'll have to dig a bit deeper and maybe talk to my instructor to figure out what's going on. Thanks all.
 

Related Threads for: Current Through a Capacitor in a RC Circuit

Replies
3
Views
3K
Replies
7
Views
903
Replies
6
Views
2K
Replies
5
Views
1K
Replies
6
Views
4K
Replies
46
Views
4K
Replies
27
Views
2K
Replies
2
Views
687
Top