Charge and voltage across circuit

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  • #1
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Homework Statement



Initially, the switch in the figure View Figure is in position A and capacitors C_2 and C_3 are uncharged. Then the switch is flipped to position B. Afterward, what are the charge on and the potential difference across each capacitor?

knight_Figure_30_70.jpg


Homework Equations



capacitor in series
Q1 = Q2 = Q3
Vbat = sum of Vc
Ceq = (1/C1 + 1/C2 + 1/C3)^-1

Capacitors in parallel
V1 = V2 = V3
Q1+Q2+Q3
Ceq = C1 +C2+C3

Q = CdeltaV



The Attempt at a Solution



if the switch is moved to position B then how is there any voltage going through?
 

Answers and Replies

  • #2
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sorry i forgot to mention

the question ask to find Q1, Q2 , Q3 and V1, V2 , V3

what i am thinking

this is my understanding so far
1) while the switch is in position A.........C1 would be charged to Q = CV = 15(100) = 1500 microC

2) once the switch is in position B..........the charge on C1 is distributed to C2 and C3 ......but there is no potential going back to the battery........
so.....the capacitors would be in series..........then Q1 = Q2 = Q3 = 1500uC

3)to find the voltage.........find the Ceq in series......[1/15 + 1/30 + 1/20]^ -1 = 6.667
use Ceq and Q to find Vc
Vc = 225

i don't think i am right
 
  • #3
208
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if c1 had 1500microC to begin with, how can each capacitor end up with 1500microC?
 
  • #4
19
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then how would you find the charge across each capacitor.......?
 
  • #5
2
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im stuck on the same question and got the Q1 wrong. mastering physics gave me the answer for q1 as 833uC. no idea how they got that. i was thinking along the same lines as you, that the c1 was fully charged 1500 uC.
 
  • #6
19
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ya i have no idea how mastering physics got Q1.......but i eventually got Q2/3..........
i used Ceq (taking all the circuits in series).........
weird question
 
  • #7
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yeah i got a friend to help me. We must be in the same class, do u go to york?
 
  • #8
1
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When the switch is flipped to position B, C1 and C2+C3 are in parallel. One thing to note is that when the switch is flipped, the original Q1 (calculated from position A is different). We will name it Q1' in position B.

The trick here is knowing that Q1' + Q_2+3 = Q1. Remember, we already know Q1. We will call this equation A.

It is also fair to say that in a parallel circuit V1 = V_2+3.. which would imply that Q1'/C1 = Q_2+3/C_2+3. We will call this equation B.

Somehow, you have to combine equation A and B and use algebra to eliminate terms. Unfortunatly, I suck at algebra. T

This is where I need help.
 
  • #9
1
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One year later but same problem...

This problem doesn't make any sense to me. As I understand the problem, C1 is being charged by the batter. Then the Battery gets disconnected and the charged capacitor charges Capacitors 2 and 3.

To me that looks like C1 becomes the new Power source with 55.5 V (which I obtained by taking the 833uC for C1 found by Prema) C2 and 3, are in series with this new Power source therefore V1=V2+V3. C2 should have 22.2V and C3 should be 33.3V (wrong though)

About the Q2 and Q3 I have no idea how to even start?

Any ideas guys?
 
  • #10
152
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...C2 and 3, are in series with this new Power source...
C2 and C3 are in series.
The combination C2 C3 is in parallel with C1.
 
  • #11
2
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C2 and C3 are in series.
The combination C2 C3 is in parallel with C1.
Actually, once the switch has been flipped, C1, C2, and C3 are all in series. Capacitors in series each share the same charge. Use C=Q/V and you're done!
 

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