anonymous
  • anonymous
C_2 and C_3 are uncharged. after the switch is closed, what are the charges on each capacitor? what are the potential difference across each capacitor?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1332439168812:dw|
anonymous
  • anonymous
so before switch is closed, Q=15uF*100V=1500uC found a similar problem online that says to divide the V in half to find individual charges and V after switch is closed. so Q_1=(15uF)100V/2=750uC Q_2+3=1500uC-750uC=750uC Q_2 should equal Q_3 so 750uC/2=375uC V_1 should equal V_2+3: 50V V_2+3=(Q_2+3)/C_2+(Q_2+3)/C_3=(750/20)+(750/30)=62.5V...not equal to 50V
anonymous
  • anonymous
BOth branches are in parallel, so voltage is the same. 100V=Voltage across Cap1=Voltage across Cap2+Voltage across Cap3 For Cap1: Q1=C1V1=(15E-6)*(1E2)=15E-4 C = 150 mC Vcap1=Vcap2+Vcap3 Q1/C1=Q2/C2+Q3/C3 Charge conservation: Q1=Q2+Q3 Then we have two equations with two unknowns easily solvable. I hope it helps.

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anonymous
  • anonymous
for Q1^^^, what is 15E-6?
anonymous
  • anonymous
15E-6 F =15 microfarads. its scientific notation. I should have specify it . Sorry.
anonymous
  • anonymous
o ok
anonymous
  • anonymous
Q1 can also =2*Q2, correct? or 2*Q3
anonymous
  • anonymous
Not, Q1=Q2+Q3 (this is charge conservation), given that we already know the value of Q1, you should use this value in the second equation: Q1/C1=150mC/15uF =100=Q2/(10E-6) + Q3/(30E-6); the other equation would be 150E-3 = Q2+Q3. You can use determinants for solving or you can also apply any method for resolution of equations.
anonymous
  • anonymous
or, i meant Q2=Q3, the cap are in series but when i solve for them, they don't equal each other. it specifically says in the text cap in series have same charge
anonymous
  • anonymous
Yes they both are not equal. before the switch closes, Cap1 was in parallel with the source of 100V. this charged Cap1 with 150mC; later, when switch closes and connected in parallel Cap1 with Cap2&Cap3, the charge must divide in proportional parts inside each cap (2&3), given that cap2&3 have different values of C (15uF & 30uF) they have different "capacity" for hold charge.
anonymous
  • anonymous
ok, so cap in series have the same charge when the charge doesn't come from one single charged cap?
anonymous
  • anonymous
Two Caps connected in series have the same charge when they have the same value of Capacitance. A capacitor cant hold more charge than its nominal value of capacitance
anonymous
  • anonymous
hmm not so according to the text and other problems i've been doing. regardless of values of C1 and C2, in series, Q should be same. here's the text "two capacitors in series have equal +-Q. potential difference in series is added though, and should equal in parallel if you're thinking of that maybe?
anonymous
  • anonymous
Mmm, you made me doubt. Does this excercise comes from a book?, does it have an answer at the end of the book?
anonymous
  • anonymous
Oh, you were right. Q2=Q3. So sorry. but this simplifies things. http://farside.ph.utexas.edu/teaching/302l/lectures/node46.html
anonymous
  • anonymous
yes, Q1=.83, Q2=Q3=.67, V1=55, V2=34, V3=22. so capacitance isn't how much charge it can hold, depends on potential difference too.
anonymous
  • anonymous
i've been googling and searching and the steps are laid out in cramster -_-. thx for helping tho
anonymous
  • anonymous
Are you sure about V1=55. Cap1 was connected in parallel with 100Vsource, so it should have the same value of V.
anonymous
  • anonymous
What is the book this excercise comes from?
anonymous
  • anonymous
physics for scientists and engineers second edition -randall d. knight. idk, but another source said initial V was suppose to divide in half
anonymous
  • anonymous
http://www.cramster.com/solution/solution/1212480
anonymous
  • anonymous
I'm checking your link. ..please wait
anonymous
  • anonymous
Yes, I re-read the theory. initial charge must be conservated in the capacitors after the switch disconnects the source from Cap1. Review the attached jpg. Best Regards.
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