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elica85
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?
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
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.
for Q1^^^, what is 15E-6?
15E-6 F =15 microfarads. its scientific notation. I should have specify it . Sorry.
Q1 can also =2*Q2, correct? or 2*Q3
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.
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
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.
ok, so cap in series have the same charge when the charge doesn't come from one single charged cap?
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
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?
Mmm, you made me doubt. Does this excercise comes from a book?, does it have an answer at the end of the book?
Oh, you were right. Q2=Q3. So sorry. but this simplifies things. http://farside.ph.utexas.edu/teaching/302l/lectures/node46.html
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.
i've been googling and searching and the steps are laid out in cramster -_-. thx for helping tho
Are you sure about V1=55. Cap1 was connected in parallel with 100Vsource, so it should have the same value of V.
What is the book this excercise comes from?
physics for scientists and engineers second edition -randall d. knight. idk, but another source said initial V was suppose to divide in half
I'm checking your link. ..please wait
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.