Potential difference Capacitance problem

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Homework Statement


http://img509.imageshack.us/img509/8805/problem5fw8.th.png [Broken]
The potential difference V = 100 V is applied to the capacitor arrangement shown in the figure. Here C1 = 10 microF, C2 = 5 microF, and C3 = 4microF. If capacitor C3 undergoes electrical breakdown (i.e. becomes equivalent to a conducting wire), what is the increase in (a) the charge on capacitor 1, and (b) the potential difference across capacitor 1?

The Attempt at a Solution


I wasn't sure how to solve it so I started doing different things:
I first found the equivalent capacitance, and with that I found the total charge to be Q_tot = 3.16 * 10^-4 C.
Then, I thought since the equivalent of C1 and C2, and C3 are in series, Q1 + Q2 = Q3, so Q_tot = 2 * Q3.
Then, from Q3 and C3, I find the potential difference across of C3 to be V3 = 39.5V, which in turn makes V1 = V2 = 60.5V.
From V1 and V2, I found the initial charges Q1 = 605 microC, and Q2 = 302.5 microC.
I'm not sure where to go from here...

I tried also to find the final potential difference across C1:
C12 = C1 + C2 = Q_tot / V_new
V_new = 21V, but it's supposed to be 79V... what am I doing wrong here?

Since Q = CV, if I plug in V = 79V, then Q1 = 790 microC, which is the answer for part (a), right? So should I try to solve part (b) first?

Any help would be greatly appreciated!

P.S. sorry for not using tex...
 
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Answers and Replies

  • #2
1,707
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you've made a mistake

Q1 + Q2 = Q3, so Q_tot = 2 * Q3.

this is the faulty assumption

it should be
Q1 + Q2 = Q3 = Q_tot
 
  • #3
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But if Q1 + Q2 = Q3, and Q_tot = Q1 + Q2 + Q3, then Q_tot = Q3 + Q3 = 2*Q3 ... right?
 
  • #4
1,707
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But if Q1 + Q2 = Q3, and Q_tot = Q1 + Q2 + Q3, then Q_tot = Q3 + Q3 = 2*Q3 ... right?
you just restated what i already said was wrong

Q_tot = Q1 + Q2 = Q3
 

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