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Charging the Capacitor

  1. Dec 5, 2004 #1
    So I have done most of this problem, but can't seem to find the final charge on that capacitor. It says that there was a switch, even though it is not in the image, and that at t=0 the switch is closed, and that the capictor had no charge initially. I solved for and know that I1 = 4.2 A, I2 = 1.4 A, and I3 = 2.8 A, where 1, 2, and 3 refer to the resistors with the same numbers. But I can't solve for the final charge on the capacitor! I know that Qf = EC, and the capacitance is 4 * 10^6, and that E = 42V, but (4*10^6)(42) wasn't right. So then I solved for the Voltage that goes through resistor 3 and got that it was 8.2V, but (4*10^6)(8.2) wasn't right either. I don't understand what the proper answer is, or how to obtain it. Your help is appreciated; thanks!

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  2. jcsd
  3. Dec 5, 2004 #2


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    If you're asking for the "final charge" on the capacitor then there should be no current flowing through the third resistor.
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