Capacitors/Resistors/time constant current

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SUMMARY

The discussion focuses on a circuit involving two capacitors (3 µF and 2 µF) charged by a 14.0 V battery and subsequently disconnected. The key equations utilized are V=IR and q=Q(1-e^-t/RC), where the time constant is critical for determining charge and current after 1.80 ms. The effective capacitance of the capacitors is essential for solving the problem, which requires calculating the remaining charge on each capacitor and the current through the resistor.

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Homework Statement



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The given pair of capacitors in the figure below are fully charged by a 14.0 V battery. The battery is disconnected and the circuit closed. Answer the following after the circuit has been closed for 1.80 ms.

Homework Equations



V=IR
q=Q(1-e^-t/RC)

The Attempt at a Solution



I know we have to at least use the second equation above but I just don't know how to start this problem.
 

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whoops! Forgot the include the questions in the post, haha:

a) How much charge remains on each capacitor (3 and 2 microF)?

b) What is the current in the resistor?

Thanks guys.
 
What is the effective capacitance, which you can use in place of the two capacitors?
 

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