Charged Capacitors - Homework Equations & Attempt at Solution

AI Thread Summary
The discussion revolves around a homework problem involving two charged capacitors connected through a resistor. The initial voltage difference across the capacitors is calculated as 10V. Participants debate the correct approach to derive the total voltage across the resistor and the individual voltages across each capacitor over time. There is a suggestion to formulate a differential equation specific to the setup to better understand the application of the voltage equations. The conversation emphasizes the importance of deriving the equations from first principles to ensure accurate application in this context.
jh001
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Homework Statement


A 10nF and 20nF capacitor is independently charged to respectively 40V and 50V. The positive sides of the two capacitors are connected, and a 1kΩ resistor is connected across the negative terminals. Find equations for total voltage VT(t) across the resistor and voltages across each capacitor V1(t) and V2(t)

Homework Equations


V(t)=V0e-t/RC

The Attempt at a Solution


It seems VT(0) = 50-40 = 10V.
Can we use VT(t)=V2(t)-V1(t), since the capacitors are connected positive to positive? In that case what value for R do we use for each capacitor?
 
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There may be some law you can appeal to, but it isn't hard to work out from first principles. (I.e., obtain a differential equation for the current.)
 
"work out from first principles" is a bit cryptic/generic and not really helpful. The equation I gave IS in effect a differential equation (well, the solution to one) Can you please elaborate?
 
jh001 said:
"work out from first principles" is a bit cryptic/generic and not really helpful. The equation I gave IS in effect a differential equation (well, the solution to one) Can you please elaborate?
Yes, it's the solution to a differential equation, but you don't know how to apply it because the equation was for a simpler set-up. I'm proposing writing out the differential equation for the set-up in the question and solving that in the same way. You can then be confident in how to apply it here. (It turns out to be just what you might have guessed.)
 
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