Two capacitors (different capacitance and innitial charge) in a series circuit

AI Thread Summary
The discussion revolves around analyzing a series circuit with two capacitors (C1 and C2) and a resistor (R), focusing on the current flow and energy dissipation under different configurations of the capacitors. In the first experiment, the positive plate of C2 is connected to the negative plate of C1, leading to a derived expression for the current I(t) based on the charges on the capacitors and the resistor's relationship. The second experiment alters the connections, requiring a new expression for I(t) while maintaining the initial charges. Participants express difficulty in evaluating the equations for Q1 and Q2, particularly in determining the initial current and time constant of the circuit. The conversation emphasizes the need to understand the equivalent capacitance and how it affects the circuit's behavior over time.
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Homework Statement


Two capacitors C1 and C2 and a resistor R are all connected in series. At t=0, a
charge Q1 resides on C1 and a charge Q2 on C2 (Q1 > Q2; C1 < C2), and the positive plate
on C2 is connected to the negative plate on C1.

a) In a first experiment, the positive plate on C2 is connected to the negative plate on
C1. Derive an expression for the time dependence of the current I(t) which flows
in the circuit. Indicate the direction of this current on a circuit diagram, and
include the sign of the charges on the plates on each of the capacitors at t=0.

b) The same circuit is used for a second experiment, with the same initial charges,
except that in this case the negative plate of C2 is connected to the negative plate
on C1. All three components are otherwise still connected in series as before.
Derive an expression for I(t), and indicate the direction of the current on a second
diagram (again indicate the sign of initial charge on each capacitor on this
diagram).

c) Hence calculate the energy dissipated in the resistor in each case.

Homework Equations


V = IR

V=Q/C for a capacitor

The Attempt at a Solution



ok, I'm stuck on the first part. firstly, the voltage gain across the two capacitors is:

V = \frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}}

And since V = IR, we have:

I = \frac{1}{R}(\frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}})

and the capacitors must each lose the same amount of charge per unit time:

\frac{dQ_{1}(t)}{dt} = \frac{dQ_{2}(t)}{dt} = -I

the current is negative since the charges are decreasing. So, we have:

\frac{dQ_{1}(t)}{dt} = \frac{dQ_{2}(t)}{dt} = \frac{-1}{R}(\frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{c_{2}})

And this is where I'm stuck. I'm not entirely sure how to go about evaluating this for Q1 and Q2.
 
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You have the initial potential that the resistor will see: the sum of the two capacitor voltages as you've shown. What initial current must flow then?

The circuit consists of a resistor and two capacitors in series. What's the equivalent capacitance? How about the circuit's time constant?

What can you do with an initial current and the time constant?
 
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