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

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SUMMARY

The discussion focuses on analyzing a series circuit consisting of two capacitors, C1 and C2, and a resistor R, with different initial charges (Q1 > Q2). The first experiment connects the positive plate of C2 to the negative plate of C1, leading to a derived expression for the current I(t) in the circuit. The second experiment connects the negative plate of C2 to the negative plate of C1, requiring a separate derivation for I(t). The energy dissipated in the resistor for both configurations is also calculated, emphasizing the importance of understanding capacitor behavior in series circuits.

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  • Understanding of capacitor behavior in series circuits
  • Knowledge of Kirchhoff's voltage law
  • Familiarity with differential equations
  • Basic concepts of electrical energy dissipation
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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.
 
Last edited:
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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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