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KaiserBrandon
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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:
[itex]V = \frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}}[/itex]
And since V = IR, we have:
[itex] I = \frac{1}{R}(\frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}})[/itex]
and the capacitors must each lose the same amount of charge per unit time:
[itex]\frac{dQ_{1}(t)}{dt} = \frac{dQ_{2}(t)}{dt} = -I [/itex]
the current is negative since the charges are decreasing. So, we have:
[itex]\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}})[/itex]
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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