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Apteronotus
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Homework Statement
I have a simple circuit consisting of a charged capacitor and a resistor connected in series to a battery.
Suppose that initially the potential across the capacitor is greater than that of the battery. (ie. [tex]\frac{q}{C}>V[/tex]).
What happens to the capacitor when the switch is closed and the circuit completed?
Homework Equations
Kirchhoff's 2nd:
[tex]V+IR-\frac{q}{C}=0[/tex]
Solving for q:
[tex]q(t)=CV+(q_0-CV)e^{(t-t_0)/RC}[/tex]
where [tex]q_0[/tex] is the initial charge on the capacitor at time [tex]t_0[/tex]
The Attempt at a Solution
I'm guessing that when the switch is closed we should expect the capacitor to discharge to a certain degree and after some time have the same potential as the battery.
But my equation [tex]q(t)=CV+(q_0-CV)e^{(t-t_0)/RC}[/tex] does not reflect this.
As [tex]t\rightarrow\infty[/tex] the charge on the capaictor [tex]q(t)\rightarrow (q_0-CV)e^{(t-t_0)/RC}\rightarrow\infty[/tex]
What am I doing wrong?