(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I have a simple circuit consisting of achargedcapacitor and a resistor connected in series to a battery.

Suppose that initially the potential across the capacitor isgreaterthan 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?

2. Relevant 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]

3. 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?

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# Homework Help: RC-circuit for charged capacitor

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