Question

3. The discharge curve for a capacitor connected in series with a resistor is shown below. (a) If the capacitance of the capacitor is 22.50 farads, what is the resistance ofthe resistor? (b) Please calculate the time constant for this circuit. After two time constants what is the potential across capacitor? (c) Please calculate the amount of charge on the capacitor 25.00 minutes after discharge has started. (b) Once capacitor is fully discharged, the charging process begun. How much time, after charging begins, is required for the potential difference across the capacitor to reach 16.50 volts? Please show detailed calculations for each step. (20 pts) 3000 25.00 15.00 5.00 000 000 3000 7000 Time (minutes)
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Answer #1

a)

V=V_{0}e^{\frac{-t}{RC}}\\ \\ lnV=lnV_{0}-\frac{t}{RC}\\ \\ \frac{t}{RC}=lnV-lnV_{0}\\ \\ RC=-\frac{t}{lnV-lnV_{0}}\\ \\ R=-\frac{t}{C(lnV-lnV_{0})}\\ \\ R=-\frac{60-0}{22.50(in40-in1.0)}\\ \\R=0.723\Omega

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b)

time\ constant\\ \\ \tau =RC\\ \\ \tau =0.723\times 22.50sec\\ \\ \tau =16.3sec

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v=V_{0}e\frac{-t}{2\tau }\\ \\ V=1\times e^{\frac{60}{2\times 16.3}}\\ \\ V=0.158V

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Q=cV

Q=22.5\times 0.158C\\ \\ Q=3.56C

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