Question 18.7: Discharging a Capacitor in an RC Circuit Goal Calculate some...
Discharging a Capacitor in an RC Circuit
Goal Calculate some elementary properties of a discharging capacitor in an R C circuit.
Problem Consider a capacitor C being discharged through a resistor R as in Figure 18.17a, page 606. (a) How long does it take for the charge on the capacitor to drop to one-fourth of its initial value? (b) Compute the initial charge and time constant. (c) How long does it take to discharge all but the last quantum of charge, 1.60 \times 10^{-19} \mathrm{C}, if the initial potential difference across the capacitor is 12.0 \mathrm{~V}, the capacitance is 3.50 \times 10^{-6} \mathrm{~F}, and the resistance is 2.00 \Omega ? (Assume an exponential decrease during the entire discharge process.)
Strategy This problem requires substituting given values into various equations, as well as a couple algebraic manipulations involving the natural logarithm. In part (a), set q=\frac{1}{4} Q in Equation 18.9
q=Q e^{-t/ R C} (18.9)
for a discharging capacitor, where Q is the initial charge, and solve for time t. For part (b), substitute into Equations 16.8
C\equiv{\frac{Q}{\Delta V}} (16.8)
and 18.8
\tau=R C (18.8)
to find the initial capacitor charge and time constant, respectively. In part (c), substitute the results of part (b) and q=1.60 \times 10^{-19} \mathrm{C} into the discharging-capacitor equation, again solving for time.
(a) How long does it take for the capacitor to drain to one-fourth its initial value?
Apply Equation 18.9:
q(t)=Q e^{-t / R C}
Substitute q(t)=Q / 4 into the preceding equation and cancel Q :
\frac{1}{4} Q=Q e^{-t / R C} \rightarrow \frac{1}{4}=e^{-t / R C}
Take natural logarithms of both sides and solve for the time t :
\begin{aligned} & \ln \left(\frac{1}{4}\right)=-t / R C \\ & t=-R C \ln \left(\frac{1}{4}\right)=1.39 R C=1.39 \tau \end{aligned}
(b) Compute the initial charge and time constant from the given data.
Use the capacitance equation to find the initial charge:
\begin{gathered} C=\frac{Q}{\Delta V} \rightarrow \quad Q=C \Delta V=\left(3.50 \times 10^{-6} \mathrm{~F}\right)(12.0 \mathrm{~V}) \\ Q=4.20 \times 10^{-5} \mathrm{C} \end{gathered}
Now calculate the time constant:
\tau=R C=(2.00 \Omega)\left(3.50 \times 10^{-6} \mathrm{~F}\right)=7.00 \times 10^{-6} \mathrm{~s}
(c) How long does it take to drain all but the last quantum of charge?
Apply Equation 18.9 , divide by Q, and take natural logarithms of both sides:
q(t)=Q e^{-t / \tau} \rightarrow e^{-t / \tau}=\frac{q}{Q}
Take the natural logs of both sides:
-t / \tau=\ln \left(\frac{q}{Q}\right) \rightarrow \quad t=-\tau \ln \left(\frac{q}{Q}\right)
Substitute q=1.60 \times 10^{-19} \mathrm{C} and the values for Q and \tau found in part (b):
\begin{aligned} t & =-\left(7.00 \times 10^{-6} \mathrm{~s}\right) \ln \left(\frac{1.60 \times 10^{-19} \mathrm{C}}{4.20 \times 10^{-5} \mathrm{C}}\right) \\ & =2.32 \times 10^{-4} \mathrm{~s} \end{aligned}
Remarks Part (a) shows how useful information can often be obtained even when no details concerning capacitances, resistances, or voltages are known. Part (c) demonstrates that capacitors can be rapidly discharged (or conversely, charged), despite the mathematical form of Equations 18.7
q=\,Q(1-\,e^{-t/R C}) (18.7)
and 18.9, which indicate an infinite time would be required.