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 18.18. (a) How long does it take the charge on the capacitor to drop to one-fourth its initial value? Answer as a multiple of \tau. (b) Compute the initial charge and time constant, and (c) the time it takes to discharge all but the last quantum of charge, 1.60 \times 10^{-19} \mathrm{C}, if the initial potential differ ence across the capacitor is 12.0 \mathrm{~V}, the capacitance is equal to 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 few algebraic manipulations involving the natural logarithm. In part (a) set q=\frac{1}{4} Q in Equation 18.9
q = Qe^{-t/RC} [18.9]
for a discharging capacitor, where Q is the initial charge, and solve for time t. In part (b) substitute into Equations 16.8
C \equiv \frac{Q}{\Delta V} [16.8]
and 18.8
\tau = RC [18.8]
to find the initial capacitor charge and time constant, respectively. In part (c) substitute the results of part (b) and the final charge q=1.60 \times 10^{-19} \mathrm{C} into the discharging-capacitor equation, again solving for time.
(a) How long does it take the charge on the capacitor to reduce 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} \quad \rightarrow \quad \frac{1}{4}=e^{t / R C}
Take natural logarithms of both sides and solve for the time t :
\begin{aligned}&\ln \left(\begin{array}{l}1 \\ \hline 4\end{array}\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:
C=\frac{Q}{\Delta V} \quad \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}
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, dividing both sides by Q:
q(t)=Q e^{-t / \tau} \quad \rightarrow \quad e^{-t / \tau}=\frac{q}{Q}
Take the natural logs of both sides and solve for t :
-t / \tau=\ln \left(\frac{q}{Q}\right) \quad \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, resis tances, 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/RC}) [18.7]
and 18.9, which indicate an infinite time would be required.