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 RC circuit.
PROBLEM Consider a capacitor C being discharged through a resistor R as in Active Figure 18.17a (page 630). (a) How long does it take the charge on the capacitor to drop to one-fourth its initial value? Answer as a multiple of τ. (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 × 10^{-19} C, if the initial potential difference across the capacitor is 12.0 V, the capacitance is equal to 3.50 × 10^{-6} F, and the resistance is 2.00 Ω? (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 = \textstyle{\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. In part (b) substitute into Equations 16.8
C\equiv\ {\frac{Q}{Δ\,{V}}} [16.8]
and 18.8
τ = 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 × 10^{-19} 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
{\textstyle\frac{1}{4}}Q=\ Q e^{-t/R C}\,\,\,\rightarrow\,\,\,\,\,\,{\textstyle\frac{1}{4}}=e^{-t/R C}
Take natural logarithms of both sides and solve for the time t:
\ln\,(\frac{1}{4})\,=\,-\,t/R C
t=-R C\ln\left({\frac{1}{4}}\right)=1.39R C=\mathrm{{1}.39~\tau}
(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=({3.50}\times{10}^{-6}\,{ F})({12.0\,}\,{ V})
Q=\mathrm{4}.20\times10^{-5}\,\mathrm{C}
Now calculate the time constant:
\tau=R C=(2.00\ \Omega)(3.50\times10^{-6}\,{F})\,=\,\mathbb{7}.00\times10^{-6}\,{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)\ \ \to\ \ \ t=-\tau\ln\left({\frac{q}{Q}}\right)
Substitute q = 1.60 × 10^{-19} C and the values for Q and τ found in part (b):
t=-(7.00\times10^{-6}\,\mathrm{s})\ln\left(\frac{1.60~\times~10^{-19}\,\mathrm{C}}{4.20~\times~10^{-5}\,\mathrm{C}}\right)
=\;2.32\times10^{-4}\,s
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 and 18.9, which indicate an infinite time would be required.