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 Figure 18.18. (a) How long does it take the charge on the capacitor to drop to one-fourth its

Question

DISCHARGING A CAPACITOR IN AN RC CIRCUIT

GOAL Calculate some elementary properties of a discharging capacitor in an [latex]R C[/latex] circuit.

PROBLEM Consider a capacitor [latex]C[/latex] being discharged through a resistor [latex]R[/latex] as in Figure [latex]18.18[/latex]. (a) How long does it take the charge on the capacitor to drop to one-fourth its initial value? Answer as a multiple of [latex] au[/latex]. (b) Compute the initial charge and time constant, and (c) the time it takes to discharge all but the last quantum of charge, [latex]1.60 imes 10^{-19} \mathrm{C}[/latex], if the initial potential differ ence across the capacitor is [latex]12.0 \mathrm{~V}[/latex], the capacitance is equal to [latex]3.50 imes 10^{-6} \mathrm{~F}[/latex], and the resistance is [latex]2.00 \Omega[/latex]. (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 [latex]q=\frac{1}{4} Q[/latex] in Equation [latex]18.9[/latex]

[latex]q = Qe^{-t/RC}[/latex] [18.9]

for a discharging capacitor, where [latex]Q[/latex] is the initial charge, and solve for time [latex]t[/latex]. In part (b) substitute into Equations [latex]16.8[/latex]

[latex]C \equiv \frac{Q}{\Delta V}[/latex] [16.8]

and [latex]18.8[/latex]

[latex] au = RC[/latex] [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 [latex]q=1.60 imes 10^{-19} \mathrm{C}[/latex] into the discharging-capacitor equation, again solving for time.

Accepted Answer

(a) How long does it take the charge on the capacitor to reduce to one-fourth its initial value?

Apply Equation 18.9:

[latex]q(t)=Q e^{-t / R C}[/latex]

Substitute [latex]q(t)=Q / 4[/latex] into the preceding equation and cancel [latex]Q[/latex] :

[latex]\frac{1}{4} Q=Q e^{-t / R C} \quad ightarrow \quad \frac{1}{4}=e^{t / R C}[/latex]

Take natural logarithms of both sides and solve for the time [latex]t[/latex] :

[latex]\begin{aligned}&\ln \left(\begin{array}{l}1 \ \hline 4\end{array} ight)=-t / R C \&t=-R C \ln \left(\frac{1}{4} ight)=1.39 R C=1.39 au\end{aligned}[/latex]

(b) Compute the initial charge and time constant from the given data.

Use the capacitance equation to find the initial charge:

[latex]C=\frac{Q}{\Delta V} \quad ightarrow \quad Q=C \Delta V=\left(3.50 imes 10{ }^{-6} \mathrm{~F} ight)(12.0 \mathrm{~V})[/latex]

[latex]Q=4.20 imes 10^{-5} \mathrm{C}[/latex]

Now calculate the time constant:

[latex] au=R C=(2.00 \Omega)\left(3.50 imes 10^{-6} \mathrm{~F} ight)=7.00 imes 10^{-6} \mathrm{~s}[/latex]

(c) How long does it take to drain all but the last quantum of charge?

Apply Equation [latex]18.9[/latex], dividing both sides by [latex]Q:[/latex]

[latex]q(t)=Q e^{-t / au} \quad ightarrow \quad e^{-t / au}=\frac{q}{Q}[/latex]

Take the natural logs of both sides and solve for [latex]t[/latex] :

[latex]-t / au=\ln \left(\frac{q}{Q} ight) \quad ightarrow \quad t=- au \ln \left(\frac{q}{Q} ight)[/latex]

Substitute [latex]q=1.60 imes 10^{-19} \mathrm{C}[/latex] and the values for [latex]Q[/latex] and [latex] au[/latex] found in part (b):

[latex]\begin{aligned}t &=-\left(7.00 imes 10^{-6} \mathrm{~s} ight) \ln \left(\frac{1.60 imes 10^{-19} \mathrm{C}}{4.20 imes 10^{-5} \mathrm{C}} ight) \&=2.32 imes 10^{-4} \mathrm{~s}\end{aligned}[/latex]

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

[latex]q = Q(1-e^{-t/RC})[/latex] [18.7]

and 18.9, which indicate an infinite time would be required.

Question 18.7: DISCHARGING A CAPACITOR IN AN RC CIRCUIT GOAL Calculate some...

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