Question 18.6: Charging a Capacitor in an RC Circuit Goal Calculate element...
Charging a Capacitor in an RC Circuit
Goal Calculate elementary properties of a simple R C circuit.
Problem An uncharged capacitor and a resistor are connected in series to a battery, as in Active Figure 18.16a. If \mathcal{E}=12.0 \mathrm{~V}, C=5.00 \mu \mathrm{F}, and R=8.00 \times 10^{5} \Omega, find (a) the time constant of the circuit, (b) the maximum charge on the capacitor, (c) the charge on the capacitor after 6.00 \mathrm{~s}, (d) the potential difference across the resistor after 6.00 \mathrm{~s}, and (\mathbf{e}) the current in the resistor at that time.
Strategy Finding the time constant in part (a) requires substitution into Equation 18.8.
\tau=R C (18.8)
For part (b), the maximum charge occurs after a long time, when the current has dropped to zero. By Ohm’s law, \Delta V=I R, the potential difference across the resistor is also zero at that time, and Kirchhoff’s loop rule then gives the maximum charge. Finding the charge at some particular time, as in part (c), is a matter of substituting into Equation 18.7.
q=Q(1-e^{-t/R C}) (18.7)
Kirchhoff’s loop rule and the capacitance equation can be used to indirectly find the potential drop across the resistor in part (d), and then Ohm’s law yields the current.
(a) Find the time constant of the circuit.
Use the definition of the time constant, Equation 18.8:
\tau=R C=\left(8.00 \times 10^{5} \Omega\right)\left(5.00 \times 10^{-6} \mathrm{~F}\right)=4.00 \mathrm{~s}
(b) Calculate the maximum charge on the capacitor.
Apply Kirchhoff’s loop rule to the R C circuit, going clockwise, which means that the voltage difference across the battery is positive and the differences across the capacitor and resistor are negative.
\text { (1) } \Delta V_{\mathrm{bat}}+\Delta V_{C}+\Delta V_{R}=0
From the definition of capacitance (Equation 16.8) and Ohm’s law, we have \Delta V_{C}=-q / C and \Delta V_{R}=-I R. These are voltage drops, so they’re negative. Also, \Delta V_{\text {bat }}= +\mathcal{E} =12.0 \mathrm{~V}.
\mathcal{E}-\frac{q}{c}-I R=0
When the maximum charge q=Q is reached, I=0. Solve Equation (2) for the charge:
\mathcal{E}-\frac{Q}{C}=0 \rightarrow Q=C \mathcal{E}
Substitute to find the maximum charge:
Q=\left(5.00 \times 10^{-6} \mathrm{~F}\right)(12.0 \mathrm{~V})=60.0 \mu \mathrm{C}
(c) Find the charge on the capacitor after 6.00 \mathrm{~s}.
Substitute into Equation 18.7:
\begin{aligned} q & =Q\left(1-e^{-t / \tau}\right)=(60.0 \mu \mathrm{C})\left(1-e^{-6.00 \mathrm{~s} / 4.00 \mathrm{~s}}\right) \\ & =46.6 \mu \mathrm{C} \end{aligned}
(d) Compute the potential difference across the resistor after 6.00 \mathrm{~s}.
Compute the voltage drop \Delta V_{C} across the capacitor at that time:
\Delta V_{C}=-\frac{q}{C}=\frac{-46.6 \mu \mathrm{C}}{5.00 \mu \mathrm{F}}=-9.32 \mathrm{~V}
Solve Equation 1 for \Delta V_{R}, and substitute:
\begin{aligned} \Delta V_{R} & =-\Delta V_{\text {bat }}-\Delta V_{C}=-12.0-(-9.32 \mathrm{~V}) \\ & =-2.68 \mathrm{~V} \end{aligned}
(e) Find the current in the resistor after 6.00 \mathrm{~s}.
Apply Ohm’s law, using the results of part (d) (remember that \Delta V_{R}=-I R here):
\begin{aligned} I & =\frac{-\Delta V_{R}}{R}=-\frac{(-2.68 \mathrm{~V})}{\left(8.0 \times 10^{5} \Omega\right)} \\ & =3.4 \times 10^{-6} \mathrm{~A} \end{aligned}
Remark In solving this problem, we paid scrupulous attention to signs. These signs must always be chosen when applying Kirchhoff’s loop rule, and must remain consistent throughout the problem. Alternately, magnitudes can be used and the signs chosen by physical intuition. For example, the magnitude of the potential difference across the resistor must equal the magnitude of the potential difference across the battery minus the magnitude of the potential difference across the capacitor.