When a capacitor is initially uncharged, it is very easy to stuff charge in. As charge builds, it repels new charge with more and more force. Due to this effect, the charging of a capacitor follows a logarithmic curve. When you pass current through a resistor into a capacitor, the capacitor eventually “fills up” and no more current flows. A typical RC circuit is shown below; when the switch is closed, the capacitor discharges with an exponentially decreasing current.

\begin{align*}I(t) = I_0 e^{\frac{-t}{\tau}} && \text{Electric current flow varies with time in a like manner} \end{align*}

#### Example

In the circuit diagram shown above, the resistor has a value of \begin{align*}100\;\Omega\end{align*}

(a): To solve the first part of the problem, we'll use the equation that gives charge as a function of time.

\begin{align*}
Q(t)&=Q_oe^{\frac{-t}{\tau}} && \text{start with the equation give above}\\
.1Q_o&=Q_oe^{\frac{-t}{\tau}} && \text{substitute } .1Q_o \text{ for } Q(t) \text{ because that's the charge at the time we want to find}\\
.1&=e^{\frac{-t}{\tau}} && \text{simplify the equation}\\
t&=-\tau\ln(.1) && \text{solve for time}\\
t&=-RC\ln(.1) && \text{substitute in the value for } \tau\\
t&=-100\;\Omega * 500\;\mu\text{F} * \ln(.1) && \text{substitute in all the known values}\\
t&=.12\;\text{s}\\
\end{align*}

(b): Solving the second part of this problem will be a two step process. We will use the capacitance and the voltage drop to determine how much charge was originally on the capacitor (\begin{align*}Q_o\end{align*}

\begin{align*}
Q_o&=CV\\
Q_o&=500\;\mu\text{F} * 12\;\text{V}\\
Q_o&=.006\;\text{C}\\
\end{align*}

Now we can plug in the time we found in part A to the equation for charge as a function of time.

\begin{align*}
Q(t)&=Q_oe^{\frac{-t}{\tau}}\\
Q(.12\;\text{s})&=.006\;\text{C}e^{\frac{-.12\;\text{s}}{100\;\Omega * 500\;\mu\text{F}}}\\
Q(.12\;\text{s})&=5.44*10^{-4}\;\text{C}\\
\end{align*}

### Review

- Design a circuit that would allow you to determine the capacitance of an unknown capacitor.
- The power supply (i.e. the voltage source) in the circuit below provides \begin{align*}10\;\mathrm{V}\end{align*}
10V . The resistor is \begin{align*}200 \Omega\end{align*}200Ω and the capacitor has a value of \begin{align*}50 \mu \mathrm{F}\end{align*}50μF .- What is the voltage across the capacitor
*immediately*after the power supply is turned on? - What is the voltage across the capacitor after the circuit has been hooked up for a long time?

- What is the voltage across the capacitor
- Marciel, a bicycling physicist, wishes to harvest some of the energy he puts into turning the pedals of his bike and store this energy in a capacitor. Then, when he stops at a stop light, the charge from this capacitor can flow out and run his bicycle headlight. He is able to generate \begin{align*}18\;\mathrm{V}\end{align*}
18V of electric potential, on average, by pedaling (and using magnetic induction).- If Mars wants to provide \begin{align*}0.5\end{align*}
0.5 A of current for 60 seconds at a stop light, how big a \begin{align*}18\;\mathrm{V}\end{align*}18V capacitor should he buy (i.e. how many farads)? - How big a resistor should he pass the current through so the RC time is three minutes?

- If Mars wants to provide \begin{align*}0.5\end{align*}
- A simple circuit consisting of a \begin{align*}39 \mu \mathrm{F}\end{align*}
39μF and a \begin{align*}10 \text{k} \Omega \end{align*}10kΩ resistor. A switch is flipped connecting the circuit to a 12 V battery.- How long until the capacitor has 2/3 of the total charge across it?
- How long until the capacitor has 99% of the total charge across it?
- What is the total charge possible on the capacitor?
- Will it ever reach the full charge in part c.?
- Derive the formula for V(t) across the capacitor.
- Draw the graph of V vs. t for the capacitor.
- Draw the graph of V vs. t for the resistor.

- If you have a \begin{align*}39 \mu \mathrm{F}\end{align*}
39μF capacitor and want a time constant of 5 seconds, what resistor value is needed?

### Review (Answers)

- Answers will vary
- a. 0 V b. 10 V
- \begin{align*}3.3 \;\mathrm{F}\end{align*}
3.3F b. \begin{align*}54 \ \Omega\end{align*}54 Ω - a. 0.43 seconds b. 1.8 seconds c. \begin{align*} 4.7 \times 10^{-4} \text{C} \end{align*}
4.7×10−4C d. No, it will asymptotically approach it. e. The graph is same shape as the Q(t) graph. It will rise rapidly and then tail off asymptotically towards 12 V. f. The voltage across the resistor is 12 V minus the voltage across the capacitor. Thus, it exponentially decreases approaching the asymptote of 0 V. - About \begin{align*}128 \text{k} \Omega \end{align*}
128kΩ