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# RC Time Constant

## A charging capacitor rapidly builds up charge at first but quickly slows down as it approaches its maximum capacity.

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Practice RC Time Constant
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RC Time Constant

Students will learn about the RC time constant and how to solve various problems involving resistor-capacitor circuits.

### Key Equations

\begin{align*}Q(t) = Q_0 e^{\frac{-t}{\tau}} && \text{Discharge rate, where } \tau= RC \end{align*}

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

Guidance
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.

#### Example 1

In the circuit diagram shown above, the resistor has a value of \begin{align*}100\;\Omega\end{align*} and the capacitor has a capacitance of \begin{align*}500\;\mu\text{F}\end{align*}. After the switched is closed, (a) how long will it be until the charge on the capacitor is only 10% of what it was when the switch was originally closed? If the capacitor was originally charged by a 12 V battery, how much charge will be left on it at this time?

##### Solution

(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*}

### Explore More

1. Design a circuit that would allow you to determine the capacitance of an unknown capacitor.
2. The power supply (i.e. the voltage source) in the circuit below provides \begin{align*}10\;\mathrm{V}\end{align*}. The resistor is \begin{align*}200 \Omega\end{align*} and the capacitor has a value of \begin{align*}50 \mu \mathrm{F}\end{align*}.
1. What is the voltage across the capacitor immediately after the power supply is turned on?
2. What is the voltage across the capacitor after the circuit has been hooked up for a long time?
3. 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*} of electric potential, on average, by pedaling (and using magnetic induction).
1. If Mars wants to provide \begin{align*}0.5\end{align*} A of current for 60 seconds at a stop light, how big a \begin{align*}18\;\mathrm{V}\end{align*} capacitor should he buy (i.e. how many farads)?
2. How big a resistor should he pass the current through so the RC time is three minutes?
4. A simple circuit consisting of a \begin{align*}39 \mu \mathrm{F}\end{align*} and a \begin{align*}10 \text{k} \Omega \end{align*} resistor. A switch is flipped connecting the circuit to a 12 V battery.
1. How long until the capacitor has 2/3 of the total charge across it?
2. How long until the capacitor has 99% of the total charge across it?
3. What is the total charge possible on the capacitor?
4. Will it ever reach the full charge in part c.?
5. Derive the formula for V(t) across the capacitor.
6. Draw the graph of V vs. t for the capacitor.
7. Draw the graph of V vs. t for the resistor.
5. If you have a \begin{align*}39 \mu \mathrm{F}\end{align*} capacitor and want a time constant of 5 seconds, what resistor value is needed?

1. .
2. a. 0 V b. 10 V
3. \begin{align*}3.3 \;\mathrm{F}\end{align*}b. \begin{align*}54 \ \Omega\end{align*}
4. a. 0.43 seconds b. 1.8 seconds c. \begin{align*} 4.7 \times 10^{-4} \text{C} \end{align*} 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.
5. about \begin{align*}128 \text{k} \Omega \end{align*}