RC Time Constant
James H Dann, Ph.D.
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Printed: January 5, 2016
AUTHOR
James H Dann, Ph.D.
CONTRIBUTORS
Chris Addiego
Antonio De Jesus López
www.ck12.org
Chapter 1. RC Time Constant
C HAPTER
1
RC Time Constant
• Define the RC time constant and solve various problems involving resistor-capacitor circuits.
Students will learn about the RC time constant and how to solve various problems involving resistor-capacitor
circuits.
Key Equations
−t
Q(t) = Q0 e τ
−t
I(t) = I0 e τ
Discharge rate, where τ = RC
Electric current flow varies with time in a like manner
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 100 Ω and the capacitor has a capacitance of 500 µF.
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.
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−t
Q(t) = Qo e τ
.1Qo = Qo e
.1 = e
−t
τ
−t
τ
start with the equation give above
substitute .1Qo for Q(t) because that’s the charge at the time we want to find
simplify the equation
t = −τ ln(.1)
solve for time
t = −RC ln(.1)
substitute in the value for τ
t = −100 Ω ∗ 500 µF ∗ ln(.1) substitute in all the known values
t = .12 s
(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 (Qo ).
Qo = CV
Qo = 500 µF ∗ 12 V
Qo = .006 C
Now we can plug in the time we found in part A to the equation for charge as a function of time.
−t
Q(t) = Qo e τ
−.12 s
Q(.12 s) = .006 Ce 100 Ω∗500 µF
Q(.12 s) = 5.44 ∗ 10−4 C
Watch this Explanation
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Explore More
1. Design a circuit that would allow you to determine the capacitance of an unknown capacitor.
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Chapter 1. RC Time Constant
2. The power supply (i.e. the voltage source) in the circuit below provides 10 V. The resistor is 200Ω and the
capacitor has a value of 50µF.
a. What is the voltage across the capacitor immediately after the power supply is turned on?
b. 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 18 V of electric potential, on average, by pedaling
(and using magnetic induction).
a. If Mars wants to provide 0.5 A of current for 60 seconds at a stop light, how big a 18 V capacitor should
he buy (i.e. how many farads)?
b. How big a resistor should he pass the current through so the RC time is three minutes?
4. A simple circuit consisting of a 39µF and a 10kΩ resistor. A switch is flipped connecting the circuit to a 12 V
battery.
a.
b.
c.
d.
e.
f.
g.
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.
5. If you have a 39µF capacitor and want a time constant of 5 seconds, what resistor value is needed?
Answers to Selected Problems
1.
2.
3.
4.
.
a. 0 V b. 10 V
3.3 Fb. 54 Ω
a. 0.43 seconds b. 1.8 seconds c. 4.7 × 10−4 C 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 128kΩ
3