Transcript RC Circuits - McMaster Physics & Astronomy Outreach

RC Circuits
- circuits in which the currents vary in time
- rate of charging a cap depends on C and R of circuit
- differential equations
Discharging a Capacitor
I
C
q
-q
Given: R, C, qo (initial charge)
R
1)
q
 IR  0
C
2)
dq
I 
dt
Find: q(t) and I(t) when switch
is closed
(Kirchhoff’s Loop Rule)
(- sign because q decreases for I > 0
That is, current in circuit equals the
decrease of charge on the capacitor)
Combine 1) and 2) to get:
I
C
q
-q
R
dq
q

dt
RC
where: q = q(t)
q(0) = qo
This is a differential equation for the function
q(t), subject to the initial condition q(0) = q0 .
We are looking for a function which is proportional
to its own first derivative (since dq/dt ~ -q).
Solution:
q(t)  qο e

t
RC
RC is called the “time constant” or “characteristic
time” of the circuit.
Units: 1 Ω x 1 F = 1 second (show this!)
Write t (“tau”) = RC, then:
q(t )  qo e
t
 
t 
(discharging)
Discharging
q
qo
q (t )  qo e
t
t,
t = 2 t,
t = 3 t,
t=
t∞,
2
t
t
t
t  RC
3
t
q ≈ 0.37 qo
= (qo/e)
q ≈ 0.14 qo
= (qo/e2)
q ≈ 0.05 qo
= (qo/e3)
q0
= (qo/e∞)
t
Draw a graph for I(t).
Example 1
A capacitor is charged up to 18 volts, and then
connected across a resistor. After 10 seconds,
the capacitor voltage has fallen to 12 volts.
a) What is the time constant RC ?
b) What will the voltage be after another 10
seconds (20 seconds total)?
A)
B)
C)
D)
8V
6V
4V
0
Charging a capacitor

C
R
C is initially uncharged, and the switch
is closed at t=0. After a long time,
the capacitor has charge Qf .
dq q
 R  0
dt C
Then,
t 

q(t)  Q f 1 - e t 


where t  RC.
Question: What is Qf equal to?
What is V(t) ? Recall, C=Q/V, so V(t)=Q(t)/C
Charging a capacitor
t 

q(t )  Q f 1  e t 


t  RC
q
Qf
t
t = 0,
q=0
t = RC,
q 0.63 Qf
t = 2 RC, q 0.86 Qf
2
t
3
t
t = 3 RC, q 0.95 Qf
etc.
t
Draw a graph of I(t). Why is I=+dq/dt this time?
Example 2
100 kΩ
12 V
2 µF
The capacitor is initially uncharged.
After the switch is closed, find:
i) Initial current
ii) Initial voltage across the resistor
iii) Initial voltage across the capacitor
iv) Time for voltage across C to reach 0.63*12V
v)
Final voltage across the resistor
vi) Final voltage across the capacitor
Solution
Example 3
The circuit below contains two resistors, R1 = 2.00 kΩ and R2 = 3.00 kΩ, and
two capacitors, C1 = 2.00 μF and C2 = 3.00 μF, connected to a battery with emf
ε = 120 V. No charge is on either capacitor before switch S is closed. Determine
the charges q1 and q2 on capacitors C1 and C2, respectively, after the switch is
closed.
Solution:
1) reconstruct the circuit so that it becomes a simple
RC circuit containing a single resistor and single capacitor
2) determine the total charge q stored in the equivalent
circuit.
“RC” Circuits
• a capacitor takes time to charge or discharge
through a resistor
• “time constant” or “characteristic time”
t
= RC
(1 ohm) x (1 farad) = 1 second