RC-1
RC Circuits
An RC circuit is a circuit with a resistor (R) and a capacitor (C). RC circuits are used to
construct timers and filters.
Example 1. Very simple RC circuit: a capacitor C, charged to an initial voltage V0 = Q0/C,
attached to a resistor R with a switch.
switch
+Q0
C
C=
R
–Q0
Q
V
⇒ V0 =
Q0
C
Close the switch at time t = 0 , so current I starts to flow. The charged capacitor is acting like a
battery: it produces a voltage difference across the resistor which drives the current through the
resistor:
At t = 0+,
C
+Q
I
–Q
R
I=−
Vacross C = Vacross R , VC = VR ,
dQ
dt
I0 =
V0
.
R
(– sign because Q is decreasing)
Q
Q
dQ
= IR ,
= −
⋅R ,
C
C
dt
dQ
1
= −
Q
dt
RC
RC = "time constant" = τ , has units of time
is a differential equation of the form
dx
= a ⋅ x , where a is a constant.
dt
This equation says: (rate of charge of x) ∝ x ⇔ exponential solution : x = x 0 exp ( a t )
d ( x 0 ⋅ ea t )
dx
Check:
=
= a ⋅ x 0 ⋅ ea t = a ⋅ x .
dt
dt
a > 0 ⇒ exponential growth ,
The solution to
dQ
1
= −
Q
dt
RC
It works!
a < 0 ⇒ exponential decay
is
⎛ t ⎞
⎛ t⎞
Q(t) = Q 0 exp ⎜ −
⎟ = C V0 exp ⎜ − ⎟
⎝ RC ⎠
⎝ τ⎠
Notice that at t = 0, the formula gives Q = Q0 .
In time τ , Q falls by a factor of exp(–1) = 1/e ≅ 0.37..
Last update: 10/8/2009
Dubson Phys1120 Notes, ©University of Colorado
RC-2
In time 2τ, Q falls by a factor of exp(–2) = (1/e)(1/e) ≅ 0.14..
⇒ Q approaches zero asymptotically, and so does V and I
Q
⎛ t ⎞
⎛ t⎞
Q(t) = Q 0 exp ⎜ −
⎟ = Q 0 exp ⎜ − ⎟
⎝ RC ⎠
⎝ τ⎠
Q0
Q0/e
t
τ
I =−
V=Q/C
V
dQ
= + 0 exp(− t / τ)
dt
R
t
t
Example 2: More complex RC circuit: Charging a capacitor with a battery.
switch
E
C
VC = Q / C
Let's use symbol E for battery voltage (E
short for emf) because there are so many
other V's in this example.
Before switch is closed, I = 0, Q = 0.
R
VR = I R
Close switch at t = 0.
Always true that E = VC + VR ,
by Kirchhoff's Voltage Law (Loop Law)
The charge Q on the capacitor and the voltage VC = Q / C across the capacitor cannot change
instantly, since it takes time for Q to build up, so ..
At t = 0+ , Q = 0 , VC = 0, E = VC + VR = VR = I R ⇒ I0 = E / R
Although Q on the capacitor cannot change instantly, the current I = dQ/dt can change instantly.
"Current through a capacitor" means dQ/dt . Even
though there is no charge ever passing between the
plates of the capacitor, there is a current going into one
Last update: 10/8/2009
+
- I
+
+
+
Dubson Phys1120 Notes, ©University of Colorado
I
RC-3
plate and the same current is coming out of the other plate, so it is as if there is a current passing
through the capacitor.
C
E
E = VR + VC
Q
C
E = IR +
I
R
E =
,
I=+
dQ
dt
dQ
Q
R +
dt
C
Qualitatively, as t ↑ , Q ↑ , VC = Q/C ↑ , VR ↓ , I = VR / R ↓. After a long time , t >> τ = RC ,
the current decreases to zero: I = 0, VC = E , Q = C E
Vc = Q / C
Analytic solution:
E
VC (t) = E [1 − exp(− t / RC) ]
Q(t) = E ⋅ C [1 − exp( − t / RC) ]
t
Things to remember:
•
Uncharged capacitor acts like a "short" ( a wire ) since VC = Q / C = 0.
•
After a long time, when the capacitor is fully charged, it acts like an "open-circuit" ( a
break the wire). We must have IC = 0 eventually, otherwise Q → ∞ , VC → ∞ .
Last update: 10/8/2009
Dubson Phys1120 Notes, ©University of Colorado