Download RC circuits Discharging a capacitor

RC circuits
Initially one has +Q0 and –Q0 on the Capacitor plates.
Thus, the initial Voltage on the Capacitor V0 = Q0/C.
What do you think happens when the switch is closed?
i
+Q0
-Q0
R = 10 Ω
C = 0.1F
Discharging a capacitor
Close the switch at t=0, then current i starts to flow.
+Q0
-Q0
V0
R
dQ
Later i (t ) = −
dt
At t=0, i0 =
i
R = 10 Ω
C = 0.1F
* Negative sign since Q
is decreasing.
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Discharging a capactiror
Voltage across C = Voltage across R
i
Vc = VR
+Q0
-Q0
C = 0.1F
dQ
Q
R
= iR = −
dt
C
dQ
1
=−
Q
dt
RC
We now need to solve this differential equation.
Discharging a capacitor
Solving the differential equation:
dQ
1
Q
=−
dt
RC
Q(t ) = Q0 e −t / RC
Check solution by taking the derivative...
1 ⎞ −t / RC
1
dQ
⎛
= Q0 ⎜ −
=−
Q
⎟e
dt
RC
⎝ RC ⎠
Also Q(t) = Q0 at t=0.
Q(t = 0) = Q0e −0 / RC = Q0
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Discharging a capacitor
Exponential Decay
Q(t)
Q(t ) = Q0 e −t / RC
After a time τ = RC, Q has dropped
by e-1 = 1/e.
Q0
After a time t = 2RC, Q has dropped
by e-2 = 1/e2
Q0/e
τ
Thus τ=RC is often called the time
constant and has units [seconds].
time
Discharging a capacitor
1 ⎞ −t / RC
1
dQ
⎛
= Q0 ⎜ −
=−
Q
⎟e
dt
RC
RC
⎝
⎠
| i (t ) |=|
dQ ⎛ Q0 ⎞ −t / RC
|= ⎜
= i0 e −t / RC
⎟e
dt ⎝ RC ⎠
Thus, the current also falls with the same exponential function.
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Clicker Question
A capacitor with capacitance 0.1F in an RC circuit is initially
charged up to an initial voltage of Vo = 10V and is then
discharged through an R=10Ω resistor as shown. The switch is
closed at time t=0. Immediately after the switch is closed, the
initial current is Io =Vo /R=10V/10Ω.
What is the current I through the resistor at time t=2.0 s?
A) 1A
C) 1/e A = 0.37A
E) None of these.
B) 0.5A
D) 1/e2 A = 0.14A
R=10Ω
C=0.1F
Answer: 1/e2 A = 0.14A. The time constant for this circuit is
RC=(10Ω)(0.10F) = 1.0 sec. So at time t=2.0 sec, two time constants have
passed. After one time constant, the voltage, charge, and current have all
decreased by a factor of e. After two time constants, everything has fallen
by e2. The initial current is 1A. So after two time constants, the current is
1/e2 A = 0.135A.
Charging a capacitor
More complex RC circuit: Charging C with a battery.
R = 10 Ω
Before switch closed i=0, and
charge on capacitor Q=0.
V=10V
C=0.0010 F
Close switch at t=0.
Try Voltage loop rule.
+ Vb + VR + VC = 0
+ Vb − iR − Q / C = 0
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Charging a capacitor
+ Vb − iR − Q / C = 0
+ Vb −
dQ
R −Q/C = 0
dt
(
dQ
V
Q
=+ b −
R RC
dt
Q(t ) = CVb 1 − e − t / RC
)
Charging a capacitor
(
Q(t ) = CVb 1 − e −t / RC
i (t ) =
)
Q
dQ
V
(t ) = b e −t / RC
dt
R
R = 10 Ω
V=10V
i=
time
dQ
dt
C=0.0010 F
time
•Although no charge actually passes between the capacitor plates, it acts just like
a current is flowing through it.
•Uncharged capacitors act like a “short”: VC=Q/C=0
•Fully charged capacitors act like an “open circuit”. Must have iC = 0 eventually,
otherwise Q
infinity.
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Clicker Question
An RC circuit is shown below. Initially the switch is open
and the capacitor has no charge. At time t=0, the switch is
closed. What is the voltage across the capacitor
immediately after the switch is closed (time = 0)?
R = 10 Ω
A) Zero
C) 5V
B) 10 V
D) None of these.
V=10V
C=0.0010 F
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