Now we will discuss circuits with batteries, resistors and
capacitors. Æ “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?
Simplest case is just a Capacitor C, charged to a Voltage
V0=Q0/C, attached to a Resistor R and a switch.
i
+Q0
R = 10 Ω
-Q0
C = 0.1F
R = 10 Ω
C = 0.1F
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Close the switch at t=0, then current i starts to flow.
+Q0
-Q0
R = 10 Ω
C = 0.1F
Voltage across C = Voltage across R
i
V0
R
dQ
Later i (t ) = −
dt
At t=0, i0 =
i
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Vc = VR
+Q0
-Q0
Q
dQ
R
= iR = −
C
dt
dQ
1
Q
=−
dt
RC
C = 0.1F
* Negative sign since Q
is decreasing.
We now need to solve this differential equation.
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Solving the differential equation:
dQ
1
Q
=−
dt
RC
Exponential Decay
Q (t ) = Q0 e −t / RC
Q0
dQ
1
⎛ 1 ⎞ −t / RC
= Q0 ⎜ −
=−
Q
⎟e
dt
RC
RC
⎝
⎠
Q (t = 0) = Q0 e
−0 / RC
After a time t = 2RC, Q has dropped
by e-2 = 1/e2
Q0/e
= Q0
τ
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Q (t ) = Q0 e − t / RC
After a time τ = RC, Q has dropped
by e-1 = 1/e.
Q(t)
Check solution by taking the derivative...
Also Q(t) = Q0 at t=0.
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Thus τ=RC is often called the time
constant and has units [seconds].
time
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1
Clicker Question
dQ
1
⎛ 1 ⎞ −t / RC
= Q0 ⎜ −
=−
Q
⎟e
dt
RC
RC
⎝
⎠
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?
dQ ⎛ Q0 ⎞ −t / RC
|= ⎜
= i0 e −t / RC
⎟e
dt ⎝ RC ⎠
| i (t ) |=|
Thus, the current also falls with the same exponential function.
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R = 10 Ω
Before switch closed i=0, and
charge on capacitor Q=0.
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.
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+ Vb −
C=0.0010 F
Close switch at t=0.
dQ
R −Q/C = 0
dt
V
dQ
Q
=+ b −
dt
R RC
+ Vb + VR + VC = 0
Try Voltage loop rule.
B) 0.5A
D) 1/e2 A = 0.14A
+ Vb − iR − Q / C = 0
More complex RC circuit: Charging C with a battery.
V=10V
A) 1A
C) 1/e A = 0.37A
E) None of these.
+ Vb − iR − Q / C = 0
(
Q(t ) = CVb 1 − e − t / RC
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(
Q(t ) = CVb 1 − e −t / RC
i (t ) =
)
R = 10 Ω
V=10V
i=
70
Although no charge actually passes between the capacitor
plates, it acts just like a current is flowing through it.
Q
dQ
V
(t ) = b e − t / RC
dt
R
)
Uncharged capacitors act like a “short”: VC=Q/C=0
time
Fully charged capacitors act like an “open circuit”. Must have
iC = 0 eventually, otherwise Q Æ infinity.
dQ
dt
C=0.0010 F
time
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2
Clicker Question
House Wiring
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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Wall socket = 3 prong plug
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“Hot” Voltage is AC (Alternating Current) which is sinusoidal
and not DC (Direct Current) which is flat with time.
“Ground”
(0 Volts)
“Hot”
(~120 Volts)
Shorter slot?
Harder to put finger in!
|V(peak)| = 170 V
V 2 = 120V
“Cold” or “Neutral”
(0 – 3 Volts)
Frequency = 1/T = 60 Hz
75
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Standard Electrical Wiring Colors
AC Voltage is more economical to distribute from power
plants to homes than DC Voltage (gigantic batteries).
In fact this decision resulted from a famous debate between
Edison and Tesla. More about this later.
Black = Hot
White = Cold
Green = Ground
(burnt black)
(like snow)
(like grass)
!!! Never assume house wiring is correct.
Always test first !!!
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3
Fuse Breakers
If |i(fuse)| > 15 Amps, breaker opens,
preventing the wires from overheating.
“Hot” Side
+/-170V
~
0V
0V
i
Does turning the Toaster switch on, increase the chances of
the fuse blowing? A) yes, B) no, C) no change.
“Hot” Side
+/-170V
i
120 VAC
Clicker Question
i
Toaster
~
120 VAC
“Cold” Side
Ground (safety feature)
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0V
0V
Toaster
i
“Cold” Side
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4