RC Circuits I
a
I
a
I
I
R C
ε
Cε
R
b
b
RC
+ + C
ε
2RC Cε
RC
- -
2RC
q = Cεe − t / RC
(
q
0
q = Cε 1 − e − t / RC
t
)
q
0
t
Resistor-capacitor circuits I
Let’s add a Capacitor to our simple circuit
I R
Recall voltage “drop” on C?
V=
ε
Q C ε − IR −
Write KVL:
Use
C
I=
dQ dt
Q =0 C
Now eqn. has only “Q”:
KVL gives Differential Equation !
dQ Q ε−R − =0 dt C
We will solve this later. For now, look at qualitative behavior…
Capacitors Circuits, Qualitative Basic principle: Capacitor resists change in Q Æ resists changes in V
• Charging (it takes time to put the final charge on) – Initially, the capacitor behaves like a wire (ΔV = 0, since Q = 0). – As current continues to flow, charge builds up on the capacitor
Æ it then becomes more difficult to add more charge Æ the current slows down – After a long time, the capacitor behaves like an open switch.
• Discharging – Initially, the capacitor behaves like a battery. – After a long time, the capacitor behaves like a wire.
UIL10, ACT 2 2A
• At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged.
a
2B
(b) I0+ = ε /2R
b
(c) I0+ = 2ε /R
long time?
(b) I∞ = ε /2R
C
ε
– What is the value of the current I∞ after a very
(a) I∞ = 0
I R
– What is the value of the current I0+ just after the switch is thrown?
(a) I0+ = 0
I
(c) I∞ > 2ε /R
R
UIL10, ACT 2 2A
• At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged.
a
(b) I0+ = ε /2R
I R
b
C
ε
– What is the value of the current I0+ just after the switch is thrown?
(a) I0+ = 0
I
R
(c) I0+ = 2ε /R
•
Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0!
•
Applying KVL to the loop at t=0+, ε −IR − 0 − IR = 0 ⇒ I = ε /2R
UIL10, ACT 2 2A
• At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. – What is the value of the current I0+ just after the switch is thrown?
a
I
I R
b
C
ε
R
(a) I0+ = 0 2B
(b) I0+ = ε /2R
(c) I0+ = 2ε /R
– What is the value of the current I∞ after a very
long time?
(a) I∞ = 0
(b) I∞ = ε /2R
(c) I∞ > 2ε /R
• The key here is to realize that as the current continues to flow, the charge on the capacitor continues to grow. • As the charge on the capacitor continues to grow, the voltage across the capacitor will increase. • The voltage across the capacitor is limited to ε ; the current goes to 0.
UI11Pf11:
The capacitor is initially uncharged, and the two switches are open.
E
3) What is the voltage across the capacitor immediately after switch S1 is closed? a) Vc = 0
b) Vc = E
c) Vc = 1/2 E
Initially: Q = 0 VC = 0 I = E/(2R)
4) Find the voltage across the capacitor after the switch has been closed for a very long time. a) Vc = 0 c) Vc = 1/2 E
b) Vc = E
Q=EC
I=0
UI11Pf11:
E
6) After being closed a long time, switch 1 is opened and switch 2 is closed. What is the current through the right resistor immediately after the switch 2 is closed? a) IR= 0 b) IR=E/(3R) c) IR=E/(2R) d) IR=E/R
Now, the battery and the resistor 2R are disconnected from the circuit, so we have a different circuit. Since C is fully charged, VC = E. Initially, C acts like a battery, and I = VC/R.
RC Circuits (Time-varying currents, charging) •
I
a
Charge capacitor:
I R
C initially uncharged; connect switch to a at t=0 Calculate current and
b
C
ε
charge as function of time.
Q ε − IR − = 0 C •Convert to differential equation for Q:
• Loop theorem ⇒
dQ I= dt
⇒
Would it matter where R is placed in the loop??
dQ Q ε =R + dt C
Charging Capacitor •
I
a
Charge capacitor:
I R
ε=R
dQ Q + dt C
b
ε
C
• Guess solution:
Q = Cε (1 − e
−t
RC
)
•Check that it is a solution:
dQ = C ε e − t / RC dt ⇒
1 ⎞ ⎛ − ⎜ ⎟ ⎝ RC ⎠
−t dQ Q −t / RC R + = +εe + ε (1 − e RC ) = ε ! dt C
Note that this “guess” fits the boundary conditions:
t = 0 ⇒Q = 0 t = ∞ ⇒ Q = Cε
Charging Capacitor •
Q = Cε (1− e
−t / RC
•
I R
)
b
C
ε
Current is found from differentiation:
dQ ε −t / RC I= = e dt R
I
a
Charge capacitor:
⇒
Conclusion: • Capacitor reaches its final charge(Q=Cε ) exponentially with time constant τ = RC. • Current decays from max (=ε /R) with same time constant.
Charging Capacitor Charge on C Q = Cε (1 − e − t / RC )
Max = Cε
RC
2RC
Cε
Q
63% Max at t = RC 0
Current I =
dQ ε − t / RC = e dt R
Max = ε /R
t
ε /R
I
37% Max at t = RC 0
t
Discharging Capacitor I
a
dQ Q R + =0 dt C
b
I R C
ε
• Guess solution:
+ +
Q = Q 0 e − t /τ = C ε e − t / RC
• Check that it is a solution: dQ = C ε e − t / RC dt
⇒
Note that this “guess” fits the boundary conditions:
1 ⎞ ⎛ − ⎜ ⎟ ⎝ RC ⎠
dQ Q − − R dt + = − ε e t / RC + ε e t / RC = 0 C
!
t = 0 ⇒ Q = Cε t =∞⇒Q=0
- -
Discharging Capacitor • Discharge capacitor:
Q = Q 0 e − t /τ = C ε e − t / RC • Current is found from differentiation: dQ ε −t / RC =− e I= dt R
⇒
Minus sign: Current is opposite to original definition, i.e., charges flow away from capacitor.
I
a b
I R + + C
ε
- -
Conclusion: • Capacitor discharges exponentially with time constant τ = RC • Current decays from initial max value (= -ε/R) with same time constant
Discharging Capacitor Cε
Charge on C
Q = C ε e − t / RC Max = Cε
RC
2RC
Q
37% Max at t = RC 0 zero
t
0
Current dQ ε I = = − e − t / RC dt R
I
“Max” = -ε/R 37% Max at t = RC
-ε /R
t
UI11Pf 11:
The two circuits shown below contain identical fully charged capacitors at t=0. Circuit 2 has twice as much resistance as circuit 1.
8) Compare the charge on the two capacitors a short time after t = 0 a) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q2
Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: τ1 = RC and τ2 = 2RC. Since τ2 > τ1 it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor 2 is bigger than that on capacitor 1.
Preflight 11:
The circuit below contains a battery, a switch, a capacitor and two resistors 10) Find the current through R1 after the switch has been closed for a long time. a) I1 = 0
b) I1 = E/R1
c) I1 = E/(R1+ R2)
After the switch is closed for a long time ….. The capacitor will be fully charged, and I3 = 0. (The capacitor acts like an open switch). So, I1 = I2, and we have a one-loop circuit with two resistors in series, hence I1 = E/(R1+R2)
Charging Cε
Q
RC
2RC
Cε
Q = Cε (1− e−t / RC )
t
0
I=
0
2RC
Q
t
ε /R
RC
Q = C ε e − t / RC
0
0
I
Discharging
dQ ε −t / RC = e dt R
t
I
-ε /R
dQ ε −t / RC =− e I= dt R
t
Power distribution systems
Conventional House wiring has 240/120V lines
Which prong is hot??
Remark; houses typically have two single phase, 120v hot lines
Electricity Costs Your electric costs are charged in units of energy, Kilowatt-hour 1 Kilowatt-Hour = 1000 watts x 3600 sec = 3.6 million joules Oahu residential charges are about 20 cents per K-H. This can be read on your AC Kilowatt-Hour meter outside your house. Electricity costs in Hawaii are high! Example 100W light bulb run for 10hrs/day for 30 days uses 0.1 KW x 10 x 30 = 30 Kilowatt-Hours and costs $6.00.
