2 rc
DESCRIPTION
I. I. I. a. I. a. R. R. b. b. +. +. C. C. e. -. -. e. RC Circuits. RC. 2 RC. C e. RC. 2 RC. C e. q. q. 0. 0. t. t. Text Reference: Chapter 26.6. Examples: 26.17,18 and 19. Today…. Calculate Charging of Capacitor through a Resistor - PowerPoint PPT PresentationTRANSCRIPT
a
b
R
C
II
q C e t RC 1 /
t
q
RC 2RC
0
C
C
a
b+
- -
R+
I I
RC Circuits
q
RC 2RC
0t
q C e t RC /
C
Today…
• Calculate Charging of Capacitor through a Resistor
• Calculate Discharging of Capacitor through a Resistor
Text Reference: Chapter 26.6 Examples: 26.17,18 and 19
Last time--Behavior of Capacitors(from Lect. 10)
• Charging
– Initially, the capacitor behaves like a wire.
– 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.
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
E
4) Find the voltage across the capacitor after the switch has beenclosed for a very long time.
a) Vc = 0 b) Vc = E
c) Vc = 1/2 E
The capacitor is initially uncharged, and the two switches are open.
Preflight 11:
Initially: Q = 0 VC = 0 I = E/(2R)
After a long time: VC = E Q = E C I = 0
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?
E
a) IR= 0
b) IR=E/(3R)
c) IR=E/(2R)
d) IR=E/R
Preflight 11:
After C is fully charged, S1 is opened and S2 is closed. Now, the battery and the resistor 2R are disconnected from the circuit. So we now 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)
• Charge capacitor:
C initially uncharged; connect switch to a at t=0
• Loop theorem
•Convert to differential equation for Q:
a
b
R
C
II
Calculate current and
charge as function of time.
dt
dQI
C
Q
dt
dQR
Would it matter where R is placed in the loop??0
QIR
C
RC Circuits(Time-varying currents)
• Guess solution:
•Check that it is a solution:
00 Qt
CQt
Note that this “guess” incorporates the
boundary conditions:
a
b
R
C
II
dQ QR
dt C
• Charge capacitor:
/ 1t RCdQC e
dt RC
)1(/ RC
tRCt ee
C
Q
dt
dQR !
(1 )t
RCQ C e
RC Circuits(Time-varying currents)
• Current is found from differentiation:
• Charge capacitor: a
b
R
C
II
/1 t RCQ C e
/t RCdQI e
dt R
Conclusion:
• Capacitor reaches its final charge(Q=C) exponentially with time constant = RC.
• Current decays from max (=/R) with same time constant.
Charging Capacitor
Q
0
C Charge on C
Max = C
63% Max at t=RC
/1 t RCQ C e
t
RC 2RC
I
0t
R /t RCdQ
I edt R
Current
Max =/R
37% Max at t=RC
1
Lecture 11, ACT 1• At t=0 the switch is thrown from position b to
position a in the circuit shown: The capacitor is initially uncharged.
– At time t=t1=, the charge Q1 on the capacitor is (1-1/e) of its asymptotic charge Qf=C.
– What is the relation between Q1 and Q2 , the
charge on the capacitor at time t=t2=2 ?
a
b
R
C
II
R
(a) Q2 < 2Q1 (b) Q2 = 2Q1 (c) Q2 > 2Q1
•From the graph at the right, it is clear that the charge increase is not as fast as linear.•In fact the rate of increase is just proportional
to the current (dQ/dt) which decreases with time.
•Therefore, Q2 < 2Q1.
Q
Q2
2Q1
Q1
• The point of this ACT is to test your understanding of the exact time dependence of the charging of the capacitor.
•So the question is: how does this charge increase differ from a linear increase?
)1( 2RCt
eCQ
• Charge increases according to:
RC Circuits (Time-varying currents)
• Discharge capacitor:
C initially charged with Q=C
Connect switch to b at t=0.
Calculate current and charge as function of time.
• Convert to differential equation for Q:
C
a
b+ +
- -
R
I I
dt
dQI 0
C
Q
dt
dQR
0C
QIR• Loop theorem
RC Circuits(Time-varying currents)
• Guess solution:
Q = Ce-t/RC
• Check that it is a solution:
/ 1e t RCdQ
Cdt RC
CQt 00 Qt
Note that this “guess” incorporates the
boundary conditions:
C
a
b+ +
- -
R0
dQ QR
dt C
I I• Discharge capacitor:
!RdQdt
QC
e et RC t RC / / 0
RC Circuits(Time-varying currents)
Conclusion: • Capacitor discharges
exponentially with time constant = RC
• Current decays from initial max value (= -/R) with same time constant
• Discharge capacitor: a
Q = Ce-t/RC
C
b+
- -
R+
I I
• Current is found from differentiation:
/t RCdQI e
dt R
Minus sign:original definitionof current “I” direction
Discharging Capacitor
Charge on C
Max = C
37% Max at t=RC
Q = Ce-t/RC
/t RCdQ
I edt R
Current
Max = -/R
37% Max at t=RC
t
Q
0
C RC 2RC
0
-/R
I
t
zero2
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
Preflight 11:
Initially, the charges on the two capacitorsare the same. But the two circuits have different time constants:1 = RC and 2 = 2RC. Since 2 > 1 it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor is bigger than that on capacitor 1.
Lecture 11, ACT 2• At t=0 the switch is connected to
position a in the circuit shown: The capacitor is initially uncharged.– At t = t0, the switch is thrown from
position a to position b.
– Which of the following graphs best represents the time dependence of the charge on C?
C
a b
R 2R
(a) (b) (c)
t
q
0
C
0 1 2 3 40
0.5
1
t/RC
Q f( )x
x
t0 t
q
0
C
0 1 2 3 40
0.5
1
t/RC
Q f ( )x
x
t 0 0 1 2 3 40
0.5
1
t/RCQ f ( )x
x
q
0
C
tt0
• For 0 < t < t0, the capacitor is charging with time constant = RC• For t > t0, the capacitor is discharging with time constant = 2RC
• (a) has equal charging and discharging time constants• (b) has a larger discharging t than a charging • (c) has a smaller discharging t than a charging
Charging DischargingRC 2RC
t
t
Q
0
C
I
0
/R
/1 t RCQ C e
/t RCdQI e
dt R
RC
t
2RC
0
-/R
I
t
Q
0
C
Q = C e -t/RC
/t RCdQI e
dt R
A very interesting RC circuitI1
I3
I2
R2C
R1
First consider the short and long term behavior of this circuit.
• Short term behavior:
Initially the capacitor acts like an ideal wire. Hence,
and
•Long term behavior:
Exercise for the student!!
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)
The circuit below contains a battery, a switch, a capacitor and two resistors
Preflight 11:
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)What is voltage across C after a long time? C is in parallel with R2 !!
VC = I1R2 = E R2/(R1+R2) < E
Very interesting RC circuit continued
I1
I3
I2
R2C
R1
Loop 1
Loop 2
• Node:
• Loop 1: 1 1 0Q
I RC
• Loop 2: 2 2 1 1 0I R I R
321 III
• Eliminate I1 in L1 and L2 using Node equation:
• Loop 1: 1 2 0Q dQ
R IC dt
• Loop 2: 2 2 1 2 0dQ
I R R Idt
eliminate I2 from this
• Final differential eqn:1 1 2
1 2
dQ Q
R dt R RC
R R
Very interesting RC circuit continued
• Try solution of the form:
– and plug into ODE to get parameters A and τ
• Final differential eqn:
1
21
21 RC
RRRR
Q
dt
dQ
time constant: parallel combination
of R1 and R2
/1)( teAtQ
• Obtain results that agree with initial and final conditions:
CRR
RR
21
212
1 2
RA C
R R
I1
I3
I2
R2C
R1
Loop 1
Loop 2
Very interesting RC circuit continued
• What about discharging?– Open the switch...
• Loop 1 and Loop 2 do not exist!
• I2 is only current
• only one loop– start at x marks the spot...
I2 R2
C
R1
2 2 20Q dQ
I R but IC dt
Different time constant for discharging
I1
I3
I2
R2C
R1
Loop 1
Loop 2
Summary• Kirchoff’s Laws apply to time dependent circuits they
give differential equations!• Exponential solutions
– from form of differential equation• time constant = RC
– what R, what C?? You must analyze the problem!
• series RC charging solution
• series RC discharging solution
Next time: Start Magnetism
Q = C e -t/RC
/1 e t RCQ C
Reading assignment: Ch. 28.1-2, 28.4Examples: 28.1,4,5 and 6