Transcript DC-Circuits-II-RC

a
DC Circuits - II
(RC)
I
I
a
I
I
R
C

C
R
b
b
RC
+ +
C

2RC
C
RC
- -
2RC
q = Ce - t / RC
(
q
0
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q =Ce1 -e -t / RC
t
)
q
0
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t
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Resistor-Capacitor circuits
I
Let’s try to add a Capacitor to
our simple circuit
I
R
Recall voltage “drop” on C?

Q
V=
C
Write loop eq:  - IR -
C
Q
=0
C
What’s wrong here?
Consider that I =
dQ
dt
and substitute. Now eqn. has only “Q”:
dQ Q
-R
- =0
dt C
Differential Equation !
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Capacitors Circuits, Qualitative
Basic principle: Capacitor resists rapid change in Q 
resists rapid changes in V
• Charging (it takes time to change the final charge)
– Initially, the capacitor behaves like a wire (V = 0, since Q = 0).
– As current starts to flow, charge builds up on the capacitor
 it then becomes more difficult to add more charge
 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.
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Question
The capacitor is initially
uncharged, and the two
switches are open.
E
What is the voltage across the capacitor immediately after
switch S1 is closed?
a) Vc = 0
b) Vc = E
Initially: Q = 0
VC = 0
I = E/(2R)
c) Vc = 1/2 E
Find the voltage across the capacitor after the switch has
been closed for a very long time.
a) Vc = 0
c)DeSiaMore
Vc = 1/2 E
b) Vc = E
Q=EC
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I=0
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Question
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
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E
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.
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RC Circuits
(Time-varying currents -- discharging)
• Discharge capacitor:
C initially charged with
Q0 = C
Q=
Connect switch to b at t = 0.
Calculate current and charge as
function of time.
Q
IR + = 0
• Loop theorem 
C
I
a
b

I
R
+ +
C
- -
• Convert to differential equation for Q:
Note: Although we know
dQ

the current is flowing off
I =+
the cap., we define it as
dt
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shown
so that …
dQ Q
R + =0
dt C
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Discharging Capacitor
I
a
dQ Q
R
+ =0
dt C
b
I
R
C

• Guess solution:
+ +
- -
Q = Q0e - t/ t = C e- t/ RC
• Check that it is a solution:
Note that this “guess”
incorporates the
boundary conditions:
dQ
1 ￦
- t/ R C ￦
= C e
￦
dt
￦ RC ￦
￦

dQ Q
R dt + = -  e t / RC +  e t / RC = 0
C
DeSiaMore
!
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t =0 ￞ Q =C
t= ￦ ￦ Q = 0
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Discharging Capacitor
• Discharge capacitor:
I
a
Q = Q0e - t/ t = C e- t/ RC
• Current is found from
differentiation:
dQ
 - t/ RC
I=
=- e
dt
R

Minus sign:
Current is opposite to
original definition,
i.e., charges flow away
from capacitor.
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b
I
R
+ +
C

- -
Conclusion:
• Capacitor discharges
exponentially with time constant
t = RC
• Current decays from initial max
value (= -/R) with same time
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constant
Discharging Capacitor
C
Charge on C
RC
2RC
Q = C e - t/ R C
Max = C
Q
37% Max at t = RC
0
zero
t
0
Current
dQ

I=
= - e- t/ RC
dt
R
I
“Max” = -/R
37% Max at t = RC
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-/R
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Question
The two circuits shown below contain identical fully charged capacitors at
t=0. Circuit 2 has twice as much resistance as circuit 1.
Compare the charge on the two capacitors a short time after t = 0
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
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Initially, the charges on the two capacitors are the
same. But the two circuits have different time
constants:
t1 = RC and t2 = 2RC. Since t2 > t1 it takes circuit 2
longer to discharge its capacitor. Therefore, at any
given time, the charge on capacitor 2 is bigger than
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that on
capacitor 1.
RC Circuits
(Time-varying currents, charging)
I
a
• Charge capacitor:
I
R
C initially uncharged; connect
switch to a at t=0
b

Calculate current and
C
charge as function of time.
• Loop theorem 
Q
 - IR - = 0
C
•Convert to differential equation for Q:
I=
DeSiaMore
dQ
dt

=R
Would it matter where R
is placed in the loop??
NO!
dQ Q
+
dt C
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Charging Capacitor
I
a
• Charge capacitor:
I
R
=R
dQ Q
+
dt C
b

• Guess solution:
C
-t
Q = C (1 - e RC )
•Check that it is a solution:
dQ
1 ￦
- t/ RC ￦
= -Ce
￦
dt
￦ RC ￦
￦

-t
dQ Q
- t/ RC
R
+ = e
+  (1 - e RC ) = 
dt C
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Note that this “guess”
incorporates the
boundary conditions:
t= 0 ￦ Q = 0
t =￞ ￞ Q =C
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Aside…
Rather than repeatedly solving the differential equation use
the general purpose solution:
-t
Q = A + Be
t
or V = A + Be
-t
t
or I = A + Be
-t
t
Where A and B are constants (with the appropriate units) and
t is the RC time constant
Determine the constants from the boundary conditions at t = 0 and t=infinity.
For example, from the last question;
t= 0 ￦ Q = 0
t =￞ ￞ Q =C
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Q (0 ) = 0 = A + B
Q ( ￦ ) = C = A
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Charging Capacitor
• Charge capacitor:
I
a
I
R
Q = C (1- e- t/ RC )
b

• Current is found from
differentiation:
dQ  - t/ RC
I=
= e
dt R
DeSiaMore
C

Conclusion:
• Capacitor reaches its final
charge(Q=C) exponentially
with time constant t = RC.
• Current decays from max
(=/R) with same time
constant.
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Charging Capacitor
Charge on C
RC
2RC
C
Q = C (1- e- t/ RC )
Max = C
Q
63% Max at t = RC
0
Current
dQ  - t/ RC
I=
= e
dt R
Max = /R
t
R
I
37% Max at t = RC
0
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t
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Question
a
• At t=0 the switch is thrown from position b to
position a in the circuit shown: The capacitor is
initially uncharged.
I
R
b

– At time t = t1 =t, 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 = 2t ?
(a) Q2 < 2Q1
I
C
R
(c) Q2 > 2Q1
(b) Q2 = 2Q1
• The point is to test your understanding of the time dependence of the
charging of the capacitor.
• Charge increases according to:
Q = C (1 - e
-t
2RC
)
• So the question is: how does this charge
increase differ from a linear increase?
• 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.
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• Therefore,
Q2 < 2Q1.
2Q1
Q2
Q1
Q
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t
t
Charging
C
RC
Discharging
2RC
C
Q = C (1- e- t/ RC )
Q
t
0
dQ  - t/ RC
I=
= e
dt R
I
DeSiaMore
0
Q
t
/R
2RC
Q = C e - t/ R C
0
0
RC
I
-/R
dQ
 - t/ RC
I=
=- e
dt
R
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t
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t
Question
I1
This circuit contains a battery,
a switch, a capacitor and two
resistors
I2
I3

C
R2
R1
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)
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Question
• At t = 0 the switch is closed in the
circuit shown. The initially
uncharged capacitor then begins
to charge.
I1
I2
I3

C
R2
R1
• What will be the voltage across the capacitor a long time after the
switch is closed?
(a) VC = 0
(b) VC =  R2/(R1+ R2)
(c) VC = 
After a long time the capacitor is completely charged, so no current
flows through it. The circuit is then equivalent to a battery with two
resistors in series. The voltage across the capacitor equals the
voltage across R2 (since C and R2 are in parallel). Either from direct
calculation, or remembering the “Voltage Divider Circuit”, VC = VR2 =
 RDeSiaMore
2/(R1+ R2).
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Question
• At t = 0 the switch is closed in the circuit
shown. The initially uncharged capacitor

then begins to charge.
– What is the charging time
constant t ?
(a) t = R1C
(b) t = ( R1 + R2 )C
I1
I2
I3
C
R1
￦ R1R2 ￦
(c) t = ￦
C
￦
￦ R 1 + R2 ￦
• An ideal voltage source contributes no resistance or capacitance
 time constant is entirely determined by C, R1, and R2.
•It might be easier to think about the circuit as if C were discharging; Imagine
that the capacitor is charged, and that the battery is replaced by a wire (which
also has no resistance or capacitance). Since the battery supplies a constant
voltage, it doesn’t affect the time constant.
• We need to find the effective resistance Reff through which the capacitor
(dis)charges. Looking at the new circuit, it is clear that the capacitor would be
(dis)charging
through both R1 www.desiamore.com/ifm
and R2, which are in parallel  their effective
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resistance is Reff = R1R2/(R1 + R2) and t = ReffC.
R2
Summary
• Kirchoff’s Laws apply to time dependent circuits they
give differential equations!
• Exponential solutions
-t
t
A
+
Be
– from form of differential equation
• time constant
t = RC
– what R, what C ?? You must analyze the problem!
SPECIAL CASES
• series RC charging solution Q = Q( t = ￦ ) 1 - e- t/ RC
(
)
• series RC discharging solution
DeSiaMore
Q = Q( t = 0) e - t/ RC
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