Download Lecture 21 - UConn Physics

Physics 1502: Lecture 21
Today’s Agenda
• Announcements:
– Chap.27 & 28
• Homework 06: due next Monday
• Induction / AC current
Faraday's Law
dS
B
N
S
v
B
S
N
v
B
B
Induction
Self-Inductance, RL Circuits
I
a
I
R
XXX
XXXX
XX
b
L
e
L/R
e1
1
VL
f( x ) 0.5
0.0183156
0
0
1
2
3
4
t
Recap from the last Chapter:
Faraday's Law of Induction
B
N
S
v
B
S
N
v
Can time varying current in
a conductor induce EMF in
in that same conductor ?
• Time dependent flux is generated by
change in magnetic field strength due
motion of the magnet
• Note: changing magnetic field can
also be produced by time varying
current in a nearby loop
dI/dt
B
Self-Inductance
• The inductance of an inductor ( a set of coils in some
geometry ..eg solenoid, toroid) then, like a capacitor, can be
calculated from its geometry alone if the device is constructed
from conductors and air.
• If extra material (eg iron core) is added, then we need to add
some knowledge of materials as we did for capacitors
(dielectrics) and resistors (resistivity)
SI UNITS for L :
Henry
• Archetypal inductor is a long solenoid, just as a pair of parallel
plates is the archetypal capacitor.
l
r
N turns
r << l
A
++++
d -----
Calculation
l
• Long Solenoid:
N turns total, radius r, Length l
r
N turns
For a single turn,
The total flux through solenoid is given by:
Inductance of solenoid can then be calculated as:
This (as for R and C) depends only
on geometry (material)
RL Circuits
• At t=0, the switch is closed and
the current I starts to flow.
a
I
I
R
b
• Loop rule:
e
Note that this eqn is identical in form to that for the RC
circuit with the following substitutions:
 RCRL:
RC:
\

L
Lecture 21, ACT 1
• At t=0 the switch is thrown from position b to
position a in the circuit shown:
1A – What is the value of the current I a long
time after the switch is thrown?
I
a
I
R
b
L
e
R
(a) I = 0
(b) I = e / 2R
(c) I = 2e / R
1B • What is the value of the current I0 immediately after the switch
is thrown?
(a) I0 = 0
(b) I0 = e / 2R
(c) I0 = 2e / R
RL Circuits
• To find the current I as a fct of
time t, we need to choose an
exponential solution which
satisfies the boundary
condition:

a
I
I
R
b
e
L
RL = R
• We therefore write:
• The voltage drop across the inductor is given by:
L
RL Circuit (e on)
Current
Max = e/R
L/R
e/
R
2L/R
I
63% Max at t=L/R
0
Voltage on L
Max = e/R
t
e
VL
37% Max at t=L/R
0
t
RL Circuits
• After the switch has been in
position a for a long time,
redefined to be t=0, it is moved to
position b.
• Loop rule:
• The appropriate initial condition is:
• The solution then must
have the form:
a
I
I
R
b
e
L
RL Circuit (e off)
Current
Max = e/R
e/R
L/R
2L/R
I
37% Max at t=L/R
0
Voltage on L
Max = -e
t
0
VL
37% Max at t=L/R
-e
t
e on
e/R
L/R
e off
2L/R
e/R
I
L/R
2L/R
I
0
t
t
0
e
VL
VL
0
0
-e
t
t
Review: RC Circuits
(Time-varying currents)
I
a
I
• Discharge capacitor:
C initially charged with Q=Ce
Connect switch to b at t=0.
Calculate current and charge
as function of time.
b
e
• Loop theorem 
• Convert to differential equation for q:

R
+ +
C
- -
Review: RC Circuits
(Time-varying currents)
I
a
I
• Discharge capacitor:
b
e
• Trial solution:
R
+ +
C
- -
q = C ee -t/RC
• Check that it is a solution:
Note that this “guess”
incorporates the
boundary conditions:

!
Review: RC Circuits
(Time-varying currents)
I
a
• Discharge capacitor:
q = C ee -t/RC
b
• Current is found from
differentiation:

I
R
+ +
C
e
- -
Conclusion:
• Capacitor discharges
exponentially with time constant
= RC
• Current decays from initial max
value (= -e/R) with same time
constant
Discharging Capacitor
Charge on C
Max = Ce
Ce
RC
2RC
q
37% Max at t=RC
0
t
0
Current
I
Max = -e/R
37% Max at t=RC
-e/R
t
Charging
Ce
RC
2RC
Ce
q
0
Discharging
RC
2RC
q
0
t
t
0
e/R
I
I
0
- e/R
t
t
Energy of an Inductor
• How much energy is stored in
an inductor when a current is
flowing through it?
• Start with loop rule:
I
a
I
R
b
e
L
• Multiply this equation by I:
•
From this equation, we can identify PL, the rate at which energy is
being stored in the inductor:
•
We can integrate this equation to find an expression for U, the
energy stored in the inductor when the current = I:

Where is the Energy Stored?
• Claim: (without proof) energy is stored in the Magnetic field
itself (just as in the Capacitor / Electric field case).
• To calculate this energy density, consider the uniform field
generated by a long solenoid:
l
• The inductance L is:
r
N turns
• Energy U:
• We can turn this into an energy density by dividing by the
volume containing the field:
Mutual Inductance
• Suppose you have two coils
with multiple turns close to
each other, as shown in this
cross-section
Coil 1
Coil 2
B
• We can define mutual inductance
M12 of coil 2 with respect to coil 1
as:
It can be shown that :
N1
N2
Inductors in Series
• What is the combined (equivalent)
inductance of two inductors in series,
as shown ?
a
a
L1
Note: the induced EMF of two inductors
now adds:
L2
b
Since:
And:
Leq
b
Inductors in parallel
• What is the combined (equivalent)
inductance of two inductors in
parallel, as shown ?
Note: the induced EMF between
points a and be is the same !
a
L1
a
L2
b
Also, it must be:
We can define:
And finally:
Leq
b
LC Circuits
• Consider the LC and RC
series circuits shown:
C R
C
L
• Suppose that the circuits are
formed at t=0 with the
capacitor C charged to a value Q. Claim is that there is a
qualitative difference in the time development of the currents
produced in these two cases. Why??
•
Consider from point of view of energy!
• In the RC circuit, any current developed will cause energy to be
dissipated in the resistor.
• In the LC circuit, there is NO mechanism for energy dissipation;
energy can be stored both in the capacitor and the inductor!
RC/LC Circuits
i
Q+++
---
i
Q+++
---
C R
0
i
0
1
t
L
LC:
current oscillates
RC:
current decays exponentially
-i
C
0
t
LC Oscillations
(qualitative)
+ +
- -
C
L

C

C
L

L

-
-
+ +
C
L
LC Oscillations
(quantitative)
• What do we need to do to turn our
qualitative knowledge into quantitative
knowledge?
• What is the frequency w of the
oscillations?
+ +
- -
C
L
LC Oscillations
(quantitative)
i
• Begin with the loop rule:
Q
+ +
- -
C
L
• Guess solution: (just harmonic oscillator!)
remember:
where:
• w0 determined from equation
• f, Q0 determined from initial conditions
• Procedure: differentiate above form for Q and substitute into
loop equation to find w0.
Review: LC Oscillations
i
• Guess solution: (just harmonic oscillator!)
Q
where:
+ +
- -
C
L
• w0 determined from equation
• f, Q0 determined from initial conditions
which we could have determined
from the mass on a spring result:
1
Lecture 21, ACT 2
• At t=0 the capacitor has charge Q0; the resulting
oscillations have frequency w0. The maximum
current in the circuit during these oscillations has
value I0 .
– What is the relation between w and w2 , the
1A frequency of oscillations when0 the initial
charge
= 2Q0 ?
(a) w2 = 1/2 w0
(b) w2 = w0
t=0
+ +
Q  Q0
- -
C
(c) w2 = 2 w0
L
Lecture 21, ACT 2
t=0
• At t=0 the capacitor has charge Q0; the
resulting oscillations have frequency
w0. The maximum current in the circuit
during these oscillations has value I0 .
+ +
Q  Q0
- -
C
1B • What is the relation between I0 and I2 , the maximum
current in the circuit when the initial charge = 2Q0 ?
(a) I2 = I0
(b) I2 = 2 I0
(c) I2 = 4 I0
L