Transcript Inductance, RL Circuits

Physics 121 - Electricity and Magnetism
Lecture 12 - Inductance, RL Circuits
Y&F Chapter 30, Sect 1 - 4
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Inductors and Inductance
Self-Inductance
RL Circuits – Current Growth
RL Circuits – Current Decay
Energy Stored in a Magnetic Field
Energy Density of a Magnetic Field
Mutual Inductance
Summary
Copyright R. Janow – Fall 2014
Induction: basics
•
Magnetic Flux:

 
dB  B  dA  B  n̂dA

B
•
Faraday’s Law: A changing magnetic
flux through a coil of wire induces an
EMF in the wire, proportional to the
number of turns, N.
•
Lenz’s Law: The current driven by an induced EMF creates
an induced magnetic field that opposes the flux change.
Eind  N
n̂
d B
dt
Bind & iind oppose changes in B
•
Induction and energy transfer: The forces on the loop
oppose the motion of the loop, and the power required to
sustain motion provides electrical power to the loop.
 
P  F  v  Fv
P  i
  Blv
•
Transformer principle: changing current i1 in primary
induces EMF and current i2 in secondary coil.
•
Generalized Faraday Law:
A
changing magnetic flux creates
non-conservative electric field.
 
   E  ds   N dB
dt
Copyright R. Janow – Fall 2014
Example: Find the electric field induced by
changing magnetic flux


 B  ds  0ienc

dB
Eind  ds  
loop
dt

Eind 
Assume: dB/dt = constant over circular region.
Find the magnitude E of the induced electric field at
points within and outside the magnetic field.
Due to symmetry and Gauss’ Law, E is tangential:


Eind   E  ds   Eds  E ds  E(2r )
For r < R:
So
B  BA  B(r 2 )
For r > R:
So
B  BA  B(R 2 )
dB
E(2r )  R 2
dt
E(2r )  r 2
dB
dt
E
r dB
2 dt
R 2 dB
E
2r dt
The magnitude of induced electric field grows
linearly with r, then falls off as 1/r for r>R
Copyright R. Janow – Fall 2014
Example: EMF generated Faraday Disk Dynamo
Conducting disk, radius R, rotates at rate w in uniform constant field B,
FLUX ARGUMENT:
Eind  
dB
dA
 B
dt
dt
areal velocity
dA = area swept out by radius vector in dq
= fraction of full circle in dq x area of circle
dq
R2
2
dA R 2 dq R 2
dA 
R 
dq



w
2
2
dt
2 dt
2
Eind  
w, q
1
BwR 2
2
+
USING MOTIONAL EMF FORMULA:
Emf induced across
conductor length ds
 

dE  v  B  ds
Moving conductor sees vXB as electric field E’
For points on rotating disk: v = wr,
Eind  
R
0

 
v  B  ds 
(Equation 29.7)

R
0
vXB = E’ is radially outward, ds = dr
Bwrdr 
1
BwR 2
2
current flows
radially out
Copyright R. Janow – Fall 2014
Self-Inductance: Analogous to inertia
•ANY magnetic flux change is resisted
•Changing current in a single coil induces a “back EMF”
Eind in the same coil opposing the current change, an
induced current iind, and a consistent induced field Bind.
N
TWO RELATED FARADAYS LAW EFFECTS:
•Mutual-induction: di1/dt in “transformer primary” induces EMF and current i2 in
“linked” secondary coil (transformer principle).
•Self-induction in a single Coil: di/dt produces “back EMF” due to Lenz &
Faraday Laws: ind opposes d/dt due to current change. Eind opposes di/dt.
• Changing current in a single coil generates magnetic field and flux
• Flux Change induces flux opposing the change, along with opposing EMF and current.
• This back EMF limits the rate of current (and therefore flux) change in the circuit
• For increasing current,
back EMF limits the
rate of increase
• For decreasing current,
back EMF sustains the
current
Inductance measures oppositon to the rate of change
ofJanow
current
Copyright R.
– Fall 2014
Definition of Self-inductance
Recall capacitance: depends only on geometry
It measures charge stored per volt
Self-inductance depends only on coil geometry
It measures flux created per ampere
number of turns
L
self-inductance
SI unit of
inductance:
N B
i
C
L
Q
V
linked flux
unit current
Joseph Henry
1797 – 1878
flux through one turn depends
on current & all N turns
cancels current dependence in flux above
2
1 Henry  1 H.  1 T.m / Ampere  1 Weber / Ampere
 1 Volt.sec / Ampere (.sec)
Why choose
this definition?
Cross-multiply……. Take time derivative
Li  N B
di
EL   L
dt
di
d B
L
N
 - EL
dt
dt
• L contains all the geometry
• EL is the “back EMF”
Another form of Faraday’s
Law!
Copyright
R. Janow – Fall 2014
Example: Find the Self-Inductance of a solenoid
Field:
+
-
L
B  μ0in where n 
Flux in just
ΦB
one turn:
RL
NA
 BA  μ0i

Apply definition of self-inductance:
 N turns
 Area A
 Length l
 Volume V = Al

•
NΦB
N2 A
2
L
 μ0
 μ0n V
•
i

N
# turns


unit Length
ALL N turns
contribute to selfflux through ONE
turn
Depends on geometry
only, like capacitance.
Proportional to N2 !
Check: Same L if you start with Faraday’s Law for B:
Eind  N
dΦB
dt
for solenoid use
Eind
B
above
2
di
 NA di   μ0N A di
 N.  μ0

 L

 dt 

dt
dt

Note: Inductance per unit length
has same dimensions as 0
L
A
 μ0N2 2


T.m
H

A. – Fall 2014
m
Copyright R. Janow
[ 0 ] 
Example: calculate self-inductance L for an ideal solenoid
N  1000 turns, radius r  0.5 m, length   0.2 m
l
L 

2
μ 0N A

L  49.4  10
Ideal inductor
(abstraction):
3
7
6
2 2
 10  π  (5  10 )
0.2
Henrys 49.4 m illi - Henrys
 Internal resistance r  0 (recall ideal battery)
 B  0 outside
 B  μ0in inside (ideal solenoid)
Non-ideal inductors have
internal resistance:
r

4π  10
L
 VIND  EL  ir  measured voltage
 Direction of ir depends on current
 Direction of E depends on di/dt
L
 If current i is constant, then induced E  0
L
Vind
Inductor behaves like a wire with resistance r
Copyright R. Janow – Fall 2014
Induced EMF in an Inductor
12 – 1: Which statement describes the current through the inductor
below, if the induced EMF is as shown?
A.
B.
C.
D.
E.
Rightward and constant.
Leftward and constant.
Rightward and increasing.
Leftward and decreasing.
Leftward and increasing.
L 
di
EL   L
dt
Copyright R. Janow – Fall 2014
Lenz’s Law applied to Back EMF
i
+
If i is increasing:
EL
Di / Dt
-

EL opposes increase in i
Power is being stored in B field of inductor
If i is decreasing:
-
EL
+
 0
dt
i
Di / Dt
dΦB

dΦB
 0
dt
EL opposes decrease in i
Power is being tapped from B field of inductor
What if CURRENT i is constant?
Copyright R. Janow – Fall 2014
Example: Current I increases uniformly from 0 to 1 A. in
0.1 seconds. Find the induced voltage (back
EMF) across a 50 mH (milli-Henry) inductance.
+
EL
i defines positive direction
-
di
dt
i
and tow ard the right
i
Apply:
EL   L
di
Substitute:
dt
EL  50 m H  10
 0 m eansthat current i is increasing
Am p
se c
Δi
Δt
  0.5 Volts

 1 Am p
0.1 s ec
 10
Am p
s ec
Negative result
means that induced
EMF is opposed to
both di/dt and i.
Copyright R. Janow – Fall 2014
Inductors in Circuits—The RL Circuit
• Inductors, sometimes called “coils”, are common circuit components.
• Insulated wire is wrapped around a core.
• They are mainly used in AC filters and tuned (resonant) circuits.
Analysis of series RL circuits:
E
•
A battery with EMF
drives a current around the loop
•
•
Changing current produces a back EMF or sustaining EMF EL in the inductor.
Derive circuit equations: apply Kirchoff’s loop rule, convert to differential
equations (as for RC circuits) and solve.
New rule: when traversing an inductor in the same
direction as the assumed current, insert:
di
EL   L
Copyright R. Janow – Fall
dt2014
Series LR circuits
+
i
a
R
E
-
b
L
i
EL
•
•
•
•
Inductance & resistance + EMF
Find time dependent behavior
Use Loop Rule & Junction Rule
Treat EL as an EMF along current
di
EL   L
dt
ALWAYS
NEW TERM FOR
LOOP RULE
Given E, R, L: Find i, EL, UL for inductor as functions of time
Growth phase, switch
to “a”. Loop equation:
di
E  iR  L
 0
dt
Decay phase, switch
to “b”, exclude E,
Loop equation:
di
 iR  L
 0
dt
• i through R is clockwise and growing: EL opposes E
• At t = 0, rapidly growing current but i = 0, EL= E
L acts like a broken wire
• As t  infinity, large stable current, di/dt  0
Back EMF EL 0, i  E / R,
L acts like an ordinary wire
• Energy is stored in L & dissipated in R
• Energy stored in L now dissipated in R
• Current through R is still clockwise, but collapsing
• EL now acts like a battery maintaining current
• Current i at t = 0 equals E / R
• Current  0 as t  infinity – energy depleted
Copyright R. Janow – Fall 2014
LR circuit: decay phase solution
R
• After growth phase equilibrium, switch from a to b, battery out
• Current i0 = E / R initially still flowing CW through R
• Inductance tries to maintain current using stored energy
• Polarity of EL reverses versus growth. Eventually EL 0
Loop Equation is :
- iR  EL  0
EL (t )   L
Substitute :
di
dt
b
EL
L
+
i
Circuit Equation:
di
R
  i
dt
L
di/dt <0
during decay,
opposite to
current
• First order differential equation with simple exponential solution
• At t = 0: large current implies large di / dt, so EL is large
• As t  infinity: current stabilizes, di / dt and current i both  0
Current decays exponentially:
i(t )  i0e  t / tL
tL  L/R
i0 
E
i0
e 1  0.37
i
R
 inductive time constant
EMF EL and VR also decay exponentially:
di
E  t / tL

e

dt
RtL
t
2t
3t
t
 t / tL
Compare to RC circuit, decay
VR  E L iR
RC Copyright
 capacitive
time cons tan t
R. Janow – Fall 2014
EL  E e
Q(t )  CEe  t / RC
LR circuit: growth phase solution
Loop Equation is :
Substitute :
E - iR  EL  0
EL (t )   L
Circuit Equation:
L di
E
i 

R dt
R
di
dt
• First order differential equation again - saturating exponential solutions
• As t  infinity, di / dt approaches zero, current stabilizes at iinf = E / R
• At t = 0: current is small, di / dt is large, back EMF opposes battery.
Current starts from zero, grows as a saturating exponential.

i(t )  iinf 1  e
tL  L/R
 t / tL

iinf 
iinf
E
R
 inductive time constant
• i = 0 at t = 0 in above equation  di/dt = E/L
fastest rate of change, largest back EMF
Back EMF EL decays exponentially
di
E

e  t / tL  E L   E e  t / tL
dt R tL
Voltage drop across resistor VR= -iR
i
1  e 1  0.63
t
2t
3t
t
Compare to RC circuit, charging


Q(t )  CE 1  e  t / RC
RC  capacitive time cons tan t
Copyright R. Janow – Fall 2014
Example: For growth phase find back EMF EL as a function of time
Use growth phase solution i(t) 
E
(1  e  t / tL )
R
At t = 0: current = 0
S
+
di E (-)(-)  t / tL

e
dt R tL



where
t
L
tL 
R
L acts like a broken wire at t = 0
L

L
R

0.1H
1Ω
 0.1 s e c
t
At t  0 : EL   E
Back EMF EL equals the battery potential
causing current i to be 0 at t = 0
iR drop across R = 0
EL
i
-
Back EMF is ~ to rate of change of current
di
EL  L
  Ee  t / tL ;
dt
L  0.1H
E  5V
E
i(t  0)  (1  e0 )  0
R
Derivative :
R  1Ω
EL
- 0.37 V0
-E
After a very long (infinite) time:
 Current stabilizes, back EMF=0
E
i 
 5A
R
 L acts like an ordinary wire at t = infinity
Copyright R. Janow – Fall 2014
Current through the battery - 1
12 – 2: The three loops below have identical inductors, resistors,
and batteries. Rank them in terms of current through the battery
just after the switch is closed, greatest first.
A.I, II, III.
B.II, I, III.
C.III, I, II.
D.III, II, I.
E.II, III, I.
I.

II.
III.
Hint: what kind of wire does L act like?
i(t )  iinf 1  e  t / tL

iinf 
E
R eq
tL  L/R eq
 inductive time constant
Copyright R. Janow – Fall 2014
Current through the battery - 2
12 – 3: The three loops below have identical inductors, resistors,
and batteries. Rank them in terms of current through the battery a
long time after the switch is closed, greatest first.
A.
B.
C.
D.
E.
I, II, III.
II, I, III.
III, I, II.
III, II, I.
II, III, I.
I.

II.
III.
Hint: what kind of wire does L act like?
i(t )  iinf 1  e  t / tL

iinf 
E
R eq
tL  L/R eq
 inductive time constant
Copyright R. Janow – Fall 2014
Summarizing RL circuits growth phase

• When t is large: i  
Inductor acts like a wire.
i  (1  e  Rt / L )
•
R
When t is small (zero), i = 0.
Inductor acts like
an open circuit.
•
The current starts from zero and
increases up to a maximum of i   / R
with a time constant given by
L
Inductive time constant
tL 
R
Compare: tC  RC Capacitive time constant
•
The voltage across the resistor is
R
VR  iR  (1  eRt / L )
•
The voltage across the inductor is
VL    VR    (1  eRt / L )   eRt / L
Copyright R. Janow – Fall 2014
Summarizing RL circuits decay phase
The switch is thrown from a to b
•
Kirchoff’s Loop Rule for growth
was:
  iR  L di  0
dt
•
Now it is:
•
di
0
dt
The current decays exponentially:
iR  L
i
R
Voltage across resistor also decays:
VR  iR   eRt / L
•
VR (V)
•
 e Rt / L
Voltage across inductor:
VL  L
di
 d eRt / L   eRt / L
 L
dt
R dt
Copyright R. Janow – Fall 2014
Energy stored in inductors
Recall: Capacitors store energy in their electric fields
UE  electrostatic potentialenergy
UE 
2
1Q
2 C
1
2
 CV 2
uE 
electrostatic energy density
derived
UE
1
2
using p-p
uE 
  E
Volume 2 0
capacitor
Inductors also store energy, but in their magnetic fields
Magnetic PE Derivation – consider power into or from inductor
dUB
di
1
Power 
 ELi  Li
 UB   dUB  L  idi  Li2
2
dt
dt
UB  magnetic potentialenergy
1 2
UB  Li
2
uB  magnetic energydensity
B2
uB 

Volume
2
UB
0
derived
using
solenoid
• UB grows as current increases, absorbing energy
• When current is stable, UB and uB are constant
• UB diminishes when current decreases. It powers
the persistent EMF during the decay phase for the inductor
Copyright R. Janow – Fall 2014
Sample problem: energy storage in magnetic field
of an inductor during growth phase
a) At equilibrium (infinite time) how much energy is stored in the coil?
i

U



E
R
1
2
(Coil acts like a w ire) 
Li
12
0.35
 34.3 A
2
3
2
 x 53 x 10
x (34.3)

2
U

L = 53
mH
E = 12 V
1
R  0.35 Ω
 31 J
b) How long (t1/2) does it take to store half of this energy?
At t1/2 : UB  21 U
i1/2 
e

i
1
2
 i (1  e
2
 t 1 / 2 / tL

2
Li1/2

1 1

2 2
 t1/2 / tL
 L  i2

i1/ 2 
i
2
)
2
1

 1  1/ 2
2
2
take natural log of both sides
t1 / 2  - tL ln(1  1 /
2 )  1.23 t
tL  L / R  53 x 10-3 / 0.35
 0.15 sec.
Copyright
R. Janow – Fall 2014
Mutual Inductance
• Example: a pair of co-axial coils
• di/dt in the first coil induces current in the
second coil, in addition to self-induced effects.
• M21 depends on geometry only, as did L and C
• Changing current in primary (i1) creates varying
flux through coil 2  induced EMF in coil 2
Definition:
number of turns
in coil 2
mutual
inductance
N2  21
M21 
i1
flux through one turn
of coil 2 due to all N1
turns of coil 1
current in coil 1
cross-multiply
M21 i1  N2 21
time derivative
M21
di1
d 21
 N2
dt
dt
di1
E2   M21
dt
• M21 contains all the
geometry
• E2 is EMF induced
in 2 by 1
Another form of Faraday’s Law!
The smaller coil radius determines how much flux is linked, so…..
di
proof not obvious
E1   M12 2
M12  M21  M
Copyright R. Janow – Fall 2014
dt
Calculating the mutual inductance M
Coil 1 (outer) is a short loop of N1 turns
(not a long Solenoid)
B1  Field inside loop 1 near center

 0i1N1
2R 1
large coil 1
N1 turns
radius R1
(primary)
small coil 2
N2 turns
radius R2
(secondary)
(assume uniform B - use arc formula with   2  )
Flux through inner flat coil 2 (one turn) depends
on smaller area A2 & B1
μ iN
Φ21  B1A2  0 1 1 R 22 (for each loop in coil 2 )
2R1
Current in Loop 1 is changing:
E2
 induced voltage in loop 2
 M21  M 
Summarizing
results for mutual
inductance:
μ0N1N2 R 22 di1
dΦ21
di1
 N2

 M21
dt
2R1
dt
dt
μ0N1N2
R 22
2R 1
E2  - M21
M21 
smaller radius (R2)
determines the linkage
di1
dt
M21 
N2Φ21
i1
N2Φ21 N1Φ12

 MCopyright
12  M R. Janow – Fall 2014
i1
i2
Air core Transformer Example: Concentric coils
OUTER COIL – IDEAL SOLENOID
• coil diameter D = 3.2 cm
• nout = 220 turns/cm = 22,000 turns/m
• current iout falls smoothly from
1.5 A to 0 in 25 ms
• Field within outer coil:
Bout   0 iout (t) nout
Bout
as function of time
 0  4  x 10-7 nout 
Nout
L out
FIND INDUCED EMF IN INNER COIL DURING THIS PERIOD
• coil diameter d = 2.1 cm = .021 m, Ain =  d2/4, short length
• Nin = 130 turns = total number of turns in inner coil
Din  flux change through one turn of inner coil during Dt
DBout Ain
 n Di
Di
Din
Ein  - Nin
 - Nin
 - Nin A in 0 out out  - M out
Dt
Dt
Dt
Dt
 1 .5
 Ein  - 4   10-7  2.2x10 4  130   (.021/ 2)2 
 75 mV
2.5 x10  3
Direction: Induced B is parallel to Bouter which is decreasing
Would the transformer work if we reverse the role of the coils?
Copyright R. Janow – Fall 2014