Lecture 27 inductors. stored energy. lr circuits

1. Lecture 27
Inductors. Stored energy.
LR circuits.

2. Induction between two coils
Change the current in coil 1
changes the B-flux through coil 2
induces an emf in coil 2

3. Mutual inductance
di1
d Φ2
µ −
ε 2 = −N2
dt
dt
ε 2 = −M21
di1
dt
M21 =
N2 ΦB 2
i1
M21 mutual inductance
indicates how large an emf in 2
due to current change in 1
Of course it works
both ways:
di
ε 1 = −M12 2
dt
N1 ΦB 1
M12 =
i2

4. Mutual inductance depends on the geometry, orientation and
materials of the coils. It can be shown that M21 = M12 ≡ M
Therefore, we have:
ε 1 = −M
di2
dt
ε 2 = −M
N2 ΦB 2 N1 ΦB 1
M=
=
i1
i2
Units: SI
Henry 1 H = 1 Wb/A
di1
dt

5. ACT: Mutual inductance
If the current in coil 1 is as shown, which
of the graphs gives the correct emf in
coil 2?
i1
t
ε2
ε2
ε2
t
A
t
B
t
C

6. Example: M for two solenoids
l
N2 turns, length l
N1 turns, length l
l
µ0N2i2
B2 = µ0n2i2 =
l
Φ12 =
M=
µ0N2i2
π R12
l
N1 Φ12
i2
µ0N2i2
N1 (
π R12 )
l
=
i2
µ0N1N2π R12
M=
l
Craig Ogilvie

7. Self-inductance
Ideal coil (no resistance)
di/dt
If current through coil changes,
→ flux through coil changes
→ in each loop there is an induced emf
→ loops “in series” ⇒ emf induced between two ends of coil
Single coil is called an inductor
The effect is called self-induction
ε = −L
di
dt
N ΦB
L=
i

8. ACT: Inductor
In the circuit shown, the voltage in the power supply is turned up, so
that the current increases. During this operation, which point, y or
z, is at a higher potential?
A. y
B. z
C. Both have the same potential
By Lenz law, induced emf tries to oppose change. Current is
increasing so emf tries to reduce it. Imagine a “battery” with
positive terminal near y.
εin

9. ACT: Inductor II
In the circuit shown, the voltage in the power supply has been on for
a long time. The switch is then opened, as shown. Right after this,
which point, a or b, is at a higher potential?
A. a
B. b
C. Both have the same potential
εin
By Lenz law, induced emf tries to oppose change. Current is
decreasing so emf tries to increase it. Imagine a “battery” with
positive terminal near a.

10. In-class example: Inductance of a solenoid
A solenoid is constructed out of 1000 loops of wire wound around a 1.0 cm 2
cross section tube that is 5 cm long. A 20-A current is run through the
solenoid. What is the solenoid’s self inductance L?
µ0Ni
B = µ0ni =
l
A. 250 H
B. 25 H
C. 2.5 H
D. 2.5×10−3 H
E. 2.5×10−5 H
length l, N turns
µ0Ni π R 2
Φ = BA =
l
µ0Ni π R 2
N(
)
NΦ
l
L=
=
i
i
2

T m
4π × 10 −7
( 1000 ) 1.0 × 10 −4 m2

÷
A 

L=
= 2.5 × 10 −3 H
5.0 × 10 −2 m
(
)
µ0N 2π R 2
L=
l
Inductance of
a solenoid

11. Inductors against di/dt
Inductor always acts to oppose current change.
→ setting up a current in a circuit with an inductor will take some
time.
→ used in circuit design to protect against rapid changes of
currents or spikes

12. RL circuits: current growth
Switch moved to position “a” at t = 0
induced
emf
Kirchhoff’s law ε − iR − L
di
=0
dt
i (t = 0 ) = 0
Qualitatively: i ( t = 0 ) = 0
ε
i (t → ∞ ) =
R
drop in potential from
top of inductor to bottom
Differential equation for i
+ initial condition
(no inductor)

13. di
ε − iR − L
=0
dt
di
R
= dt
ε
L
−i
R
ε
 −i
− ln  R
 ε

 R

÷ R
÷= t
÷ L
÷

 R 
R
ln  1 − i ÷ = − t
ε 
L

i
ε/R
ε
i =
R
t
ε
i (t ) =
R
R

− t
1 − e L


t

−
1 − e τ


If L is large, τ is large, i.e. current grows slowly.

÷
÷


L
÷ τ =
÷
R


14. In-class example: RL circuit
You have 1000 Ω resistor and you want to build an LR series circuit
that will go from 0 to 0.9 V0/R in 1 ms once the switch is closed. What
value should L have?
R
V0
A. 1 H
B. 0.43 H
L
ε
i (t ) =
R
0.9 = 1 − e
τ =−
C. 0.1 H
D. 0.043 H
E. 0.01 H
t

−
1 − e τ


τ =
L
R
−
1 ms

÷
÷

1 ms
τ
ln ( 0.1 )
= 0.43 ms
L = R τ = 0.43 H

15. ACT: Three circuits
All batteries, inductors, resistors are identical.
Rank the circuits according to the current through battery just
after switch is closed
1
2
A. i1 > i2 > i3 initially
B. i2 > i3 > i1 initially
C. i3 > i2 > i1 initially
3
Initially i = 0 through inductors.
⇒ i1 = 0
R2 < R3 ⇒ i2 > i3

16. DEMO:
ACT: Current decay
RL circuit
The switch has been in position “a” for
a long time, then it is moved to position
“b” at t = 0. What is the graph for
current in the resistor?
induced
emf
i
i
A
B
t
t
−iR − L
di
=0
dt
 di

< 0÷

 dt

i
→
di
R
=− i
dt
L
C
D
i
t
i (t ) = i0e
t
t
−
τ
τ =
L
R

17. Inductors store energy
i
t
After the switch is opened, the inductor drives the
current through the resistor.
The energy dissipated in the resistor must have
been stored in the inductor.

18. Energy stored in an inductor
To establish current (from i = 0 to i = I ):
di
Voltage across inductor: V = L
dt
(ignoring signs)
1) rate of energy supplied to inductor P = Vi = Li
di
dt
2) increase of energy within inductor dU = Pdt = Li di
I
3) energy stored within inductor U = ∫ dU = L ∫ idi
0
1
U = LI 2
2

19. Using the stored energy
•
•
•
•
DEMO:
Spark
Energy is stored in L after switch is moved to “a” position
Switching to “b” releases energy stored in inductor
Time it takes for i = I to i = 0 can be very short
Energy released can cause an arc across switch contacts
– spark plug, current was stored in ignition coil (an inductor)
– pulling plug from wall-socket, spark (wires as the inductor)