Lecture 27Inductors. Stored energy.

LR circuits.

Induction between two coils

Change the current in coil 1

changes the B-flux through coil 2

induces an emf in coil 2

Mutual inductance

22 2

dN

dt

12 21

diM

dt

1didt

M21 mutual inductanceindicates how large an emf in 2due to current change in 1

2 221

1

BNM

i

Of course it works both ways:

21 12

diM

dt 1 1

122

BNM

i

Mutual inductance depends on the geometry, orientation and materials of the coils. It can be shown that

21 12M M M

Therefore, we have:

12

diM

dt 2

1di

Mdt

2 2 1 1

1 2

B BN NM

i i

Units: SI Henry 1 H = 1 Wb/A

ACT: Mutual inductance

t

i1

ε2

t t

ε2

If the current in coil 1 is as shown, which of the graphs gives the correct emf in coil 2?

t

ε2

A B C

Example: M for two solenoids

Craig Ogilvie

0 2 22 0 2 2

N iB n i

l

20 2 212 1

N iR

l

20 1 2 1N N R

Ml

l

N2 turns, length l

N1 turns, length l

20 2 21 11 12

2 2

( )N iN RN

Mi i

l

l

Self-inductance

Ideal coil (no resistance)

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

di/dt

diLdt

The effect is called self-induction

Single coil is called an inductorBN

Li

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. yB. zC. 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” withpositive terminal near y.

in

ACT: Inductor II

A. aB. bC. Both have the same

potential

By Lenz law, induced emf tries to oppose change. Current is decreasing so emf tries to increase it. Imagine a “battery” withpositive terminal near a.

in

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?

In-class example: Inductance of a solenoid

A solenoid is constructed out of 1000 loops of wire wound around a 1.0 cm2 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?

A. 250 HB. 25 HC. 2.5 HD. 2.510−3 HE. 2.510−5 H

length l, N turns

00

NiB ni

l

20Ni R

BAl

20( )Ni R

NNLi i

l

27 4 2

32

T m4 10 1000 1.0 10 mA 2.5 10 H5.0 10 m

L

2 20N R

Ll

Inductance of a solenoid

Inductors against di/dt

Inductor always acts to oppose current change.

used in circuit design to protect against rapid changes of currents or spikes

setting up a current in a circuit with an inductor will take some time.

RL circuits: current growth

Switch moved to position “a” at t = 0

Kirchhoff’s law

0diiR Ldt

drop in potential fromtop of inductor to bottom

inducedemf

Differential equation for i+ initial condition

0 0i t

0 0i t Qualitatively: (no inductor)i t

R

0diiR Ldt

di R dt

LiR

lni RR t

LR

ln 1 R Ri tL

1RtLi e

R

1t

i t eR

LR

ε/R

i

t

If L is large, is large, i.e. current grows slowly.

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?

A. 1 HB. 0.43 HC. 0.1 HD. 0.043 HE. 0.01 H

LRV0

1t

i t eR

LR

1 ms0.9 1 e

1 ms 0.43 ms

ln 0.1

0.43 HL R

ACT: Three circuits

All batteries, inductors, resistors are identical. Rank the circuits according to the current through battery just after switch is closed

A. i1 > i2 > i3 initiallyB. i2 > i3 > i1 initiallyC. i3 > i2 > i1 initially

1 32

R2 < R3 i2 > i3

Initially i = 0 through inductors. i1 = 0

ACT: Current decay

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?

i

t

i

t

i

t

i

t

A B C D

inducedemf

0 ( ) t Li t i e

R

di R idt L

0 diiR Ldt

0 didt

DEMO:RL circuit

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.

Energy stored in an inductor

1) rate of energy supplied to inductor diP Vi Lidt

2) increase of energy within inductor dU Pdt Li di

3) energy stored within inductor 0

IU dU L idi

To establish current (from i = 0 to i = I ):

Voltage across inductor:(ignoring signs)

diV Ldt

212U LI

Using the stored energy

• 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)

DEMO:Spark