Self Inductance• Consider a solenoid L, connect it to a battery

• Area A, lengthl, N turns • What happens as you close the switch?• Lenz’s law – loop resists change in magnetic

field• Magnetic field is caused by the current• “Inductor” resists change in current

E +–

0NIB

0

1B

NIA

B

L

d

dt

E 1B

dN

dt

20N IAd

dt

20N A dI

dt

dI

Ldt

E 20L N A

A

l

Ch. 32

Warmup 18

Inductors• An inductor in a circuit is denoted by this

symbol:• An inductor satisfies the formula:• L is the inductance

• Measured in Henrys (H)

dIL

dtE

1 H 1 V s/A

L

Kirchoff’s rules for Inductors:• Assign currents to every path, as usual• Kirchoff’s first law is unchanged• The voltage change for an inductor is L (dI/dt)

• Negative if with the current• Positive if against the current

• In steady state (dI/dt = 0) an inductor is a wire

+–

LE

I

What is Kirchoff’s law for the loop shown?A) E + L (dI /dt) = 0 B) E – L (dI /dt) = 0 C) None of the aboveD) I don’t know Kirchoff’s law for switches

0 EdI

Ldt

dI

dt LE

I tL

E

Ex- (Serway 32-9) The current in a 90.0 mH inductor changes with time as I = t2 - 6t (in SI units). Find the magnitude of the induced emf at (a) t = 1.00 s and (b) t = 4.00 s. (c) At what time is the emf zero.

Solve on Board

Warmup 18

Energy in Inductors• Is the battery doing work on the inductor?

I V P+–

LEdI

ILdt

• Integral of power is work done on the inductor

U dtPdI

IL dtdt

L IdI 212 LI k

• It makes sense to say there is no energy in inductor with no current21

2U LI• Energy density inside a solenoid?

2 20

2

N AIU

0NIB

20L N A

Uu

A

2 20

22

N I

2

02

B

2

02

Bu

• Just like with electric fields, we can associate the energy with the magnetic fields, not the current carrying wires

RL Circuits• Circuits with resistors (R) and inductors

(L) L

E = 12 V

+–

In the steady state, with the switch closed, how much current flows through R2? How much current flows through R2 the moment after we open the switch?A) 0 A B) 6 A C) 3 AD) 2 A E) None of the above

R1 =

2

• In the steady state, the inductor is like a wire• Both ends of R2 are at the same potential: no current through R2

• The remaining structure had current I = E/R1 = 6 A running through it

6 A6 A

I = 6 A

R2 =

4

• Now open the switch – what happens?• Inductors resist changes in current, so the current

instantaneously is unchanged in inductor• It must pass through R2

I = E /R1 = 6 A

6 A

RL Circuits (2)• What happens after you open the switch?

• Initial current I0

• Use Kirchoff’s Law on loop• Integrate both sides of the equation

L

E = 12 V

+–R1 =

2

I

R =

4

0dI

L RIdt

dI RI

dt L

dI Rdt

I L dI R

dtI L

ln constantRt

IL

Rt LI e

0tI I e

L R

RL Circuits (3)• Where did the energy in the inductor go?• How much power was fed to the resistor?

L

E

+–

R

0tI I e

L R

• Integrate to get total energy dissipated

2 20 02

Rt LLRI e

R

20 2

LRI

R

2102U LI • It went to the resistor

• Powering up an inductor:• Similar calculation

1 tI eR

E

2RIP 2 20

tR LRI e

0

U dt

P 2 20

0

Rt LRI e dt

Sample Problem

L =

4.0

mH

I R

0tI I e L R

An inductor with inductance 4.0 mH is discharging through a resistor of resistance R. If, in 1.2 ms, it

dissipates half its energy, what is R?

210 02U LI 21

2U LI 102 U 21

04 LI2 21

02fI I0

10.707

2

I

I

0

0.707tIe

I ln 0.707 0.347t

0.347

t 31.2 10 s

0.00346 s0.347

LR

.00400 H

.00346 sR 1.16 R

Concept Question

L

E = 10 V

+–

The circuit at right is in a steady state. What will the voltmeter read as soon as the switch is opened?A) 0.l V B) 1 V C) 10 VD) 100 V E) 1000 V

R1 =

10

• The current remains constant at 1 A• It must pass through resistor R2

• The voltage is given by V = IR

R2 =

1 k

• Note that inductors can produce very high voltages

• Inductance causes sparks to jump when you turn a switch off

I =

1 A

1 A 1000 V IR

1000 VV

+–

Loop has unin-tended inductance

V

CT-1- When the switch is closed, the current through the circuit exponentially approaches a value I=E/R. If we repeat this experiment with an inductor having twice the number of turns per unit length, the time it takes for the current to reach a value of I/2

A increases. B. decreases. C. is the same.

CT-2 When the switch in the circuit below is closed, the brightness of the bulb

+ -

R

Bulb

L A. Starts off at its brightest and then dims. B. Slowly reaches its maximum brightness. C. Immediately reaches it maximum, constant brightness. D. Something else.

Assume inductor has no resistance

Inductors in series and parallelL1• For inductors in series, the

inductors have the same current

• Their EMF’s add

L2

1

dIL

dtE 2

dIL

dt 1 2

dIL L

dt

1 2L L L

• For inductors in parallel, the inductors have the same EMF but different currents

L1

L2

11

dIL

dtE

22

dIL

dtE

1 2dI dIdI

dt dt dt

1 2L L

E EL

E

1 2

1 1 1

L L L

Parallel and Series - Formulas

Capacitor Resistor Inductor

Series

Parallel

Fundamental Formula

1 2R R R

1 2

1 1 1

R R R 1 2C C C

1 2

1 1 1

C C C 1 2L L L

1 2

1 1 1

L L L

QV

C V IR

L

dIL

dtE

Warmup 19

CT - 3- The primary coil of a transformer is connected to a battery, a resistor, and a switch. The secondary coil is connected to an ammeter. When the switch is thrown closed, the ammeter shows

A. zero current. B. a nonzero current for a short instant. C. a steady current. D. Something else.

LC Circuits• Inductor (L) and Capacitor (C)• Let the battery charge up the capacitorNow flip the switch• Current flows from capacitor through inductor• Kirchoff’s Loop law gives:• Extra equation for capacitors:

+–

EC

L

Q

0Q C V C EI

0Q

C dI

Ldt

dQI

dt

dIQ CL

dt d dQ

CLdt dt

2

2

1d QQ

dt CL

• What function, when you take two deriva-tives, gives the same things with a minus sign?

• This problem is identical to harmonicoscillator problem

cos

sin

Q t

Q t

0 cosQ Q t

LC Circuits (2)• Substitute it in, see if it works

C

L

Q

I

0 cosQ Q t

0 sindQ

Q tdt

2

202

cosd Q

Q tdt

2

2

1d QQ

dt CL

20 0

1cos cosQ t Q t

CL

1

CL

• Let’s find the energy in the capacitor and the inductor

dQI

dt 0 sinQ t

2

2C

QU

C

2

20 cos2C

QU t

C

212LU LI 2 2 21

02 sinLQ t

2

20 sin2L

QU t

C

20 2C LU U Q C

Energy sloshes back and forth

Warmup 19

Frequencies and Angular Frequencies• The quantity is called the angular frequency• The period is the time T you have to wait for it to repeat• The frequency f is how many times per second it repeats

2T

1

CL 0 cosQ Q t T

1f T

2 f

WFDD broadcasts at 88.5 FM, that is, at a frequency of 88.5 MHz. If they generate this with an inductor with L = 1.00 H,

what capacitance should they use?

2 f 6 12 88.5 10 s 8 15.56 10 s 2 1LC

2

1C

L

28 1 6

1

5.56 10 s 10 H

3.23 pFC

Ex – Serway (32-49) A 1.00 µF capacitor is charged by a 40 V power supply. The fully-charged capacitor is then discharged through a 10.0 mH inductor. Find the maximum current in the resulting oscillations.

Solve on Board

RLC Circuits• Resistor (R), Inductor (L), and Capacitor (C)• Let the battery charge up the capacitorNow flip the switch• Current flows from capacitor through inductor• Kirchoff’s Loop law gives:• Extra equation for capacitors:

+–

EC

L

Q

I

0Q dI

L RIC dt

dQI

dt 2

20

Q dQ d QR L

C dt dt

• This equation is hard to solve, but not impossible• It is identical to damped, harmonic oscillator

20 cosRt LQ Q e t

R

2

2

1

4

R

LC L