Lecture 12 Electromagnetic Oscillations and Alternating Current Chp. 33

• Cartoon -. Opening Demo - Warm-up problem

• Physlet

• Topics

– LC Circuit Qualitatively – Electrical and Magnetic energy oscillations– Alternating current– \Pure R and L, circuti– Series RLC circuit– Power and Transfomers

• Demos– LR circuit

– Series LRC circuit

dAnBm ˆ⋅=•r

φ

AdBrr

⋅=dAB θcos=

θn̂

B

€

ε =−dΦ

dt= −d(BAcosθ)

dt= −BA

dcosθ

dt= BAsinθ

dθ

dt

= BAω sinθ but θ =ωt so dθ

dt=ω

€

ε =BAω sinωt

€

ε =εm sinωt

Where is the rotational angular frequency of the generator

f and f= 60 Hz

Axis of rotationCoil of wire

€

ε =εm sin(ωt −φ)

Instantaneous voltage

Amplitude

Angular frequency

time Phase constat

phase

€

εm

€

ε

€

t −φ

Phasor diagram

€

ε =εm sinωt

€

i = I sin(ωt)

€

vL = Ldi

dt= LωIcos(ωt)

€

di

dt=ωIcos(ωt)

€

vL = LωIcos(ωt)

€

vR=Ri

€

R=VRI

€

XL=VLI

=Lω

VR=RI VL= XLIL or VL= (LI since I=IL

L = 4.22mH

€

=2πf

€

f =1000Hz

€

ε =vR + vL

Impedance Z: New quantity for AC circuits. This is analogous to resistance in DC circuits

€

Z = R2 + (ωL)2

€

I =εmZ

€

I =εm

R2 + (ωL)2

€

XL =ωL

RL Circuit Example

Suppose εm = 100 volts, f=1000 Hz, R=10 Ohms, L=4.22 mH,Find XL, Z, I, VR, and Vl.

€

XL =ωL = 6.28 ×1000 × 0.00422H = 26.5Ω

€

Z = 102 + (26.5)2 = 28.3Ω

€

Z=R2

+(ωL)2

€

I =εmZ

=100

28.3= 3.53A

€

VR = RI =10 × 3.53 = 35.3v

€

VL = XLI = 26.5 × 3.53 = 93.5v

Power in AC circuits

€

P = i2R = (I sin(ωt))2R

Instantaneous power doesn’t mean anything

Need to average over time or one period of the sine wave

€

Pavg = 12π Rdθ

0

2π

∫ (I sin(θ))2 = 12π RI

2 sin2

0

2π

∫ θdθ = RI2 1

2= (

I

2)2R

Note

€

Irms =I

2

€

Pavg = Irms2 R

Averaging over a sine curve

Calculate Power lost in resistor from example

€

Pavg = Irms2 R

€

Irms =I

2=

3.53A

1.414= 2.50A

€

Pavg = (2.50A)210 = 62.5Watts

To calculate power produced by the generator you need to take account of the phase difference between the voltage and the current. In general you can write:

€

Pavg = εrmsIrms cosφ

For an inductor P = 0 because the phase difference between current through the inductor and voltage across the inductor is 90 degrees

Series LRC circuit

€

ε =vR + vC + vL

€

ε =εm sinωt

€

i = I sin(ωt −φ)

VL

VC

VR

€

I =εm

R2 + (XL − XC )2=εmZ

XL=L

XC=1/(C)

€

Z = R2 + (ωL −1

ωC)2

€

R2 + (XL − XC )2

€

tanφ =ωL −

1

ωCR

ELI the ICE man

Resonance

€

XL = XC

ωL =1

ωC

ω =1

LC

Series LRC demo

10 uF 4.25 mH

€

f =1

6.28 LC=

1

6.28 4.25 ×10−3H ×10−6Ff = 2442Hz

Series LCR circuit