Electromagnetic Oscillations and Alternating Current

Electromagnetic Oscillations Electromagnetic Oscillations andand

Alternating CurrentAlternating Current

Chapter 33Chapter 33

Oscillations in an LC Circuit

We will discover that charge sloshes back and forth. As this happens the current goes one way then the other.

Analogy: a block moving on a spring. Here total energy (kinetic + potential) is constant.

For the LC circuit total energy (electric + magnetic) is constant.

+ +i

+++

+

i

i

i

i

i

LC Circuit

time

€

U = UB + UE =1

2LI2 +

1

2

q2

C

dU

dt=

d

dt(1

2LI2 +

1

2

q2

C) = 0

∴ LIdI

dt+

q

C

dq

dt= 0 = L(

dq

dt)

d2q

dt 2+

q

C

dq

dt

∴ Ld2q

dt 2+

1

Cq = 0

Analyzing an LC Circuit

Total energy in the circuit:

Differentiate : No change in energy

€

U = UB + UE =1

2LI2 +

1

2

q2

C

dU

dt=

d

dt(1

2LI2 +

1

2

q2

C) = 0

∴ LIdI

dt+

q

C

dq

dt= 0 = L(

dq

dt)

d2q

dt 2+

q

C

dq

dt

∴ Ld2q

dt 2+

1

Cq = 0

Total energy in the circuit:

Differentiate : No change in energy

Analyzing an LC Circuit

€

U = UB + UE =1

2LI2 +

1

2

q2

C

dU

dt=

d

dt(1

2LI2 +

1

2

q2

C) = 0

∴ LIdI

dt+

q

C

dq

dt= 0 = L(

dq

dt)

d2q

dt 2+

q

C

dq

dt

∴ Ld2q

dt 2+

1

Cq = 0

Total energy in the circuit:

Differentiate : No change in energy

Analyzing an LC Circuit

€

U = UB + UE =1

2LI2 +

1

2

q2

C

dU

dt=

d

dt(1

2LI2 +

1

2

q2

C) = 0

∴ LIdI

dt+

q

C

dq

dt= 0 = L(

dq

dt)

d2q

dt 2+

q

C

dq

dt

∴ Ld2q

dt 2+

1

Cq = 0

Total energy in the circuit:

Differentiate : No change in energy

The charge sloshes back andforth with frequency = (LC)-1/2

The charge sloshes back andforth with frequency = (LC)-1/2

€

d2q

dt 2+ ω2q = 0

ω2 = 1LC

q = qp cosωt

Analyzing an LC Circuit

€

q = qp cosωt with ω = LC( )− 1

2

Current

€

I =dq

dt= −qpω sinωt

Current is maximum when charge is zero, and vice versa.

Energy:

€

UC =Q2

2C=

qp2

2Ccos2 ωt

UB =1

2LI2 =

1

2Lqp

2ω2 sin2 ωt =1

2Lqp

2 1

LC

⎛

⎝ ⎜

⎞

⎠ ⎟sin2 ωt =

qp2

2Csin2 ωt

UB + UC =qp

2

2C= const.

Analyzing an LC Circuit

RLC Circuit: Damped Oscillations

R

LC

The change here is that energy is dissipated in the resistor:

€

d

dtUB + UE( ) = −I2R

A similar analysis gives current and charge that continue to oscillate but with amplitudes that decay exponentially:

€

q = qpe−Rt / 2L cos ′ ω t( )

′ ω = ω2 − (R /2L)2

Alternating Current Circuits

is the angular frequency (angular speed) [radians per second].

Sometimes instead of we use the frequency f [cycles per second]

Frequency f [cycles per second, or Hertz (Hz)] f

V = VP sin (t - v ) I = IP sin (t - I )

An “AC” circuit is one in which the driving voltage andhence the current are sinusoidal in time.

v

V(t)

t

Vp

-Vp

Vp and Ip are the peak current and voltage. We also use the

“root-mean-square” values: Vrms = Vp / and Irms=Ip /

v and I are called phase differences (these determine whenV and I are zero). Usually we’re free to set v=0 (but not I).

2 2

Alternating Current Circuits

V = VP sin (t - v ) I = IP sin (t - I )

v

V(t)

t

Vp

-Vp

Vrms

I/

I(t)

t

Ip

-Ip

Irms

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.2

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2 f=377 s -1.

2

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2 f=377 s -1.

So V(t) = 170 sin(377t + v).Choose v=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

2

Resistors in AC Circuits

VR

~EMF (and also voltage across resistor): V = VP sin (t)Hence by Ohm’s law, I=V/R:

I = (VP /R) sin(t) = IP sin(t) (with IP=VP/R)

V and I“In-phase”

V

t

I

This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(C) the Capacitive Reactance

Capacitors in AC Circuits

V~

C Start from: q = C V [V=Vpsin(t)]Take derivative: dq/dt = C dV/dtSo I = C dV/dt = C VP cos (t)

I = C VP sin (t + /2)

The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference).

V

t

I

V and I “out of phase” by 90º. I leads V by 90º.

Capacitor Example

V

~

CA 100 nF capacitor isconnected to an AC supply of peak voltage 170V and frequency 60 Hz.

What is the peak current?What is the phase of the current?What is the dissipated power?

Again this looks like IP=VP/R for aresistor (except for the phase change).

So we call XL = L the Inductive Reactance

Inductors in AC Circuits

LV = VP sin (t)Loop law: V +VL= 0 where VL = -L dI/dtHence: dI/dt = (VP/L) sin(t).Integrate: I = - (VP / L cos (t)

or I = [VP /(L)] sin (t - /2)

~

Here the current lags the voltage by 90o.

V

t

I

V and I “out of phase” by 90º. I lags V by 90º.

Inductor Example

L~V

A 10 mH inductor isconnected to an AC supply of peak voltage 10V and frequency 50 kHz.

What is the peak current?What is the phase of the current?What is the dissipated power?

Circuit element

Resistance or

ReactanceAmplitude Phase

Resistor R VR= IP RI, V in phase

Capacitor Xc=1/C VC=IP XcI leads V by 90°

InductorXL=L

VL=IP XcI lags V by

90°

Phasor Diagrams

Vp

Ip t

Resistor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

Phasor Diagrams

Vp

Ip t

Vp

Ip

t

Resistor Capacitor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

Phasor Diagrams

Vp

Ip t

Vp

Ip

t

Vp Ip

t

Resistor Capacitor Inductor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.