LC

LCCircuits0

0

t

V

V

C

L

x ..,0 r1nr1

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

0

0

t

t

UB

UE

x ..,0 r1nr1

0 2 4 60

0.5

1

f( )x

x

x ..,0 r1nr1

0 2 4 60

0.5

1

f( )x

xPhysics 122 Lecture 24G. Rybka

Exam 3• Thursday• Covers RC circuits through Faraday’s law

Energy in the Magnetic Field“Power” accounting in a LR circuit...

dtdILIRIεI 2 += Loop rule x I…

dtdILI

dtdUPL == Rate of energy flow into L

U dU LIdIIU

= = ∫∫00

Total energy flow

… energy storedU LI=12

2

B

But where is it “Stored”?• Claim: energy is stored in the Magnetic field itself• Consider the uniform field inside a long solenoid:

l

r

N turns• The inductance L is:

• Get energy density by dividing by the volume containing the field:

• Stored Energy U:

IlNB 0µ=

AlNL2

0µ=

0

22

2

02

21

21

21

µµ

BAlIAlNLIU =⎟⎟

⎠

⎞⎜⎜⎝

⎛==

0

2

21µB

volU

AlUuM ===

Energy in the Electric Field

Recall & Compare

Work needed to add charge to capacitor...

dW dq V dq qC

= = ⎛⎝⎜ ⎞

⎠⎟( )

22Q

0CV21

CQ21qdq

C1W === ∫ … total work ...

+++ +++

-­ -­ -­ -­ -­ -­

20E2

1volumeWu ε== … energy density

Recall: C = ε0A/d & V = Ed

0

2

magneticB21uµ

=20electric E

21u ε=

Energy Density:

RC/LC Circuits

RC or LR: current decays exponentially

C R

-it

0

0

i

Q

0 1 2 3 40

0.5

1

t/RC

Q f( )x

x

+++-­ -­ -­

LC

LC: current oscillates

i

0

0 t

i

Q

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

+++-­ -­ -­

LC Oscillations(qualitative)

Þ

Ü

ß

LC+ +- -

i = 0

Q Q= + 0

LC+ +

- -i = 0

Q Q= − 0

LC

i i= − 0

LC

Ý

i i= + 0

Q = 0

LC Oscillations(qualitative)

0i

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

Q0

x ..,0 r1nr1

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

t

0

di__dt

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

t

0

VC

x ..,0 r1nr1

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

0

VL

0 2 4 61

0

11.01

1.01

f( )x

6.280 x

How do these change if L has a finite R?

Cos()

-­Sin()

-­Cos()

-­Cos()

Cos()

ClickerAt t=0, the capacitor in the LC circuit shown has a total charge Q0. At t = t1, the capacitor is uncharged.

– What is the value of Vab, the voltage across the inductor at time t1?

(a) Vab < 0 (b) Vab = 0 (c) Vab > 0

LC

LC

+ +- -

Q = 0Q Q= 0

t=0 t=t1

a

b

• Vab is the voltage across the inductor, but it is also the voltage across the capacitor!

• Charge on the capacitor is zero, à VC = 0

• When Q = 0 on capacitor, I is maximum through inductor • and dI/dt is zero then so VL = 0 makes sense

ClickerAt t = 0, the capacitor in the LC circuit shown has a total charge Q0. At t = t1, the capacitor is uncharged.

What is the relation between UL1, the energy stored in the inductor at t = t1, and UC1 , the energy stored in the capacitor at t = t1?

(a) UL1 < UC1 (b) UL1 = UC1 (c) UL1 > UC1

LC

LC

+ +- -

Q = 0Q Q= 0

t=0 t=t1

a

b

At t = t1, the charge on the capacitor is zero.

U QCC112

20= = U LI Q

CL1 12 0

212 2

0= = >

At t = t1, the current is a maximum.

LC Oscillations(L with finite R)

• If L has finite Resistance, then– energy will be dissipated in R and– the oscillations will become damped.

Q

0

t

R = 0

x ..,0 r1nr1

0 5 101

0

1

f( )x

xt

0

Q

R ¹ 0

r1 10 n 100x ..,0 r1nr1

0 5 101

0

1

f( )x

x

Quick checkpoint reviewAt time t = 0 the capacitor in the circuit below is fully charged with Qmax, and the current through the circuit is 0.

• What is the potential difference across the inductor at t = 0?• Ans: VL = Qmax / C (same as capacitor)

• What is the potential difference across the inductor when current is maximum?• Ans: 0

• How much energy is stored in C when I is max?• Ans: U = 0 (it’s all in the inductor)

LC Oscillations(quantitative)

• Begin with the loop rule:

• Guess solution: (harmonic oscillator)

where: • ω0 determined from equation

• φ, Q0 determined from initial conditions

• Procedure: differentiate above form for Q and substitute into loop equation to find ω0

LC+ +- -

i

QLd Qdt

QC

2

2 0+ =

Q Q t= +0 0cos( )ω φ 2

2

dtxdmkx =−In mechanics

LC Oscillations(quantitative)

• General solution:

• Differentiate twice:

• Substitute into loop eqn:Þ

∴

LC+ +- -Q Q t= +0 0cos( )ω φ

Ld Qdt

QC

2

2 0+ =dQdt

Q t= − +ω ω φ0 0 0sin( )

d Qdt

Q t2

2 020 0= − +ω ω φcos( )

( ) ( )L Q tCQ t− + + + =ω ω φ ω φ0

20 0 0 0

1 0cos( ) cos( ) − + =ω02 1 0L

C

LC1

0 =ω

Clicker

(a) ω2 = 1/2 ω0 (b) ω2 = ω0 (c) ω2 = 2 ω0

LC

+ +- -Q Q= 0

t=0

• Q0 determines the amplitude of the oscillations (initial condition)

• The frequency is determined by the circuit parameters (L,C) only

At t = 0 the capacitor has charge Q0;; the resulting oscillations have frequency ω0. The maximum current in the circuit during these oscillations has value I0 .

– What is the relation between ω0 and ω2 , the frequency of oscillations when the initial charge = 2Q0 ?

Clicker• At t = 0 the capacitor has charge Q0;; the resulting oscillations have frequency ω0. The maximum current in the circuit during these oscillations has value I0 .

What is the relation between I0 and I2 , the maximum current in the circuit when the initial charge = 2Q0 ?

(a) I2 = I0 (b) I2 = 2 I0 (c) I2 = 4 I0

LC

+ +- -Q Q= 0

t=0

• The initial charge determines the total energy : U0 = Q02/2C

• Maximum current occurs when Q = 0

• When all the energy is in the inductor: U = 1/2 LIo2• Doubling initial charge quadruples total energy.

• Implies maximum current must double

LC OscillationsDoes solution conserve energy?

)(cos21)(

21)( 0

220

2

φω +== tQCC

tQtUEEnergy in E field:

)(sin21)(

21)( 0

220

20

2 φωω +== tQLtLitUBEnergy in B field:

LC1

0 =ω

)(sin21)( 0

220 φω += tQ

CtUB

CQtUtU BE 2

)()(20=+

YES !!

Quick checkpoint reviewThe capacitor charged such that the top plate has a charge +Q0 and the bottom plate -­Q0. At time t=0, the switch is closed …

• What is the value of the capacitor C?• Ans: (500)2 x L = 1 / C• C = 10-­3 F

• Which of the following plots best represents the energy in the inductor as a function of time starting just after the switch is closed?

Energy is always POSITIVE (proportional to Square of current)

Energy Plotted vs Time

t0

UE

x ..,0 r1nr1

0 2 4 60

0.5

1

f( )x

x

0t

UB

x ..,0 r1nr1

0 2 4 60

0.5

1

f( )x

x

U t U t QCE B( ) ( )+ = 02

2

ClickerAt t = 0 the current flowing through the circuit is 1/2of its maximum value.

– Which of the following is a possible value for the phase φ, when the charge on the capacitor is described by: Q(t) = Q0cos(ωt + φ).

(a) φ = 30° (b) φ = 45° (c) φ = 60°

LC+ +- -

i

Q

• We are given a form for the charge on the capacitor as a function of time, but we need to know the current as a function of time.

φ)tsin(ωQωdtdQI(t) 000 +−==

• At t = 0, the current is given by: sinφQω)0I( 00−=

• The maximum value of the current is: 00max QωI =

• Therefore, the phase angle must be given by:21sinφ ±= Þ °±= 30φ

ClickerAt t = 0 the current flowing through the circuit is 1/2 of its maximum value.

Which of the following plots best represents UB, the energy stored in the inductor as a function of time?

(a) (b) (c)

LC+ +- -

i

Q

x ..,0r1nr1

0 2 4 60

0.5

1

f( )x

x

0 2 4 60

0.5

1

f( )x

x

x ..,0 r1nr1

0 2 4 60

0.5

1

f( )x

x

00

UB

time

00

UB

time

00

UB

time

Energy stored in the inductor proportional to the CURRENT SQUARED.

If the current at t = 0 is 1/2 its maximum value, the energy stored in the inductor will be 1/4 of its maximum value!!

LCR DampingFor your interest, we do not derive here, but only illustrate the

following behavior

t

0

Q

0

Q

t

LC+ +- -

R

r1 10 n 100x ..,0 r1nr1

0 5 101

0

1

f( )x

xr1 10 n 100x ..,0 r1

nr1

0 5 101

0

1

f( )x

x

R R= 04

R R= 0

β =RL2

Q Q e tto= +−

0β ω φcos( ' )

ω'o LCRL

= −⎛

⎝⎜

⎞

⎠⎟

14

2

2

In a LRC circuit, ω depends also on R !