LC Circuits LC Oscillations

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LC Circuits • Consider the LC and RC series circuits shown: ++++ ++++ – At t=0 the capacitor is C R C L charged to a value of Q ------

• Is there is a qualitative difference in the time development of the currents produced in these two cases. Why?? • Consider from point of view of energy – RC circuit • Resistor will dissipate energy from any current – (Ideal) LC circuit • NO mechanism for energy dissipation – Energy can be stored in the capacitor and/or in the inductor 11/3/16 3

LC Oscillations I • Kirchoff’s loop rule dI Q ++++ Q C L VL +VC = L + = 0 dt C - - - - d 2Q Q L 2 + = 0 dt C d 2x • Same form as mass on a spring m 2 +kx = 0 – Solution must be of the same form dt

d 2Q 1 • Divide by L and rearrange 2 = − Q dt LC • General solution Q Q cos t = 0 (ω +φ) 11/3/16 4

1 LC Oscillations I Q = Q cos ωt +φ 0 ( ) ++++ • Q and φ from initial conditions Q C L 0 - - - - • Differentiate to find current dQ = I = −ωQ0 sin(ωt +φ) dt 2 d Q dI 2 2 = = −ω Q0 cos(ωt +φ) dt dt • Substitute into loop equation 1 L 2Q cos( t ) Q cos( t ) 0 (−ω 0 ω +φ )+ ( 0 ω +φ ) = C 1 1 −ω2L + = 0 ⇒ ω = C LC 11/3/16 5

r1 x 0, .. r1 n 1.01 1 LC Oscillationsr1 x 0, .. r1 Q Q cos t n Q = 0 (ω +φ) 1 V1.01C f( x)00

f( x0) 0 1.01 1 1.01 1 0 2 4 6 0 x 6.28 I 1.01 1 0 2 4 6 1 f( x)00 1.01V 0 x 6.28 I = −ωQ sin(ωt +φ) L 0

f( x) 0 1.01 1 0 1.01 1 0 2 4 6 dI 0 x 6.28

dt 1.01 1 0 2 4 6 f( x) 00 t dI 0 x 6.28 = −ω2Q cos(ωt +φ) • Voltages are opposite dt 0 – Capacitor and inductor traversed 1.01 1 in “opposite” directions 0 2 4 t 6 0 x 6.28

2 Example 1 t=0 t=t1 + + Q = Q C L Q = 0 C L o L L - -

• At 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 VL, the voltage across the inductor at time t1?

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

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Example 2

• At t=0 the capacitor has charge Q0; the t=0 resulting oscillations have frequency + + Q = Q C L ω. The maximum current in the circuit o - - 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 = 2I0 (c) I2 = 4I0

3 Example 3

• At t=0 the capacitor has charge Q0; the t=0 resulting oscillations have frequency + + Q = Q C L ω . The maximum current in the circuit o 0 - - during these oscillations has value I0.

– What is the relation between ω0 and ω2, the frequency of oscillations when the

initial charge = 2Q0?

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

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LC Oscillations: Energy Check

• Oscillation frequency from loop equation 1 ω0 = LC • Q0 and φ found from initial conditions

– In Example 1, assumed initial values for the charge (Qi) and current (0)

• Q0 = Qi, φ = 0 • Question: Does this solution conserve energy?

2 1 Q (t) 1 2 2 UE (t) = = Q0 cos (ωt +φ) 2 C 2C

1 2 1 2 2 2 UB (t) = Li (t) = Lω0Q0 sin (ωt +φ) 2 2 11/3/16 13

4 r1 Energyx 0Check, .. r1 n 1 • Energy in capacitor U E 1 2 2 UE (t) = Q0 cos (ωt +φ) 2C f( x) 0.5 • Energy in inductor

1 2 2 2 UB (t) = Lω Q0 sin (ωt +φ) r1 00 2 x 0, .. r10 2 4 6 n t

1 ⇒ x ω = 1 LC U B 1 2 2 UB (t) = Q0 sin (ωt +φ) 2C f( x) 0.5 • Therefore 2 Q0 UE (t) +UB (t) = 2C 00 0 2 t 4 6 x

UB versus UE

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5 Example 4

• At t=0, the capacitor in the LC circuit shown has a total t=0 t=t + + 1 charge Q0. At t = t1, the Q = Qo C L Q = 0 C L 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

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LC Oscillations with Finite R

• If L has finite R – Energy will be dissipated in R

r1 – Oscillations will become dampedr1 r1 10 n 100 x 0, .. r1 x 0, .. r1 n n 1 1 Q Q

f( x) 00 f( x)00

1 1 0 t5 10 0 t5 10 x x R = 0 R ≠ 0

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6 Driven Oscillations

1 0 5 10 x

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Driven Oscillations

• LC circuits are natural oscillators 1 R ω re so n a n ce = in absence of R + + LC Q C - - L • Real LC circuits r1 r1 10 n 100 x 0, .. r1 – Must account for the resistance of the inductorn • Damps out the oscillations 1 t f( x) 0 • Can we modify the circuit to sustain oscillations without damping? 1 – Sinusoidally varying emf source (AC generator) 0 5 10 x – Energy supplied at rate dissipated by R

– Not limited to driving the circuit at ωresonance

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7 AC Circuits: Series LCR

• Statement of problem: R • Given ε = εmsinω t , find I(t). – Everything else will follow C L

• Procedure: Start with loop equation? ε d 2Q Q dQ L 2 + +R = εm sinωt dt C dt

• Solve with algebra • Or phasors

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Phasors: LCR • Given: ε = ε sinωt I m Q = − m cos(ωt −φ) ω • Assume: I = I sin(ωt −φ) ⇒ m dI = Imωcos(ωt −φ) dt

V = RI = RI sin(ωt −φ) R m ω Q 1 I m X L ⇒ V = = − I cos(ωt −φ) C C ωC m dI VL = L = ωLIm cos(ωt −φ) dt φ ε m φ • From these equations, we φ can draw the phasor diagram I X at the right. m C Im R 11/3/16 22

8 Phasors:LCR ImXL

I R Im(XL-XC) m φ ε m φ ε φ ⇒ m φ X − X tan L C I X I R φ = m C m R 2 1 ε2 = I 2 R2 + X − X X ≡ ωL X ≡ m m ( ( L C ) ) L C ωC ⇓ Impedance εm εm Im = = 2 2 2 Z Z ≡ R + X − X R2 X X ( L C ) +( L − C )

Lagging & Leading

• Phase φ between the current and the driving emf depends on the relative magnitudes of XL and XC. ε X − X 1 I = m tanφ = L C X ≡ ωL X ≡ m Z R L C ωC XL X Z L XL Z φ φ R R R Z XC XC XC

XL > XC XL < XC XL = XC φ > 0 φ < 0 φ = 0 I I I LAGS LEADS IN PHASE with

Vapplied Vapplied Vapplied

9 Resonance

• For fixed R,C,L the current Im will be a maximum at the resonant frequency ω0 – Makes the impedance Z purely resistive ε ε I = m = m m Z 2 R2 X X +( L − C )

• Reaches maximum when XL = XC – Frequency at which condition is achieved 1 1 ωL = ω = ωC LC • Resonant frequency = natural frequency of the LC circuit • Current and driving voltage are in phase X − X tanφ = L C = 0 11/3/16 R 25

Power in LCR Circuit

• Power supplied by the emf in a series LCR circuit depends on the frequency ω.

– Maximum power supplied at the resonant frequency ω0 • Instantaneous power (for some frequency, ω) delivered at time t is given by: P(t) = ε(t)I(t) = ε sinωt I sin(ωt −φ) = I 2(t)R ( m )( m ) • Average power delivered in a cycle – Most useful quantity considered here 〈P(t)〉= ε I 〈sinωt sin(ωt −φ)〉 m m 1 〈P(t)〉= VmIm cosφ 2

10 Power in LCR Circuit

• Often rewritten in terms of rms values:

1 1 ε ≡ ε I ≡ I ⇒〈P(t)〉= ε I cosφ rms m rms m rms rms 2 2

• Phase depends on the values of L, C, R, and ω and therefore... • Power delivered depends on the phase, φ, the “power factor” X − X Z R tanφ = L C = cosφ = R R Z