Physics -272 Lecture 19

LC Circuits RLC Circuits

AC Circuit Theory LC Circuits

• Consider the RC and LC series circuits shown: ++++ ++++ C C ---- R ---- L • Suppose that the circuits are formed at t=0 with the capacitor charged to value Q. There is a qualitative difference in the time development of the currents produced in these two cases. Why?? • Consider what happens to the energy! • In the RC circuit, any current developed will cause energy to be dissipated in the resistor. • In the LC circuit, there is NO mechanism for energy dissipation; energy can be stored both in the capacitor and the inductor! Energy in the Electric and Magnetic Fields

Energy stored in a capacitor ... +++ +++ E 1 ------U= C V 2 2 1 u= ε E 2 … energy density ... electric2 0

Energy stored in an inductor …. B 1 U= LI 2 2 1 B2 u = … energy density ... magnetic 2 µ0 RC/LC Circuits I I Q+++ Q+++ C L C ------R

RC: LC: current decays exponentially current oscillates I I

0

0 0 t 0 t LC Oscillations (qualitative)

I = 0 I = − I 0 + + C C - - L ⇒⇒⇒ L

Q = +Q0 Q = 0

⇑⇑⇑ ⇓⇓⇓

I = 0 I = +I 0 -- C C L L ⇐⇐⇐ + +

Q=0 Q = −Q0 Multi-part Clicker t=0 t=t 1 • At t=0, the capacitor in the LC circuit shown has a total charge a Q . At t = t , the capacitor is + + 0 1 = = uncharged. QQ 0 L Q 0 L - - C C – What is the value of Vab =Vb-Va, 2A the voltage across the inductor b at time t1?

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

2B – 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 Multi-part clicker t=0 t=t 1 • At t=0, the capacitor in the LC circuit shown has a total charge a Q . At t = t , the capacitor is + + 0 1 = = uncharged. QQ 0 L Q 0 L - - C C – What is the value of Vab =Vb-Va, 2A the voltage across the inductor b at time t1?

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

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

• Since the charge on the capacitor is zero, the voltage across the capacitor is zero! t=0 t=t 1 • At t=0, the capacitor in the LC circuit shown has a total charge a Q . At t = t , the capacitor is + + 0 1 = Q = 0 uncharged. QQ 0 L L - - C C – What is the relation between UL1, 2B the energy stored in the inductor b at t=t1, and UC1, the energy stored in the capacitor at t=t1?

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

• At t=t1, the charge on • At t=t1, the current is a the capacitor is zero. maximum.

Q 2 1 Q 2 U = 1 = 0 U = LI 2 = 0 > 0 C1 2C L1 2 1 2C Clicker problem:

At time t = 0 the capacitor is fully charged

with Qmax , and the current through the circuit is 0.

2) What is the potential difference across the inductor at t = 0?

a) VL = 0 b) VL = Qmax /C c) VL = Qmax /2C

3) What is the potential difference across the inductor when the current is maximum?

a) VL = 0 b) VL = Qmax /C c) VL = Qmax /2C LC Oscillations (mechanical analogy, for R=0) ω I • What is the oscillation frequency 0? • Begin with the loop rule: + + 2 Q C d Q Q - L L + = 0 - dt 2 C

• Guess solution: (just harmonic oscillator!) remember: d2 x Q = Q cos( ωt +φ) m+ kx = 0 0 dt 2 where φ, Q0 determined from initial conditions

• Procedure: differentiate above form for Q and substitute into loop equation to find ω. 1 • Note: Dimensional analysis


ω = LC LC Oscillations (quantitative) • General solution: + + C - - L Q = Q0 cos(ωt +φ)

• Differentiate: dQ d 2Q Q = −ω Q sin( ω t + φ) L + = 0 dt 0 dt 2 C d 2Q = −ω 2Q cos( ωt + φ) dt 2 0 • Substitute into loop eqn: 1 2 1 L(−ω 2Q cos( ωt +φ))+ ()Q cos( ωt +φ) = 0 ⇒⇒⇒ − ω L + = 0 0 C 0 C Therefore, which we could have determined 1 from the mechanical analogy to SHO: k /1 C 1 ω = ω = = = LC m L LC Multi-part clicker problem t=0

• At t=0 the capacitor has charge Q0; the resulting oscillations have frequency ω0. The maximum + + current in the circuit during these oscillations has Q = Q 0 L - - value I0. C – What is the relation between ω and ω , the 3A 0 2 frequency of oscillations when the initial charge = 2Q0? (a) ω2 = 1/2 ω0 (b) ω2 = ω0 (c) ω2 = 2 ω0

3B – 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 Clicker problem t=0

• At t=0 the capacitor has charge Q0; the resulting oscillations have frequency ω0. The maximum + + current in the circuit during these oscillations has Q = Q 0 L - - value I0. C – What is the relation between ω and ω , the 3A 0 2 frequency of oscillations when the initial charge = 2Q0? (a) ω2 = 1/2 ω0 (b) ω2 = ω0 (c) ω2 = 2 ω0

• Q0 determines the amplitude of the oscillations (initial condition) • The frequency of the oscillations is determined by the circuit parameters (L, C), just as the frequency of oscillations of a mass on a spring was determined by the physical parameters (k, m)! Clicker problem t=0

• At t=0 the capacitor has charge Q0; the resulting oscillations have frequency ω0. The maximum + + current in the circuit during these oscillations has Q = Q 0 L - - value I0. C – What is the relation between I and I , the 3B 0 2 maximum current in the circuit when the initial charge = 2Q0?

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

• The initial charge determines the total energy in the circuit: 2 U0 = Q0 /2 C • The maximum current occurs when Q=0 ! 2 • At this time, all the energy is in the inductor: U = 1/2 LI o • Therefore, doubling the initial charge quadruples the total energy. • To quadruple the total energy, the max current must double! Clicker question:

The current in a LC circuit is a sinusoidal oscillation, with frequency ω.

5) If the inductance of the circuit is increased, what will happen to the frequency ω?

a) increase b) decrease c) doesn’t change

6) If the capacitance of the circuit is increased, what will happen to the frequency?

a) increase b) decrease c) doesn’t change LC Oscillations Energy Check 1 • Oscillation frequency ω = has been found from the loop equation. LC

• The other unknowns ( Q0, φφφ ) are found from the initial conditions. E.g., in our original example we assumed initial values for the charge (Qi) and current (0). For these values: Q0 = Qi, φφφ = 0.

• Question: Does this solution conserve energy? 1 Q 2 (t) 1 U (t) = = Q 2 cos 2 (ω t + φ) E 2 C 2C 0 1 1 U t)( = Li 2 t)( = Lω 2Q2 sin 2 (ω t +φ) B 2 2 0 Energy Check

Energy in Capacitor UE 1 U (t) = Q 2 cos 2 (ωt + φ) E 2C 0

Energy in Inductor

1 2 2 2 UB t)( = Lω Q0 sin (ω t +φ) 0 2 t ⇒ ⇒ 1 ⇒ ⇒ ω = LC UB 1 U (t) = Q 2 sin 2 (ω t + φ) B 2C 0

Q2 Therefore, U t)( +U t)( = 0 0 E B 2C t Inductor-Capacitor Circuits Solving a LC circuit problem; Suppose ωωω=1/sqrt(LC)=3 and given the initial conditions, Q(t = 0)= 5C I()t = 0 =15 A

Solve find Q 0 and φφφ000,,, to get complete solution using,

Q(t = 0)= 5 = Q0 cos (0 +φ0 )

I()()()t = 0 = 15 = −Q0ω sin 0 +φ0 = −3Q0 sin 0 +φ0 and we find,

 15 2 ()5 2 + −  = Q 2 []sin 2 ()()φ + cos 2 φ = Q 2 , Q = 5 2  3  0 0 0 0 0  15  φ = inv .tan − , φ = −45 o 0  5⋅3 0 Remember harmonic oscillators !!

The following are all equally valid solutions

Q(t)= Q0 cos (ωt +φ0 )

Q()()t = Q0 sin ωt +φ1 ()()()()()( Q t = Q0 cos ωt cos φ0 − sin ωt sin φ0 ) Q()()()t = Acos ωt + Bsin ωt Inductor-Capacitor-Resistor Circuit

Q d 2Q 0 = + RI + L C dt 2 d 2Q dQ Q 0 = L + R + dt 2 dt C

Solution will have form of

Q(t)= Ae −αt cos (ω't +φ)

If, 1 R2 > LC 4L2 Inductor-Capacitor-Resistor Circuit 3 solutions, depending on L,R,C values

Very important !!

2 1 R2 1 R2 1 R = < > 2 2 LC 4L2 LC 4L LC 4L Inductor-Capacitor-Resistor Circuit Solving for all the terms Q(t)= Ae −αt cos (ω't +φ)  R  − t  1 R2  = Ae  2L  cos  − t +φ   2   LC 4L  R 1 R2 α = and ω'= − 2L LC 4L2 1 R2 Solution for underdamped circuit; > LC 4L2 For other solutions, use starting form, solve for λλλ and λλλ′λ′′′, Q(t)= Ae −λt + Be −λ't Application of magnetic induction : “smart” traffic lights

Traffic light in California

Another version with two loops Application of magnetic induction

Mechanical ignition in a car

Magnetic energy from ignition coil is used to fire the automotive spark plug. 2 –part Clicker question: The current in a LC circuit is a sinusoidal oscillation, with frequency ω.

I) If the inductance of the circuit is increased, what will happen to the frequency ω? a) increases b) decreases c) doesn’t change

II) If the capacitance of the circuit is increased, what will happen to the frequency?

a) increases b) decreases c) doesn’t change 1 ω = LC Energy Check for LC circuits

Energy in Capacitor UE 1 U (t) = Q 2 cos 2 (ωt + φ) E 2C 0

Energy in Inductor

1 2 2 2 UB t)( = Lω Q0 sin (ω t +φ) 0 2 t ⇒ ⇒ 1 ⇒ ⇒ ω = LC UB 1 U (t) = Q 2 sin 2 (ω t + φ) B 2C 0

Q2 Therefore, U t)( +U t)( = 0 0 E B 2C t Inductor -Capacitor (LC) Circuit Example Solving a LC circuit problem; Suppose ωωω=1/sqrt(LC)=3 and given the initial conditions, Q(t = 0)= 5C I()t = 0 =15 A

Solve find Q 0 and φφφ000,,, to get complete solution using,

Q(t = 0)= 5 = Q0 cos (0 +φ0 )

I()()()t = 0 = 15 = −Q0ω sin 0 +φ0 = −3Q0 sin 0 +φ0 and we find,

 15 2 ()5 2 + −  = Q 2 []sin 2 ()()φ + cos 2 φ = Q 2 , Q = 5 2  3  0 0 0 0 0  15  φ = inv .tan − , φ = −45 o 0  5⋅3 0 Remember harmonic oscillators !!

The following are all equally valid solutions

Q(t)= Q0 cos (ωt +φ0 )

Q()()t = Q0 sin ωt +φ1 ()()()()()( Q t = Q0 cos ωt cos φ0 − sin ωt sin φ0 ) Q()()()t = Acos ωt + Bsin ωt Resistor-Inductor-Capacitor (RLC) Circuit Q d 2Q 0 = + RI + L C dt 2 d 2Q dQ Q 0 = L + R + dt 2 dt C

Solution will have form of

Q(t)= Ae −αt cos (ω't +φ)

If, 1 R2 > LC 4L2 Inductor-Capacitor-Resistor Circuit 3 types of solutions, depending on L,R,C values

Very important !! This is just like the damped SHO

2 1 R2 1 R2 1 R = < > 2 2 LC 4L2 LC 4L LC 4L Inductor-Capacitor-Resistor Circuit

Q(t)= Ae −αt cos (ω't +φ)  R  − t  1 R2  = Ae  2L  cos  − t +φ   2   LC 4L  R 1 R2 α = and ω'= − 2L LC 4L2 1 R2 Solution for underdamped circuit; > LC 4L2 For other solutions, use starting form, solve for λλλ and λλλ′λ′′′, Q(t)= Ae −λt + Be −λ't Alternating Currents (Chap 31)

We next study circuits where the battery is replaced by a sinusoidal voltage or current source.

v(t) = V0 cos (ωt) or ti )( = I 0 cos (ωt)

The circuit symbol is,

An example of an LRC circuit connected to sinusoidal source is,

Important: I(t) is same throughout – just like the DC case. Alternating Currents (Chap 31.1) Since the currents & voltages are sinusoidal, their values change over time and their average values are zero .

A more useful description of sinusoidal currents and voltages are given by considering the average of the square of this quantities . We define the RMS (root mean square), which is the square root of the average of , 2 2 i (t) = (I 0 cos (ωt)) 1 I 2 i 2 (t) = ()I cos ()ωt 2 = I 2 ()1+ cos ()2ωt = 0 0 0 2 2

2 I 0 I RMS = i (t) = 2 V Similarly: 2 0 VRMS = v (t) = 2 Alternating Currents ; Phasors A convenient method to describe currents and voltages in AC circuits is “Phasors”. Since currents and voltages in circuits with capacitors & inductors have different phase relations, we introduce a phasor diagram. For a current, i = I cos (ωt)

We can represent this by a vector rotating about the origin. The angle of the vector is given by ωt and the magnitude of the current is its projection on the X-axis.

If we plot simultaneously currents & voltages of different components we can display relative phases .

Note this method is equivalent to imaginary numbers approach where we take the real part (x-axis projection) for the magnitude Alternating Currents: Resistor in AC circuit

A resistor connected to an AC source will have the voltage, vR, and the current across the resistor has the same phase. We can represent the current phasor and the voltage phasor with the same angle.

vR = VR cos (ωt)= iR = I cos (ωt)R

and VR = IR (just like DC case) Phasors are rotating 2 dimensional vectors Resistor in AC circuit; I & V versus ω t ω

I(t)=Icos( ωωωt) V(t)=RIcos( ωωωt)

ωωωt →→→ a b c d e f

NB: for a resistor voltage is in phase with current Alternating Currents: Capacitor in AC circuit

In a capacitor connected to an AC current source the voltage lags behind the current by 90 degree. We can draw the current phasor and the voltage phasor behind the current by 90 degrees. dq i = = I cos ()ωt dt q 1 I I v = = idt = sin ()ωt q = sin ()ωt Find voltage: C C ∫ ωC ω  1  V = I   MAX ωC  Alternating Currents ; Capacitor in AC circuit 1 We define the capacitive reactance, X , as X = C C ωC I  1  V MAX = = I  = I X Like: V = IR cap ωC  ωC  C R But frequency dependent

We stated that voltage lags by 90 deg., so equivalent solution is

I I v = cos ()ωt − 90 = [cos ωt cos 90 + sin ωt sin 90 ] ωC ωC I = sin ωt ωC Capacitor in AC circuit; I & V versus ω t ω i(t)=Icos( ωt)

v(t)=(I/ ωC)sin( ωt) a b c d e f ωωωt →→→

Note voltage lags 90 deg. Behind current

V(t)=(I/ ωC)sin( ωt)= (I/ ωC)cos( ωt-π/2) Alternating Currents: Inductor in AC circuit

In an inductor connected to an AC current source, the voltage will lead the current by 90 degrees. We can draw the current phasor and the voltage phasor ahead of the current by 90 degrees. di i = I cos ()ωt and V = L = −IL ω sin (ωt) dt VMAX = IL ω Define inductive reactance, X , as X = ωL L L Like: V = IR MAX R Vind = IωL = I(ωL)= I X L But frequency dependnt Inductor in AC circuit; I & V versus ω t ω i(t)=Icos( ωωωt)

v(t)= - IL ωωωsin( ωωωt) a b c d e f ωωωt →→→

Draw phasor diagram for each point Note voltage is 90 deg. ahead of current v(t)=IL ωωωsin( ωωωt)= IL ωωωcos( ωωωt + πππ/2) AC summary

i(t) v(t) X V Phasor ω ω resistor I cos t VR cos t R IR in phase ω ω ω capacitor I cos t VC sin t 1/ C IX C lags ω ω ω inductor I cos t -VLsin t L IX L leads