W12D2 RC, LR, and Undriven RLC Circuits; Experiment 4

Home , LC circuit, RC circuit, RLC circuit, RL circuit

Today’s Reading Course Notes: Sections 11.4-11.6

Announcements

Math Review Tuesday May 1 9pm-11 pm in 32-082

PS 9 due Tuesday Tuesday May 1 at 9 pm in boxes outside 32-082 or 26-152

Next Reading Assignment W12D3 Course Notes: Sections 12.8-12.9

Outline

Experiment 4: Part 1 RC and LR Circuits

Simple Harmonic Oscillator

Undriven RLC Circuits

Experiment 4: Part 2 Undriven RLC Circuits

1 Sign Conventions - Battery

Moving from the negative to positive terminal of a battery increases your potential

ΔV = V − V = +ε b a

Moving from the positive to negative terminal of a battery decreases your potential

ΔV = V − V = −ε b a

Sign Conventions - Resistor

Moving across a resistor in the direction of current decreases your potential

ΔV = V − V = −IR b a

Moving across a resistor opposite the direction of current increases your potential

ΔV = V − V = +IR b a

Sign Conventions - Capacitor Moving across a capacitor from the negatively to positively charged plate increases the electric potential

ΔV = V − V = +Q / C b a

Moving across a capacitor from the positively to negatively charged plate decreases the electric potential

ΔV = V − V = −Q / C b a

2 (Dis)Charging a Capacitor 1. When the direction of current flow is toward the positive plate of a capacitor, then

dQ I = + dt

2. When the direction of current flow is away from the positive plate of a capacitor, then

dQ I = − dt

LR Circuits

Inductors in Circuits Inductor: Circuit element with self-inductance Ideally it has zero resistance

Symbol:

3 Non-Ideal Inductors

Non-Ideal (Real) Inductor: Not only L but also some R

dI In direction of current: ε = −L − IR dt

Sign Conventions - Inductor

Moving across an inductor in the direction of current contributes

Moving across an inductor opposite the direction of current contributes dI ε = +L dt

Kirchhoff’s Modified 2nd Rule

  dΦ Δ V = − E ⋅ d s = + B ∑ i ∫ d t i dΦ V B 0 ⇒ ∑Δ i − = i d t If all inductance is ‘localized’ in inductors then our problems go away – we just have: d I V L 0 ∑Δ i − = i d t 12

4 Steps For Setting Up Circuit Equations for Circuit with N loops and M junctions

1. Simplify resistors in series/parallel 2. Assign current in every branch 3. Choose circulation direction for N-1 loops 4. Assign charges to each side of capacitor. 5. Determine relation between current and charge for each branch containing a capacitor 6. Write M-1 current conservation equations for junctions 7. Write N-1 loop equations.

Group Problem: RC and RL Circuits

For each of the circuits shown. above, at t = 0 the switch is in position a and then a very long time later compared to the characteristic time constant of the circuit, the switch is flipped to position b. Find the differential equation associated to each circuit when the (i) switch is in position a (ii) switch is in position b. 14

Group Problem: LR Circuit

For the circuit shown in the figure the currents through the two bottom branches as a function of time (switch closes at t = 0, opens at t = T>>L/R). State the values of current (i) just after switch is closed at t = 0+ (ii) Just before switch is opened at t = T-, (iii) Just after switch is opened + at t = T 15

5 RC Circuit Charging dQ 1 = − (Q − Cε) dt RC Solution to this equation when switch is closed at t = 0: −t /τ Q(t) = Cε(1− e ) I(t) = I e−t /τ 0 τ = RC : time constant

RC Circuit: Discharging dQ 1 = − Q dt RC Solution to this equation when switch is closed at t = 0 Q(t) = Q e−t / RC o I(t) = (Q / RC)e−t /τ o time constant: τ = RC

RL Circuit: Increasing Current dI R ⎛ ε ⎞ = − I − ⇒ dt L ⎝⎜ R⎠⎟ ε I(t) = (1− e−t /( L/ R) ) R Solution to this equation when switch is closed at t = 0:

ε I(t) = (1− e−t /τ ) R

τ = L / R : time constant

(units: seconds)

6 RL Circuit: Decreasing Current Switch is flipped from position a to position b at t = 0:

dI −L + −IR = 0 dt

Solution to this equation.

I(t) = I e−t /( L/ R) 0

τ = L / R : time constant

(units: seconds)

Measuring Time Constant

−(t /τ ) Pick a point 1 with y (t ) = y e 1 1 1 0 1 Find point 2 such that y (t ) = y (t )e− 2 2 1 1 τ ≡ t − t By definition 2 1 then

(t )/ (t / ) y (t ) = y (t + τ ) = y e− 1 +τ τ = y e− 1 τ e−1 = y (t )e−1 2 2 2 1 0 0 1 1 2) In the lab you will plot semi-log and fit curve (make sure you exclude data at both ends) −(t /τ ) ln( y(t) / y0 ) = lne = −(t / τ ) ⇒ y(t) = y e−(t /τ ) ⇒ 0 ln y(t) = ln y − (t / τ ) ⇒ τ = −1/slope 0 20

Experiment 4: RC and RL Circuits

7 Mass on a Spring: Simple Harmonic Motion

Demonstration Mass on a Spring: Simple Harmonic Motion Mass on a Spring (C 2)

Mass on a Spring

(1) (2) What is Motion? d 2 x F = −kx = ma = m dt 2 d 2 x (3) (4) m 2 + kx = 0 dt Simple Harmonic Motion x(t) = x cos(ω t + φ) 0 0

x0: Amplitude of Motion k ω0 = = Angular frequency f: Phase (time offset) m

8 Simple Harmonic Motion 1 1 Period = → T = frequency f Amplitude (x ) 2π 2π 0 Period = → T = angular frequency ω

x(t) = x cos(ω t + φ) 0 0 −π Phase Shift (φ) = 2 25

Concept Question: Simple Harmonic Oscillator

Which of the following functions x(t) has a second derivative which is proportional to the negative of the function d 2 x 2 ∝ −x? dt 1 1. x(t) = at 2 2 t /T 2. x(t) = Ae −t /T 3. x(t) = Ae ⎛ 2π ⎞ 4. x(t) = Acos t ⎜ T ⎟ ⎝ ⎠

Concept Question Answer: Simple Harmonic Oscillator

Answer 4. By direct calculation, when

⎛ 2π ⎞ x(t) = Acos t ⎜ T ⎟ ⎝ ⎠ dx(t) ⎛ 2π ⎞ ⎛ 2π ⎞ = − Asin t dt ⎜ T ⎟ ⎜ T ⎟ ⎝ ⎠ ⎝ ⎠

2 2 d 2 x(t) ⎛ 2π ⎞ ⎛ 2π ⎞ ⎛ 2π ⎞ = − Acos t = − x(t) dt 2 ⎜ T ⎟ ⎜ T ⎟ ⎜ T ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

9 Analog: RLC Circuit

Inductors are like masses (have inertia) Capacitors are like springs (store/release energy) Batteries supply external force (EMF)

Charge on capacitor is like position, Current is like velocity – watch them resonate

Group Problem: LC Circuit

1. Set up the circuit above with capacitor, inductor, resistor, and battery.

2. Let the capacitor become fully charged.

3. Throw the switch from a to b.

4. Find the differential equation associated to the circuit 29

Mass on a Spring: Energy (1) Spring (2) Mass (3) Spring (4) Mass

dx x(t) = x cos(ω t + φ) = vx (t) = −ω0 x0 sin(ω0t + φ) 0 0 dt Energy has 2 parts: (Mass) Kinetic and (Spring) Potential

2 1 ⎛ dx⎞ 1 K = m = kx2 sin2 (ω t + φ) Energy 2 ⎝⎜ dt ⎠⎟ 2 0 0 sloshes back 1 1 U = kx2 = kx2cos2 (ω t + φ) and forth s 2 2 0 0 30

10 Analog: LC Circuit

Mass doesn’t like to change velocity Kinetic energy associated with motion 2 dv d x 1 2 F = ma = m = m 2 ; K = mv dt dt 2 Inductor doesn’t like to have current change Energy associated with current 2 dI d q 1 2 ε = −L = −L 2 ; U B = LI dt dt 2 F → ε; v → I; m → L 31

Analog: LC Circuit

Spring doesn’t like to be compressed/extended Potential energy associated with compression 1 F = −kx; E = kx2 2 Capacitor doesn’t like to be charged (+ or -) Energy associated with stored charge Q 1 Q2 ε = ; U E = C 2 C

F → ε; x → Q; v → I; m → L; k → C −1 32

LC Circuit

It undergoes simple harmonic motion, just like a mass on a spring, with trade-off between charge on capacitor (Spring) and current in inductor (Mass). Equivalently: trade-off between energy stored in electric field and energy stored in magnetic field.

11 Energy stored in electric field

Energy stored in magnetic field

Energy stored in electric field

Energy stored in magnetic field

Concept Question: LC Circuit

Consider the LC circuit at right. At the time shown the current has its maximum value. At this time:

1. the charge on the capacitor has its maximum value. 2. the magnetic field is zero. 3. the electric field has its maximum value. 4. the charge on the capacitor is zero.

Concept Q. Answer: LC Circuit

Answer: 4. The current is maximum when the charge on the capacitor is zero

Current and charge are exactly 90 degrees out of phase in an ideal LC circuit (no resistance), so when the current is maximum the charge must be identically zero.

12 Concept Question: LC Circuit

In the LC circuit at right the current is in the direction shown and the charges on the capacitor have the signs shown. At this time,

1. I is increasing and Q is increasing. 2. I is increasing and Q is decreasing. 3. I is decreasing and Q is increasing. 4. I is decreasing and Q is decreasing.

Concept Q. Answer: LC Circuit

Answer: 2. I is increasing; Q is decreasing

With current in the direction shown, the capacitor is discharging (Q is decreasing).

But since Q on the right plate is positive, I must be increasing. The positive charge wants to flow, and the current will increase until the charge on the capacitor changes sign. That is, we are in the first quarter period of the discharge of the capacitor, when Q is decreasing and positive and I is increasing and positive.

LC Circuit: Simple Harmonic Oscillator Q dI dQ − L = 0 ; I = − C dt dt d 2Q 1 2 + Q = 0 ⇒ dt LC

Simple harmonic oscillator: Q(t) = Q cos(ω t + φ) Charge: 0 0 Angular frequency: ω = 1/ LC 0 Amplitude of charge oscillation: Q 0 Phase (time offset): φ

13 LC Oscillations: Energy

Notice relative phases

2 Q2 ⎛ Q ⎞ 1 1 ⎛ Q2 ⎞ U = = 0 cos2 ω t U = LI 2 = LI 2 sin2 ω t = 0 sin2 ω t E 2C ⎜ 2C ⎟ 0 B 2 2 0 0 ⎜ 2C ⎟ 0 ⎝ ⎠ ⎝ ⎠ 2 2 Q 1 2 Q0 U = U E + U B = + LI = 2C 2 2C Total energy is conserved !! 40

LC Circuit Oscillation Summary

Adding Damping: RLC Circuits

14 RLC Circuit: Energy Changes

Q dI Include finite resistance: + I R + L = 0 C dt dQ Multiply by I = ⇒ dt Q dQ dI + I 2 R + LI = 0 ⇒ C dt dt d ⎡ Q2 1 ⎤ + L I 2 = − I 2 R dt ⎢ 2C 2 ⎥ ⎣ ⎦ Decrease in stored d energy is equal to Joule 2 (U E + U B )= − I R heating in resistor dt

Damped LC Oscillations

Resistor dissipates energy and system rings down over time. Also, frequency decreases: −( R/ 2 L)t ω = 1/ LC > R / 2L Q(t) = Q0e cos(ω 't) 0 2 2 ω ' = ω0 − (R / 2L) 44

Experiment 4: Part 2 Undriven RLC Circuits

15 Appendix: Experiment 4: Part 2 Undriven RLC Circuits

Group Problem

Problem: LC Circuit

Consider the circuit shown in the figure. Suppose the switch that has been connected to point a for a long time is suddenly thrown to b at t = 0. Find the following quantities:

(a) the frequency of oscillation of the circuit. (b) the maximum charge that appears on the capacitor. (c) the maximum current in the inductor. (d) the total energy the circuit possesses as a function of time t.

Follow-Up Concept Question: LR Circuits

16 Concept Question: Inserting a Core

When you insert the iron core into a current carrying coil does the time constant ?

1. Increase,

2. decrease,

3. or stay the same.

Concept Q. Ans.: Inserting a Core

Answer 1. When you insert the iron core into the coil, the magnetic field increases in the core hence the inductance increases. Therefore the time constant increases.