RC, LR, and Undriven RLC Circuits; Experiment 4

Outline

Experiment 4: Part 1 RC and LR Circuits

Undriven RLC Circuits

Experiment 4: Part 2 Undriven RLC Circuits

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

1 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)

RL Circuit: Decreasing Current

dI −L + −IR = 0 ⇒ dt I(t) = I e−t /( L/ R) 0 Solution to this equation when switch is opened at t = 0:

ε I(t) = (1− e−t /τ ) R

τ = L / R : time constant

(units: seconds)

2 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

t t By definition τ ≡ 2 − 1 then

(t )/ (t / ) y (t ) y (t ) y e− 1 +τ τ y e− 1 τ e−1 y (t )e−1 2 2 = 2 1 + τ = 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 7

Experiment 4: RC and RL Circuits

Follow-Up Concept Question: LR Circuits

3 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.

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( t ) = 0 ω0 + φ

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

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

x(t) x cos( t ) = 0 ω0 + φ −π Phase Shift (φ) = 2 13

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 2 d 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 ' # &

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 15

5 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

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 17

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 18

6 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. What happens? 19

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.

Energy stored in electric field

Energy stored in magnetic field

Energy stored in electric field

Energy stored in magnetic field

7 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 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.

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( t ) Charge: = 0 ω0 + φ Angular frequency: 1/ LC ω0 = Amplitude of charge oscillation: Q 0 Phase (time offset): φ

8 LC Oscillations: Energy

Notice relative phases

2 2 2 Q ! Q $ 2 1 1 " Q % U = = 0 cos 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 !! 25

LC Circuit Oscillation Summary

Adding Damping: RLC Circuits

9 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 1/ LC R / 2L Q(t) = Q0e cos(ω 't) ω0 = > 2 2 ω ' = ω0 − (R / 2L) 29

Experiment 4: Part 2 Undriven RLC Circuits (Section 5 on Survey)

10 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. 31

Concept Question: Expt. 4 In today’s lab the battery turns on and off. Which circuit diagram is most representative of our circuit? 1. 2.

3. 4.

Concept Question: LC Circuit

The plot shows the charge 1.0Q 1.0I 0 T Current 0 on a capacitor (black curve) lag Charge

0.5Q 0.5I and the current through it 0 0 (red curve) after you turn 0.0Q 0.0I off the power supply. If you 0 0

-0.5Q -0.5I put a core into the inductor 0 0 Charge on Capacitor what will happen to the time through Capacitor Current -1.0Q -1.0I 0 0 0 40 80 120 TLag? Time (mS) 1. It will increase 2. It will decrease 3. It will stay the same 4. I don’t know 33

11 Concept Question: LC Circuit

1.0Q 1.0I 0 T Current 0 If you increase the lag Charge 0.5Q 0.5I resistance in the circuit 0 0

0.0Q 0.0I what will happen to rate 0 0 of decay of the pictured -0.5Q -0.5I 0 0 amplitudes? Charge on Capacitor Current through Capacitor Current -1.0Q -1.0I 0 0 0 40 80 120 Time (mS) 1. It will increase (decay more rapidly) 2. It will decrease (decay less rapidly) 3. It will stay the same 4. I don’t know