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

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W12D2 RC, LR, and Undriven RLC Circuits; Experiment 4 W12D2 RC, LR, and Undriven RLC Circuits; Experiment 4 Today’s Reading Course Notes: Sections 11.4-11.6 1 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 2 Outline Experiment 4: Part 1 RC and LR Circuits Simple Harmonic Oscillator Undriven RLC Circuits Experiment 4: Part 2 Undriven RLC Circuits 3 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 = Vb − Va = −ε 4 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 = Vb − Va = +IR 5 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 = Vb − Va = −Q / C 6 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 7 LR Circuits 8 Inductors in Circuits Inductor: Circuit element with self-inductance Ideally it has zero resistance Symbol: 9 3 Non-Ideal Inductors Non-Ideal (Real) Inductor: Not only L but also some R = dI In direction of current: ε = −L − IR dt 10 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 11 Kirchhoff’s Modified 2nd Rule dΦ Δ V = − E ⋅ d s = + B ∑ i ∫ d t i dΦ ⇒ Δ V − B = 0 ∑ i d t i 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. 13 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 16 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 −t /τ I(t) = (Qo / RC)e time constant: τ = RC 17 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) 18 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) 19 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 By definition τ ≡ t − t then 2 1 (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 y0 − (t / τ ) ⇒ τ = −1/slope 20 Experiment 4: RC and RL Circuits 21 7 Mass on a Spring: Simple Harmonic Motion 22 Demonstration Mass on a Spring: Simple Harmonic Motion Mass on a Spring (C 2) 23 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 x : Amplitude of Motion k 0 ω = = Angular frequency f: Phase (time offset) 0 m 24 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 ∝ −x? dt 2 1 1. x(t) = at 2 2 2. x(t) = Aet /T −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 28 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 v (t) x sin( t ) x(t) = x0 cos(ω0t + φ) = x = −ω0 0 ω0 + φ 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. 33 11 Energy stored in electric field Energy stored in magnetic field Energy stored in electric field Energy stored in magnetic field 34 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. 35 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. 36 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.
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