Electricity & Magnetism Lecture 21

Electricity & Magnetism Lecture 21

Uncertainties DC Meters! Digital error of at least ±1 digit in the rightmost digit.! Analog 'calibra:on' error is specified by a percentage of the reading, o?en about 1%! These two sources of error add (not in quadrature). For example if the error is specified as 5%±1 digit then a reading of 1.02 V would have error of ±(.05+.01)=±.06 V

Uncertainties Oscilloscope! Digital measurements subject to both digital and analog errors as for DMM (If unknown use ±(3%+1digit).! For measurements with cursors, move the cursors and note the minimum increment of the reading! If you read the scale by eye, use the same type es:mates as you use when reading a ruler: o?en ±½ smallest division but use discre:on for very large or small divisions, and also take into account the line width and possible noise on the signal.

Your Comments If we get stuck during the lab prac:cal, can we ask ques:ons? Does the average power change when the circuit goes to resonance? help make sense of all the equa:ons!!! do we use phase diagrams to figure all of this stuff out? ABSOLUTELY We will review again today.. What topics are going to be on Midterm 2? The concept of Q. The concept is important Where did all this weird math come from? oot mean squares? The brackets? It s all about average values of oscilla:ng quan::es fro noe of hte hw porblmes yuo spelt evrything as if yuo wree drunk. hmmm Up through AC Circuits, esonance and Power SmartPhysics Unit 21, Activity Guide 27

Practical Test Hints You will have! 4 banana plug wires, ~4 alligator clips! 1 Scope Probe (x1/x10)! 1 BNC wire! adaptors: Tee, BNC-Male Banana, BNC-Female Banana! proto-board (Use it!)! scope, func:on generator, DMM! 4 bameries and holder or DC power supply

PHASOS AE THE KEY! FOMULAS AE NOT! STAT WITH PHASO DIAGAM ε max C L DEVELOP FOMULAS FOM THE DIAGAM!! I max ε max = I max Z V = Projec:on along Ver:cal ω I max I max (X L X C ) X L = ωl X C = 1/ωC I max f X L - X C

Peak AC Problems Ohms Law for each element NOTE: Good for PEAK values only) V gen = I max Z V esistor = I max V inductor = V Capacitor = ε max C L I max Typical Problem A generator with peak voltage 15 volts and angular frequency 25 rad/sec is connected in series with an 8 Henry inductor, a 0.4 mf capacitor and a 50 ohm resistor. What is the peak current through the circuit? X L X C

Peak AC Problems Ohms Law for each element NOTE: Good for PEAK values only) V gen = I max Z V esistor = I max ε max C L V inductor = V Capacitor = I max Typical Problem A generator with peak voltage 15 volts and angular frequency 25 rad/sec is connected in series with an 8 Henry inductor, a 0.4 mf capacitor and X a 50 ohm resistor. What is the peak current through the circuit? L Which element has the largest peak voltage across it? A) Generator E) All the same. B) Inductor C) esistor D) Capacitor X C

Peak AC Problems Ohms Law for each element NOTE: Good for PEAK values only V gen = I max Z V esistor = I max V inductor = V Capacitor = ε max C L I max Typical Problem A generator with peak voltage 15 volts and angular frequency 25 rad/sec is connected in series with an 8 Henry inductor, a 0.4 mf capacitor and a 50 ohm resistor. What is the peak current through the circuit? What happens to the impedance if we decrease the angular frequency to 20 rad/sec? A) Z increases B) Z remains the same C) Z decreases (X L X C ) : (200 100) (160 125) X L X C X L Z 25 Z 20 X C

esonance Light-bulb Demo

esonance in AC circuits ω 0 : Frequency at which voltage across inductor and capacitor cancel is independent of ω X L increases with ω X C increases with 1/ω Impeance Z = at resonance Z XL X C ω 0 frequency is minimum at resonance esonance: X L = X C

Off esonance Z In general U max = max energy stored ΔU = energy dissipated in one cycle at resonance

Off esonance

CheckPoint 2 I max esonance: X L = X C Z = 70 53 35 I max Same since doesn't change 18 0 Case 1 Case 2 1

CheckPoint 4 I max 50 Voltage in second circuit will be twice that of the first because of the 2L compared to L. 38 25 13 0 1 I max Case 1 Case 2

CheckPoint 6 I max 40 The peak voltage will be greater in circuit 2 because the value of X C doubles. 30 20 10 0 1 I max Case 1 Case 2

CheckPoint 8 I max I max The voltage across the inductor and the capacitor are equal when at resonant frequency, so there is no lag or lead. Case 1 Case 2

CheckPoint 8 I max I max The voltage across the inductor and the capacitor are equal when at resonant frequency, so there is no lag or lead. Case 1 Case 2

Power P = IV instantaneous always true! Difficult for Generator, Inductor and Capacitor because of phase! esistor I,V are always in phase! P = IV = I 2 Average Power Inductor and Capacitor = 0 ( < sinωt cosωt > = 0 ) esistor <I 2 > = <I 2 > = ½ I 2 peak MS = oot Mean Square I peak = I rms 2 < I 2 > = I rms 2

Power Line Calculation If you want to deliver 1500 Watts at 100 Volts over transmission lines w/ resistance of 5 Ohms. How much power is lost in the lines?! Current Delivered: I = P/V = 15 Amps! Loss = IV (on line) = I 2 = 15 15 5 = 1125 Watts! If you deliver 1,500 Watts at 10,000 Volts over the same transmission lines. How much power is lost?! Current Delivered: I = P/V = 0.15 Amps! Loss = IV (on line) = I 2 = 0.125 Watts hxps://en.wikipedia.org/wiki/electric_power_transmission

Transformers Applica:on of Faraday s Law! Changing EMF in Primary creates changing flux! Changing flux, creates EMF in secondary Efficient method to change voltage for AC. Power Transmission Loss = I 2 Power electronics

Follow-Up from Last Lecture Consider the harmonically driven series LC circuit shown. V max = 100 V I max = 2 ma V Cmax = 113 V (= 80 sqrt(2)) The current leads generator voltage by 45 o (cos = sin = 1/ 2) L and are unknown. How should we change ω to bring circuit to resonance? A) decrease ω B) increase ω C) Not enough info V C L Original ω V L At resonance (ω 0 ) At resonance X L = X C V L φ V C V V V C V V X L increases X C decreases ω increases

More Follow-Up Consider the harmonically driven series LC circuit shown. V max = 100 V I max = 2 ma V Cmax = 113 V (= 80 2) The current leads generator voltage by 45 o (cos = sin = 1/ 2) L and are unknown. V C L By what factor should we increase ω to bring circuit to resonance? i.e. if ω 0 = fω, what is f? A) B) C) D) If ω is increased by a factor of f: At resonance X L = X C X L increases by factor of f X C decreases by factor of f X L f 15 2 X C (1/ f ) 40 2 Note: f is not frequency here!

Current Follow-Up C Consider the harmonically driven series LC circuit shown. V max = 100 V I max = 2 ma V Cmax = 113 V (= 80 2) The current leads generator voltage by 45 o (cos = sin = 1/ 2) L and are unknown. V L What is the maximum current at resonance? ( I max (ω 0 ) ) A) B) C) At resonance X L = X C

Phasor Follow-Up Consider the harmonically driven series LC circuit shown. V max = 100 V I max = 2 ma V Cmax = 113 V (= 80 2) The current leads generator voltage by 45 o (cos = sin = 1/ 2) L and are unknown. V C L What does the phasor diagram look like at t = 0? (assume V = V max sin(ωt) ) V L X V L A) B) C) D) V C V V V = V max sin(ωt) V is horizontal at t = 0 (V = 0) V V C V V L X V C V V V L X V V C V V L < V C if current leads generator voltage