Course Updates. Reminders: 1) Assignment #11 posted, due next Wednesday. 2) Quiz # 5 today (Chap 29, 30) 3) Complete AC Circuits

ourse Updates http://www.phys.hawaii.edu/~varner/phys7-spr10/physics7.html eminders: 1) Assignment #11 posted, due next Wednesday ) Quiz # 5 today (hap 9, 30) 3) omplete A ircuits

ast time: ircuits with phasors where... X ω 1 X ω The phasor diagram gives us graphical solutions for φ and I: ε m I X φ I ε I X tanφ X X I X I X ε m ε m I ( ( ) ) + X X ( X X ) IZ ε m I + Z φ I + ( ) X X

series circuit; Summary of instantaneous urrent and voltages V V V I IX IX () t I cos( ωt) () t Icos( ωt) i v v v ε ( t 1 ω () t IX cos( ωt 90) I cos( ωt 90) () t IX cos( ωt + 90) Iω cos( ωt + 90) ) v () t IZ cos( ωt + φ) ε cos( ωt + φ) V tanφ ad V V ω 1/ ω Z m ( X ) + ( X - X )

agging & eading The phase φ between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances. I ε Z m tan φ X X X ω 1 X ω X Z X X φ φ Z X X Z X X > X φ > 0 current AGS applied voltage X < X φ < 0 current EADS applied voltage X X φ 0 current IN PHASE WITH applied voltage

Impedance, Z From the phasor diagram we found that the current amplitude I was related to the drive voltage amplitude ε m by ε m Im Z Z is known as the impedance, and is basically the frequency dependent equivalent resistance of the series circuit, given by: Impedance Triangle ε IZ φ I I ( ) m Z + X X Im X X or Z Z φ cos( φ) X X Note that Z achieves its minimum value () when φ 0. Under this condition the maximum current flows in the circuit.

esonance For fixed,, the current I will be a maximum at the resonant frequency ω which makes the impedance Z purely resistive (Z ). εm i.e., εm I m Z + ( X X ) reaches a maximum when: X X This condition is obtained when: 1 ω ω ω 1 Note that this resonant frequency is identical to the natural frequency of the circuit by itself! At this frequency, the current and the driving voltage are in phase: X X tan φ 0

esonance Plot the current versus ω, the frequency of the voltage ε source: m I m 0 0 ω ω For ω very large, X >> X, φ 90, I m 0 For ω very small, X >> X, φ -90, I m 0 Example: vary V100 v ω1000 rad/s 00, 500, 000 ohm H 0.5 μ

Question 1 onsider a general A circuit containing a resistor, capacitor, and inductor, driven by an A generator. As the frequency of the circuit is either raised above or lowered below the resonant frequency, the impedance of the circuit. a) always increases b) only increases for lowering the frequency below resonance c) only increases for raising the frequency above resonance

Question onsider a general A circuit containing a resistor, capacitor, and inductor, driven by an A generator. At the resonant frequency, which of the following is true? a) The current leads the voltage across the generator. b) The current lags the voltage across the generator. c) The current is in phase with the voltage across the generator.

Impedance Z sqrt( + (X -X ) ) At resonance, (X -X ) 0, and the impedance has its minimum value: Z As frequency is changed from resonance, either up or down, (X -X ) no longer is zero and Z must therefore increase. hanging the frequency away from the resonant frequency will change both the reductive and capacitive reactance such that X -X is no longer 0. This, when squared, gives a positive term to the impedance, increasing its value. By definition, at the resonance frequency, I max is at its greatest and the phase angle is 0, so the current is in phase with the voltage across the generator.

Question 3 Fill in the blank. This circuit is being driven its resonance frequency. a) above b) below c) exactly at

Question 4 The generator voltage the current. a) leads b) lags c) is in phase with

At resonance, X X. Here, X > X Therefore, need to reduce X ω and increase X 1/(ω) Therefore, lower ω!! From diagram, ε leads I (rotation ccw)

Power in circuit () t I() t cos( ω t) ; v ( t) V ( ωt + φ) i ad cos The instantaneous power delivered to -- is: () t i() t v () t V cos ( ωt + φ) I cos( t) P ad ω We can use trig identities to expand the above to, P () V[ cos( ωt) cos( φ) sin( ωt) sin( φ) ] I cos( ωt) VI cos ( ωt) cos( φ) VI sin( ωt) cos( ωt) sin( φ) P() t VI cos ( ωt) cos( φ) VI sin( ωt) cos( ωt) sin( φ) P t ave P ave VI V cos 1 V I ( ωt) cos( φ) VI cos( φ) 1 () VI cos( φ) cos( φ) P t MS I MS cos ( φ)

Power in circuit, continued 1 P ave P t () VI cos( φ) V I cos( φ) MS General result. V MS is voltage across element, I MS is current through element, and ϕ is phase angle between them. Example; 100Watt light bulb plugged into 10V house outlet, Pure resistive load (no and no ), ϕ 0. P V P I rms I rms rms ave P V V ave rms rms V 10 100 100 10 rms 144Ω 0.83A MS Note; 10V house voltage is rms and has peak voltage of 10 170V Question: What is P AVE for an inductor or capacitor?

Question 5 If you wanted to increase the power delivered to this circuit, which modification(s) would work? a) increase b) increase c) increase, d) decrease e) decrease,

Question 5 If you wanted to increase the power delivered to this circuit, which modification(s) would work? a) increase b) increase or c) increase, d) decrease e) decrease,

Question 6 Would using a larger resistor increase the current? a) yes b) no

Power ~ Icosφ ~ (1/Z)(/Z) (/Z ) To increase power, want Z to decrease: : decrease X decrease : increase X decrease : decrease Z decrease Since power peaks at the resonant frequency, try to get X and X to be equal. Power also depends inversely on, so decrease to increase Power.

Power Pt () ε I cosφ rms ( I ) rms power factor rms { Z Summary ε rms + 1 ε m ( ) X X I rms tanφ 1 X I m X Driven Series ircuit: esonance condition esonant frequency ω 1

For next time Homework #11 posted, due next Wed. Quiz now: Faraday s aw, Inductance and Inductors, circuits