Physics 1 ecture 1 esonance and power in A circuits Physics 1 ecture 1, Slide 1
I max X X = w I max X w e max I max X X = 1/w I max I max I max X e max = I max Z I max I max (X -X ) f X -X Physics 1 ecture 1, Slide
Peak A Problems 07 Ohms aw for each element PEAK values V gen = I max Z V esistor = I max V inductor = I max X V apacitor = I max X Problem A generator with peak voltage 15 volts and angular frequency 5 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 X w 00 1 100 w Z X X 11 Vgen Imax 0.13A Z X X Physics 1 ecture 1, Slide 3
Peak A Problems 1 Ohms aw for each element V gen = I Z Z X X V esistor = I V inductor = I X V apacitor = I X Typical Problem A generator with peak voltage 15 volts and angular frequency 5 rad/sec is connected in series with an 8 Henry inductor, a 0.004 Farad capacitor and a 50 ohm resistor. What is the peak current through the circuit? Which element has the largest peak voltage across it? A) Generator E) All the same. B) Inductor ) esistor D) apacitor X X V I X max w 1 w max X X X X w 00 1 100 w Z X X 11 Vgen Imax 0.13A Z Physics 1 ecture 1, Slide 5
Peak A Problems 14 Ohms aw for each element V gen = I Z Z X X V esistor = I V inductor = I X V apacitor = I X Typical Problem A generator with peak voltage 15 volts and angular frequency 5 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 0 rad/sec? A) Z increases B) Z remains the same X X w 1 w X X Z 5 Z 0 ) Z decreases (X -X ) : (00-100) (160-15) X X Physics 1 ecture 1, Slide 6
esonance ight-bulb Demo Physics 1 ecture 1, Slide 7
esonance Impedance Frequency at which voltage across inductor and capacitor cancel is independent of w esonance in A ircuits X increases with w X w Z = at resonance X increases with 1/w 1 X w Z Z ( X X ) is minimum at resonance X w 0 X frequency esonance: X = X w0 1 Physics 1 ecture 1, Slide 8 10
heckpoint 1a onsider two circuits with identical generators and resistors. Both circuits are driven at the resonant frequency. ircuit II has twice the inductance and 1/ the capacitance of circuit I as shown above. ompare the peak voltage across the resistor in the two circuits A. V I > V II B. V I = V II. V I < V II I max X I max X esonance: X = X Z = Same since doesn't change I max I max X ase 1 I max ase I max X Physics 1 ecture 1, Slide 9
heckpoint 1b onsider two circuits with identical generators and resistors. Both circuits are driven at the resonant frequency. ircuit II has twice the inductance and 1/ the capacitance of circuit I as shown above. ompare the peak voltage across the inductor in the two circuits A. V I > V II B. V I = V II. V I < V II I max X I max X I max I max I max X Voltage in second circuit will be twice that of the first because of the compared to ase 1 ase I max X Physics 1 ecture 1, Slide 10
heckpoint 1c onsider two circuits with identical generators and resistors. Both circuits are driven at the resonant frequency. ircuit II has twice the inductance and 1/ the capacitance of circuit I as shown above. ompare the peak voltage across the capacitor in the two circuits A. V I > V II B. V I = V II. V I < V II I max X I max X I max I max I max X The peak voltage will be greater in circuit because the value of X doubles. ase 1 ase I max X Physics 1 ecture 1, Slide 11
heckpoint 1d onsider two circuits with identical generators and resistors. Both circuits are driven at the resonant frequency. ircuit II has twice the inductance and 1/ the capacitance of circuit I as shown above. At the resonant frequency, which of the following is true? A. urrent leads voltage across the generator B. urrent lags voltage across the generator. urrent is in phase with voltage across the generator I max X I max X I max I max I max X The voltage across the inductor and the capacitor are equal when at resonant frequency, so there is no lag or lead. ase 1 ase I max X Physics 1 ecture 1, Slide 1
Quality factor Q Z In general U Q U max U max = max energy stored U = energy dissipated in one cycle at resonance Physics 1 ecture 1, Slide 13
arger Q, sharper resonance Physics 1 ecture 1, Slide 14
Series circuit 1. The current leads generator voltage by 45 o. V ~. Assume V = V max sinwt What does the phasor diagram look like at t = 0? (A) X (B) () X (D) X V = V max sinwt V is horizontal at t = 0 (V = 0) V V V V V < V if current leads generator voltage Physics 1 ecture 1, Slide 15
Series circuit The current leads generator voltage by 45 o V ~ How should we change w to bring circuit to resonance? (A) decrease w (B) increase w () Not enough info Original w At resonance (w 0 ) At resonance X = X f X increases X decreases Increase w Physics 1 ecture 1, Slide 16
Series circuit 1. The current leads generator voltage by 45 o. V ~. Suppose X = X at frequency w. By what factor should we increase w to bring circuit to resonance w 0? w 0 w 0 w w w 0 w w 0 w (A) (B) () (D) 4 X X 1 1 1 w w w0 w Physics 1 ecture 1, Slide 17
onsider the harmonically driven series circuit shown. At some frequency w, X = X = V max = 100 V = 50 k Series circuit V ~ If we change w, what is the maximum current that can flow? I max ma I max ma (A) (B) () Imax ma At resonance X = X Z I 100 ( w ) V ma 50 x10 max max 0 3 Physics 1 ecture 1, Slide 18
Instantaneous power P(t) = I(t)V(t) P(t) varies in time. We want the average power. Average, over one cycle, of any periodic function : P t T 1 T 0 P t dt ircuit element: P t T 1 T 0 V t I t dt For harmonically-varying functions like sin wt or cos wt, T = π/ω. Physics 1 ecture 1, Slide 19
For or, voltage and current are / out of phase: I sin wt, V cost wt T 1 P, t sin wt cos wtdt 0 T 0 Average power for is not zero since V and I are in phase. T 1 1 peak sin w peak T 0 P t I I t dt I 1 e peak 1 I I Z e peak peak peak cosf Z Define root mean square (MS) f 1 I I peak IMS 1 I I.707 I MS peak peak X -X
Average power dissipated in for circuit 1 Power I peakepeak cosf Z f X -X Maximum power delivered when f = 0 at resonance! I max w 0 I max = e max At resonance, V V w I w V V I 0 max 0 max Q I max / w 0 w 0 = 1/w 0 f = 0
Power Transmission 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? urrent Delivered: I = P/V = 15 Amps oss from power line = I = 15*15 * 5 = 1,15 Watts! If you deliver 1500 Watts at 10,000 Volts over the same transmission lines. How much power is lost? urrent Delivered: I = P/V = 0.15 Amps oss from power line = I = 0.15*0.15*5 = 0.113 Watts DEMO Physics 1 ecture 1, Slide
How do we transform 10,000 V into 10 V for your toaster? Transformer Time-varying voltage in primary causes time-varying B in iron-core. Flux per turn = B (Area) = B Iron core maintains same B throughout. Primary coil captures flux P = N P B Secondary coil captures flux S = N S B 700 V / 40 V Faraday s aw: V P = d P /dt = N P B V S = d S /dt = N S B V V p s N N p s Physics 1 ecture 1, Slide 3