Lecture 23. Impedance, Resonance in R-C-L Circuits. Preparation for the Final Exam

Μέγεθος: px

Εμφάνιση ξεκινά από τη σελίδα:

Download "Lecture 23. Impedance, Resonance in R-C-L Circuits. Preparation for the Final Exam"

Ἀβραάμ Παπαδάκης
8 χρόνια πριν
Προβολές:

1 Lecture 3. Impedance, Resonance in R-C-L Circuits (a) Start earlier! Preparation for the Final Exam (b) Review the concepts (lectures + textbook) and prepare your equation sheet. Think how you can use every equation on your sheet, what types of problems can be solved with these equations. (c) Work on practice exams. (d) Review all HW and Iclicker questions. (e) Go over the end-of-chapter problems (you don t need to solve them, just check that you know how to approach them). At the Exam (a) Make sure you understand the problem, read the problem formulation carefully. Make a drawing!!! If you remain uncertain raise your hand and ask the proctors. (b) Get the units right. It is easy to eliminate the answers with wrong units. This applies to formulas too.

2 Resistor X R = R Reactance (recap) V = IR = IX R Capacitor X C = V t = I t i X C Inductor X L = ωl V t = I t ix L AC (cos + φ ) driven circuits!

3 V t I t RLC Impedance Impedance is a measure of how much the circuit impedes the flow of current. The impedance is a complex number (time-independent phasor), it relates timedependent phasors V(t) and I(t). = e ii = 0 V t I t V t = I t II I φ φ V RR V rrr e iωt = I rrr e iii ii e ii V rrr = I rrr e ii e ii V rrr = I rrr all terms are real Reactances: X R = R X C = Impedances: R = R C = i V is the reference phasor X L = ωl L = iωl 3

4 V t I t C circuit Can we plug a -µf capacitor into a wall outlet (ω = π 60 rrr s, V rrr = 0V) if the circuit breakers can take 5A? V rrr = I rrr C = I rrr C = X C = = Ω = 650Ω π I rrr = V rrr X c = 0V 650Ω = 45mm This current is sufficiently small. The primary concern is the voltage rating of the capacitor, which should be around 00V. Current (reference phasor) Voltage V t = I t C = I t i ωc 4

5 V t I t L circuit What happens when we plug a -H inductor into a wall outlet? V rrr = I rrr L = I rrr L = X L = = π 60 Ω = 377Ω I rrr = V rrr X L = 0V 377Ω = 0.3A Again, the current won t blow a circuit breakers. The inductor must be designed to carry 0.3A without overheating or saturating the iron core. Current (reference phasor) Voltage V t = I t L = I t iil 5

6 Series R-C circuit = R + C = R ix c = e ii = R + X c φ = aaaaaa III RRR = aaaaaa X C R II φ R RR V rrr = I rrr R + X c = I rrr R + ωc V t = I t = I(t) R ix c φ φ V 0 e iii = I 0 ei φ e ii 6

7 R-C circuits: Example II φ R RR e ii = R i tan φ = φ = aaaaaa III RRR = aaaaaa X C R R = 0 3 = II I Note that φ is negative (as it should be for the RC circuits). φ φ V ii RR V(t) = I(t) V 0 e iii = I 0 e i φ e ii 7

8 = R ix c = R + X c Low-Pass Filter Goal: to suppress high-frequency (f > f 0 ) components in the spectrum of a signal. V ooo V ii = V ii = I R ix c X c R + X c = ωc R + V ooo = I ix c V ii = I R + X c V ooo = IX c = ωrc + = ωτ RR + I V R V ii V C = V ooo Output power: V ooo V ii = ωτ RR + ωτ RR ω τ RR ω τ RR = two times Cutoff frequency: ω 0 = πf 0 = RR We want to suppress the high-frequency (f > 0kkk) components in the output of an audio amplifier with the output resistance 00 Ω. What capacitance do you need? C = πf 0 R = π0 4 F = 60nn 00 8

9 Series R-L-C Circuits For R, C, and L in series: = R + ix L ix c = R + i I = = R + V rrr = I rrr = I rrr R + iii L V(t) = I(t) R + i iii C? 9

10 Series R-L-C circuits: Example V L = iωll V rrr L =. 80 = 76V V R V rrr R =. 40 = 88V V s V rrr C =. 0 = 4V V C = i I V rrr = I rrr = I rrr R + V rrr = = 0V 0

11 Series R-L-C circuits: Example An R-L-C series circuit with an inductance of 0.9H, a resistance of 44 Ω, and a capacitance of 7.7 µf carries an rms current of 0.446A with a frequency of 39Hz.. What is the impedance of the circuit? ω = 455 rrr/s = R + ix L ix c = R + i. What is the phase angle? 0 = R + = 339Ω II φ φ V RR tan φ = ωl ωc R = arctan rrr I 3. What is the rms voltage of the source? V rrr = I rrr 0 = 0.446A 339Ω = 5V 4. What average power is delivered by the source? cos 0.77 = power factor for this circuit P aa = V rrr I rrr cos φ = = 48.6W - average rate at which electrical energy is converted to thermal energy in the resistor

12 Parallel R-L-C Circuit: Example P aa = V rrr I rrr cos φ = R + i = R + = R + i = R i R + II φ φ I V RR tan φ = /R I rrr = V rrr = 3 4 = V rrr R + cos φ = + ttt φ = 4 5 = = 50 A P aa = V rrr I rrr cos φ = = 00W

13 I Series Resonance in the R-L-C circuits For R, C, and L in series: = = R + I = V = V = R + ix L ix c = R + i R + Resonance condition: = ω 0 = LL - resonance frequency At ω =ω 0 minimum (real) impedance, max current. Note that at ω =ω 0, V C and V L can be greater than V. ω > ω 0 ω < ω 0 ω = ω 0 3

14 I Parallel Resonance in the R-L-C circuits = R + iii + ωc i = R i + i = X L X C R + i = R + ωc ωl At the resonance frequency ω 0 = LL is at its minimum ω 0 L is a short ω C is a short I = V = V R + min at = Note that at ω =ω 0, I C and I L can be greater than I. R = Ω, C = F, L = H, and V = V 6

15 Transformer Φ B - the flux per turn ℇ p = N p dφ B dd ℇ s = N s dφ B dd ℇ p ℇ s = N p N s For an ideal transformer (R s = R p = 0): Energy conservation: V p V s = N p N s V p I p = V s I s I s I p = N p N s V p V s I p = N p N s V s I s V p I p = N p N s R - as if the source had been connected directly to a resistance N p N s R impedance transformation Using mutual inductance M = L p L s : M Φ s I p = N sφ B I p M di p dd = N dφ B s dd ℇ s = M di p dd 7

16 Sloppy formulation Example ℇ s = M di p dd = I sr s M = I sr s di p dd = 0.4 cos 377t cos 377t =.55mm ℇ s = M di p dd = M 6t = = 0.8V 8

17 Next time: Lecture 4. Electromagnetic Waves,

Σχετικά έγγραφα

EE101: Resonance in RLC circuits

EE101: Resonance in RLC circuits EE11: Resonance in RLC circuits M. B. Patil mbatil@ee.iitb.ac.in www.ee.iitb.ac.in/~sequel Deartment of Electrical Engineering Indian Institute of Technology Bombay I V R V L V C I = I m = R + jωl + 1/jωC