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winning Objectives - Ch 35 To utilize a phasor examination to investigate AC circuits.
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Ch – 35 AC Circuits

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Reading Quiz – Ch. 35 1. The examination of AC circuits utilizes a pivoting vector called a : a. unit circle vector b. phasor c. emf vector d. sinusoidal vector 2. In a capacitor, the pinnacle current and pinnacle voltage are connected by the a. capacitive resistance. b. capacitive reactance. c. capacitive impedance. d. capacitive inductance.

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Learning Objectives – Ch 35 To utilize a phasor investigation to dissect AC circuits. • To comprehend RC channel circuits. • To comprehend the arrangement RLC circuit and reverberation.

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Alternating current - Circuits controlled by a sinusoidal emf are Power plants create swaying emf and streams. Steam or falling water turning a turbine, which thusly, causes a loop of wire in an attractive field.

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ε = ε 0 cos ω t ε 0 is pinnacle emf ω is precise recurrence in rads/s or ω = 2 π f, where f is the recurrence in Hz (cycles every second) or s - 1 .

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Phasor Diagram A phasor is a vector that pivots counterclockwise around the root at rakish recurrence ω . The length of the phasor(radius) speaks to the pinnacle estimation of the amount. ω t is a stage point. On the off chance that there are more than one phasor in the graph, there can be various stage edges.

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Phasor Diagram The immediate esteem is the genuine esteem you would gauge at time t . This esteem is never more noteworthy than ε 0 . The quick esteem can be spoken to as the projection of the phasor vector onto the level hub of the circle.

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Here is a graphical representation along the t hub:

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Stop to Think The size of the momentary estimation of the emf spoke to by this phasor is: Increasing Decreasing Constant Need to know t

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Stop to Think The extent of the immediate estimation of the emf spoke to by this phasor is: Increasing, since it is voyaging ccw

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Resistor Circuits Use of lowercase for prompt qualities, e.g. v R , i R Uppercase for pinnacle values v R = V R cos ω t i R = I R cos ω t In an AC circuit, resistor voltage and current sway in stage .

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Resistor Circuits Representation of resistor current and voltage on a phasor graph. Vectors turn at a similar recurrence and have a similar stage edge with the root Cannot look at sizes, since units are distinctive.

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Capacitor Circuits For the capacitor circuit appeared at right: v C = ε = V C cos ω t i = dq/dt and q = Cv C : i C = - ω CV C sin ω t, which we compose as: i C = ω CV C cos ( ω t + π/2)

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Capacitor Circuits - ICE v C = V C cos ω t i C = ω CV C cos ( ω t + π/2) The result of this is the capacitor voltage and current don\'t waver in stage. The present leads voltage by π/2 rads, or by T/4. ICE

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Capacitor Circuits - ICE Current achieves crest esteem I C the moment the capacitor is completely released and v C =0. The current is zero the moment the capacitor is completely charged.

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Capacitive Reactance – Relationship between pinnacle current and voltage in a capacitor I C = ω CV C , in spite of the fact that they don\'t occur in the meantime. Characterize the capacitive reactance X C = 1/ω C, then: I C = V C/X C This is similar to Ohm\'s Law for DC.

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Properties of Capacitive Reactance Only utilized for pinnacle values i C DOES NOT EQUAL v C/X C Dependent on emf recurrence, not at all like resistance, which is a property of the resistor free of circuit recurrence. diminishes as recurrence increments. at high frequencies, X C approaches 0 and the capacitor demonstrations like a wire.

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RC Filter Circuits resistor and capacitor are in arrangement: I C = I R at low frequencies, X C will be substantial, constraining I. Since V R = IR, voltage crosswise over resistor will little. At high frequencies, X C will be little, so the capacitor will act more like a perfect wire, with almost no voltage drop.

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Analyzing a RC circuit Draw the current phasor. All circuit components in arrangement have a similar current whenever. Edge is self-assertive. Current is in stage with V R , so attract that phasor stage with I. Current leads V c by 90 0 , so draw the capacitor voltage phasor 90 0 behind (i.e. clockwise.

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Analyzing a RC circuit Kirchoff\'s circle law says v R + v C = ε , for the prompt qualities. The expansion of pinnacle qualities is a vector expansion. Accordingly ε 0 is drawn as the resultant vector as appeared. ε = ε 0 cos ω t; the point amongst emf and x-hub is ω t.

Slide 21

Analyzing a RC circuit ε 0 2 = V R 2 + V C 2 This relationship is for pinnacle values. Top streams are identified with pinnacle voltages by: V R = IR V C = IX C ε 0 2 = (R 2 + 1/ω 2 C 2 )I 2

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Crossover Frequency For low frequencies, where X C >>R, the circuit is basically capacitive. A heap in parallel with the capacitor will get the greater part of the potential contrast and along these lines the power.

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Crossover Frequency For high frequencies, where X C <<R, the circuit is basically resistive. A heap in parallel with the capacitor will be obstructed from getting any power.

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Crossover Frequency Crossover recurrence is found where V R and V C are equivalent: ω c = 1/RC f c = ω c/2 π

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Filters Low pass channel (best): associate the yield in parallel to capacitor at frequencies well underneath ω c the flag is transmitted with little misfortune. at frequencies well above ω c , the flag is lessened

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Filters High pass channel: interface the yield in parallel to resistor, as in the lower graph at frequencies well underneath ω c the flag is constricted at frequencies well above ω c , the flag is transmitted with little misfortune.

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Inductor Circuits An inductor is a gadget that delivers a uniform attractive field when a present goes through it. A solenoid is an inductor. |∆V L | = L dI/dt

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Inductor circuits (ELI) v L = ε = V L cos ω t Where V L = ε 0 , and v L = L di L/dt For i L , prompt current: di L = (v L/L) dt di = V L cos ω t dt L Integrating, we get: I L = V L/ω L sin ω t, which we compose as: i L = V L/ω L cos ( ω t - π/2) i L = I L cos ( ω t - π/2)

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Inductor circuit - ELI The math is like that of a capacitor: i L = I L cos ( ω t - π/2) and i C = I C cos ( ω t + π/2) The distinction is that the current in the circuit slacks the inductor voltage by 90 0 or T/4: ELI

Slide 30

Inductive reactance We can characterize the inductive reactance to be X L = ω L, then: I L = V L/X L (legitimate for pinnacle estimations of I, V just) Compare to: X C = 1/ω C both reactances are recurrence subordinate. inductive reactance increments with recurrence. capacitive reactance diminishes with recurrence.

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The Series RLC circuit This circuit goes about as both a low pass and a high pass channel in the meantime. It just permits flag to go from a restricted scope of frequencies.

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The Series RLC circuit Two imperative perceptions: The immediate current through every one of the 3 components is the same at a given time: i = i R + i C + i L The entirety of the prompt voltages mean the emf at a given time: ε = v R + v C + v L

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Analyzing a RLC circuit Draw the current phasor. All circuit components in arrangement have a similar current whenever. Edge is subjective. Current is in stage with V R , so attract that phasor stage with I. Current leads V c by 90 0 (ICE), so draw the capacitor voltage phasor 90 0 behind (i.e. clockwise). Current slacks V L by a similar sum (ELI) so draw it ahead.

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RLC circuit investigation Kirchoff\'s circle law says v R + v C + v L = ε , for the quick values. The expansion of the pinnacle qualities is a vector. expansion. In this way ε 0 is drawn as the resultant vector as appeared. V C and V L are in inverse bearings thus can be spoken to as the vector (V L – V C or the other way around).

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RLC circuits The length of the emf phasor is the hypotenuse of a right triangle: ε 0 2 = V R 2 + (V L - V C ) 2 This relationship is for pinnacle values.

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Phase point amongst I and ε 0 If V L > V C then the emf drives current: i = I cos( ω t – φ ) where ω t is the edge between the emf and the level pivot. On the off chance that V C > V L then phasor chart would be beneath level pivot and emf slacks current: i = I cos[ ω t – (- φ )] or i = I cos( ω t + φ )

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RLC Circuits ε 0 2 = V R 2 + (V L - V C ) 2 Peak streams are identified with pinnacle voltages by: V R = IR V C = IX C ε 0 2 = [R 2 + (X L - X C ) 2 ]I 2 Taking the square foundation of both sides…

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Impedance The impedance of a RLC circuit is characterized as: Z = and has units of ohms. Ohm\'s Law for air conditioning circuits can be composed as: I = ε 0/Z This is for pinnacle values as it were.

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Impedance The impedance of a RLC circuit is characterized as: Z = and has units of ohms. Ohm\'s Law for air conditioning circuits can be composed as: I = ε 0/Z This is for pinnacle values as it were.

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Phase point, returned to From the chart on the right we see that: tan φ = (V L - V C )/V R tan φ = (X L - X C )I/IR φ = tan - 1 (X L - X C )/R φ

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Phase edge, returned to The phasor outline at the right demonstrates a situation where the present slacks emf by: φ = tan-1 (X L - X C )/R We can express the pinnacle resistor voltage as: V R = ε 0 cos φ Resistor voltage sways in stage with emf just if φ =0 rads, i.e there are no capacitors or inductors in the circuit.

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Resonance Frequency Z = I = ε 0/Z In a RLC circuit, current will be constrained at low recurrence by XC = 1/ω C being substantial and at high recurrence by XL = ω L being extensive. Current will be most extreme when impedance, Z is minimized. Any thoughts what

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