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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

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.

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.

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.

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 .

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.

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.

Here is a graphical representation along the t hub:

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

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

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 .

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.

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)

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

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.

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.

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.

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.

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.

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.

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

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.

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.

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

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

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.

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

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)

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

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.

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.

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

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.

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).

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.

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 )

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…

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.

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.

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

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.

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