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AC Circuits - PowerPoint PPT Presentation

w. AC Circuits. Lecture Outline. Driven Series LCR Circuit: General solution Resonance condition Resonant frequency “Sharpness of resonance” = Q Power considerations Power factor depends on impedance Transformers Voltage changes Faraday’s Law in action gives induced primary current.

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Presentation Transcript

AC Circuits

• Driven Series LCR Circuit:

• General solution

• Resonance condition

• Resonant frequency

• “Sharpness of resonance” = Q

• Power considerations

• Power factor depends on impedance

• Transformers

• Voltage changes

• Faraday’s Law in action gives induced primary current.

• Power considerations

Text Reference: Chapter 33.4-6

• R: V in phase with i

• C: V lags i by 90°

• L: V leads i by 90°

y

y

w

x

Phasors

Þ

Þ

• A phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I) and which rotates counterclockwise in a 2-d plane with angular velocity w. Recall uniform circular motion:

Þ

The projections of r (on the vertical y axis) execute sinusoidal oscillation.

i

w

i

0

wt

t

w

i

i

0

wt

w

i

i

wt

0

Phasors for L,C,R

ß

C

L

e

~

• Assume a solution of the form:

Series LCRAC Circuit

• Back to the original problem: the loop equation gives:

• Here all unknowns, (im,f) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: e = em sinwt.

• To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram.

R

C

L

e

~

w

Phasors: LCR

• Assume:

Þ

• From these equations, we can draw the phasor diagram to the right.

Þ

• This picture corresponds to a snapshot at t=0. The projections of these phasors along the vertical axis are the actual values of the voltages at the given time.

R

C

L

e

~

The unknowns (im,f) can now be solved for graphically since the vector sum of the voltages VL + VC + VR must sum to the driving emf e.

Phasors: LCR

• The phasor diagram has been relabeled in terms of the reactances defined from:

Ohms ->

(c)

(b) w large

(a) w small

Lecture 20, ACT 3

• A driven RLC circuit is connected as shown.

• For what frequencies w of the voltage source is the current through the resistor largest?

w

• A driven RLC circuit is connected as shown.

• For what frequencies w of the voltage source is the current through the resistor largest?

(c)

(b) w large

(a) w small

• This is NOT a series RLC circuit. We cannot blindly apply our techniques for solving the circuit. We must think a little bit.

• However, we can use the frequency dependence of the impedances (reactances) to answer this question.

• The reactance of an inductor = XL = wL.

• The reactance of a capacitor = XC = 1/(wC).

• Therefore,

• in the low frequency limit, XL® 0 and XC® ¥.

• Therefore, as w ® 0, the current will flow mostly through the inductor; the current through the capacitor approaches 0.

• in the high frequency limit, XL® ¥ and XC® 0.

• Therefore, as w ® ¥, the current will flow mostly through the capacitor, approaching a maximum imax = e/R.

Phasors:LCR

Þ

ß

x

imXL

em

f

imR

imXC

“Full Phasor Diagram”

“ Impedance Triangle”

Phasors:Tips

• This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis.

• Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when i=0).

From this diagram, we can also create a triangle which allows us to calculate the impedance Z:

imXL

the loop eqn

w

em

f

imR

XL

imXC

Z

XL - XC

f

R

XC

We have found the general solution for the driven LCR circuit:

Phasors:LCR

XL

Z

XL

XL

Z

f

f

R

R

R

Z

XC

XC

XC

The phase f between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances.

XL > XC

f > 0

current

LAGS

applied voltage

XL < XC

f < 0

current

applied voltage

XL = XC

f = 0

current

IN PHASE

applied voltage

(c) impossible

(b) decrease w

(a) increase w

• Which of the following phasors represents the current i at t=0?

1B

(c)

(b)

(a)

i

i

i

Conceptual Question

• The series LCR circuit shown is driven by a generator with voltage e = emsinwt. The time dependence of the current i which flows in the circuit is shown in the plot.

• How should w be changed to bring the current and driving voltage into phase?

(c) impossible

(b) decrease w

(a) increase w

XL

Therefore,

Therefore, the phasor diagram must look like this:

i

e

XC

Lecture 21, ACT 1

e

• The series LCR circuit shown is driven by a generator with voltage e = emsinwt. The time dependence of the current i which flows in the circuit is shown in the plot.

• How should w be changed to bring the current and driving voltage into phase?

• From the plot, it is clear that the current is LEADING the applied voltage.

• To bring the current into phase with the applied voltage, we need to increase XL and decrease XC.

• Increasing w will do both!!

(c) impossible

(b) decrease w

(a) increase w

1B

(c)

(b)

(a)

Lecture 21, ACT 1

• The series LCR circuit shown is driven by a generator with voltage e = emsinwt. The time dependence of the current i which flows in the circuit is shown in the plot.

• How should w be changed to bring the current and driving voltage into phase?

• Which of the following phasors represents the current i at t=0?

w

i

i

w

i

• The sign of i is correct at t=0.

• However, it soon will become negative!

• nope

• This one looks just right!!

• f = -30 °

• The projection of i along the vertical axis is negative here.

• no way jose

reaches a maximum when:

Resonance

• For fixed R,C,L the current im will be a maximum at the resonant frequencyw0 which makes the impedance Zpurely resistive.

the frequency at which this condition is obtained is given from:

Þ

• Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself!

• At this frequency, the current and the driving voltage are in phase!

R=Ro

im

R=2Ro

0

2wo

0

w

Resonance

The current in an LCR circuit depends on the values of the elements and on the driving frequency through the relation

Suppose you plot the current versus w, the source voltage frequency, you would get:

Power in LCR Circuit

• The power supplied by the emf in a series LCR circuit depends on the frequency w. It will turn out that the maximum power is supplied at the resonant frequency w0.

• The instantaneous power (for some frequency, w) delivered at time t is given by:

• The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle.

• To evaluate the average on the right, we first expand the sin(wt-f) term.

sinwtcoswt

(Product of even and odd function = 0)

0

-1

2p

0

wt

+1

sin2wt

1/2

0

0

-1

2p

0

wt

Power in LCR Circuit

• Expanding,

• Taking the averages,

• Generally:

• Putting it all back together again,

Power in LCR Circuit

• This result is often rewritten in terms of rms values:

• Power delivered depends on the phase, f, the“power factor”

• phase depends on the values of L, C, R, and w

• therefore...

• Power, as well as current, peaks at w = w0. The sharpness of the resonance depends on the values of the components.

• Recall:

• Therefore,

We can write this in the following manner (you can do the algebra):

…introducing the curious factors Q and x...

The Q factor

A parameter “Q” is often defined to describe the sharpness of resonance peaks in both mechanical and electrical oscillating systems. “Q” is defined as

where Umax is max energy stored in the system and DU is the energy dissipated in one cycle

For RLC circuit, Umax is (e.g.)

And losses only in R, namely

This gives

And for completeness, note

FWHM

R=Ro

<P>

R=2Ro

0

2wo

0

w

For Q > few,

Power in RLC

I

2A

(c) QII > QI

(b) QII = QI

(a) QII < QI

• What is the relation between PI and PII , the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency?

2B

(c) PII > PI

(b) PII = PI

(a) PII < PI

Conceptual Question 2

• Consider the two circuits shown where CII = 2CI.

• What is the relation between the quality factors, QI and QII , of the two circuits?

I

2A

(c) QII > QI

(b) QII = QI

(a) QII < QI

• We know the definition of Q:

\

Doubling C decreases Q by sqrt(2)!

Lecture 21, ACT 2

• Consider the two circuits shown where CII = 2 CI.

• What is the relation between the quality factors, QI and QII , of the two circuits?

• At first glance, it looks like Q is independent of C.

• At second glance, we see this cannot be true, since the resonant frequency wo depends on C!

Doubling C decreases wo by sqrt(2)!

Doubling C increases the width of the resonance!

(c) QII > QI

(b) QII = QI

(a) QII < QI

2B

(c) PII > PI

(b) PII = PI

(a) PII < PI

Lecture 21, ACT 2

II

I

• Consider the two circuits shown where CII = 2CI.

• What is the relation between the quality factors, QI and QII , of the two circuits?

• What is the relation between PI and PII , the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency?

• At the resonant frequency, the impedance of the circuit is purely resistive.

• Since the resistances in each circuit are the same, the impedances at the resonant frequency for each circuit are equal.

• Therefore, The power delivered by the generator to each circuit is identical.

Make Vin big

Power Transmission

• How can we transport power from power stations to homes? Why do we use “high tension” lines?

• At home, the AC voltage obtained from outlets in this country is 120V at 60Hz.

• Transmission of power is typically at very high voltages ( eg ~500 kV)

• Transformers are used to raise the voltage for transmission and lower the voltage for use. We’ll describe these next.

• But why?

• Calculate ohmic losses in the transmission lines:

• Define efficiency of transmission:

• Note for fixed input power and line resistance, the inefficiency µ 1/V2

With Vin=735kV, e = 80%.

The efficiency goes to zero quickly if Vin were lowered!

Example: Quebec to Montreal

1000 km Þ R= 220W

suppose Pin = 500 MW

Transformers

iron

• The AC current in the primary circuit creates a time-varying magnetic field in the iron

~

V1

e

V2

• This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils.

(secondary)

(primary)

• The iron is used to maximize the mutual inductance. We assume that the entire flux produced by each turn of the primary is trapped in the iron.

No resistance losses

All flux contained in iron

• The primary circuit is just an AC voltage source in series with an inductor. The change in flux produced in each turn is given by:

• The change in flux per turn in the secondary coil is the same as the change in flux per turn in the primary coil (ideal case). The induced voltage appearing across the secondary coil is given by:

• Therefore,

• N2 > N1Þ secondary V2 is larger than primary V1 (step-up)

• N1 > N2Þ secondary V2 is smaller than primary V1 (step-down)

• Note: “no load” means no current in secondary. The primary current, termed “the magnetizing current” is small!

R

Ideal Transformers

• What happens when we connect a resistive load to the secondary coil?

• Flux produced by primary coil induces an emf in secondary

• emf in secondary produces current i2

• This current produces a flux in the secondary coil µ N2i2, which opposes the original flux -- Lenz’s law

• This changing flux appears in the primary circuit as well; the sense of it is to reduce the emf in the primary...

• However, V1 is a voltage source.

• Therefore, there must be an increased current i1 (supplied by the voltage source) in the primary which produces a flux µN1i1 which exactly cancels the flux produced by i2.

• With a resistive load in the secondary, the primary current is given by:

This is the equivalent resistance seen by the source.

3A

3B

(c) 480 V

(c) 32 A

(b) 120 V

(b) 16 A

(a) 30 V

(a) 8 A

Lecture 21, ACT 3

• The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary.

• If V1 = 120 V, what is the potential drop across the resistor R ?

If 960 W are dissipated in the resistor R, what is the current in the primary ?

3A

(c) 480 V

(b) 120 V

(a) 30 V

Lecture 21, ACT 3

• The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary.

• If V1 = 120 V, what is the potential drop across the resistor R?

The ratio of turns, (N2/N1) = (200/50) = 4

The ratio of secondary voltage to primary voltage is equal to the ratio of turns, (V2/V1) = (N2/N1)

Therefore, (V2/V1) = 480 V

3A

3B

(c) 480 V

(c) 32 A

(b) 120 V

(b) 16 A

(a) 30 V

(a) 8 A

Lecture 21, ACT 3

• The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary.

• If V1 = 120 V, what is the potential drop across the resistor R ?

The ratio of turns, (N2/N1) = (200/50) = 4 The ratio (V2/V1) = (N2/N1). Therefore, (V2/V1) = 480 V

• If 960 W are dissipated in the resistor R, what is the current in the primary ?

Gee, we didn’t talk about power yet….

But, let’s assume energy is conserved…since it usually is around here

Therefore, 960 W should be produced in the primary

P1 = V1 I1 implies that 960W/120V = 8 A