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# ch - 35 ac circuits - PowerPoint PPT Presentation

Ch – 35 AC Circuits . Reading Quiz – Ch. 35. 1. The analysis of AC circuits uses a rotating vector called a : a. unit circle vector b. phasor c. emf vector d. sinusoidal vector 2. In a capacitor, the peak current and peak voltage are related by the a. capacitive resistance.

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1. The analysis of AC circuits uses a rotating vector called a :

a. unit circle vector

b. phasor

c. emf vector

d. sinusoidal vector

2. In a capacitor, the peak current and peak voltage are related by the

a. capacitive resistance.

b. capacitive reactance.

c. capacitive impedance.

d. capacitive inductance.

• To use a phasor analysis to analyze AC circuits.

• To understand RC filter circuits.

• To understand the series RLC circuit and resonance.

• Power plants produce oscillating emf and currents.

• Steam or falling water turning a turbine, which in turn, causes a coil of wire in a magnetic field.

ε are = ε0 cos ωt

ε0 is peak emf

ω is angular frequency in rads/s or

ω = 2πf, where f is the frequency in Hz (cycles per second) or

s-1.

• A phasor is a vector that rotates counterclockwise around the origin at angular frequency ω.

• The length of the phasor(radius) represents the peak value of the quantity.

• ωt is a phase angle. If there are more than one phasor in the diagram, there can be multiple phase angles.

• The instantaneous value is the actual value you would measure at time t. This value is never greater than ε0.

• The instantaneous value can be represented as the projection of the phasor vector onto the horizontal axis of the circle.

The magnitude of the instantaneous value of the emf represented by this phasor is:

• Increasing

• Decreasing

• Constant

• Need to know t

The magnitude of the instantaneous value of the emf represented by this phasor is:

• Increasing, since it is traveling ccw

• Use of lowercase for instantaneous values, e.g. vR, iR

• Uppercase for peak values

vR = VR cos ωt

iR = IR cos ωt

• In an AC circuit, resistor voltage and current oscillate in phase.

• Representation of resistor current and voltage on a phasor diagram.

• Vectors rotate at the same frequency and have the same phase angle with the origin

• Cannot compare magnitudes, since units are different.

• For the capacitor circuit shown at right:

vC = ε = VC cos ωt

i = dq/dt and q = CvC:

iC = - ωCVC sin ωt, which we write as:

iC = ωCVC cos (ωt + π/2)

vC = VC cos ωt

iC = ωCVC cos (ωt + π/2)

The consequence of this is that the capacitor voltage and current do not oscillate in phase.

The current leads voltage by π/2 rads, or by T/4.

ICE

• Current reaches peak value IC the instant the capacitor is fully discharged and vC =0. The current is zero the instant the capacitor is fully charged.

Capacitive Reactance – Relationship between peak current and voltage in a capacitor

• IC = ωCVC, although they don’t happen at the same time.

• Define the capacitive reactanceXC = 1/ ωC, then:

• IC = VC/XC

• This is analogous to Ohm’s Law for DC.

Properties of Capacitive Reactance and voltage in a capacitor

• Only used for peak values

iC DOES NOT EQUAL vC/XC

• Dependent on emf frequency, unlike resistance, which is a property of the resistor independent of circuit frequency.

• decreases as frequency increases.

• at very high frequencies, XC approaches 0 and the capacitor acts like a wire.

RC Filter Circuits and voltage in a capacitor

• resistor and capacitor are in series: IC = IR

• at low frequencies, XC will be large, limiting I.

• Since VR = IR, voltage across resistor will small.

• At high frequencies, XC will be small, so the capacitor will act more like an ideal wire, with very little voltage drop.

Analyzing an RC circuit and voltage in a capacitor

• Draw the current phasor. All circuit elements in series have the same current at any time. Angle is arbitrary.

• Current is in phase with VR, so draw that phasor in phase with I. Current leads Vc by 900, so draw the capacitor voltage phasor 900 behind (i.e. clockwise.

Analyzing an RC circuit and voltage in a capacitor

• Kirchoff’s loop law says vR + vC = ε, for the instantaneous values. The addition of peak values is a vector addition. Therefore ε0 is drawn as the resultant vector as shown.

• ε = ε0 cos ωt; the angle between emf and x-axis is ωt.

Analyzing an RC circuit and voltage in a capacitor

• ε02 = VR2 + VC2

• This relationship is for peak values.

• Peak currents are related to peak voltages by:

• VR = IR

• VC = IXC

• ε02 = (R2 + 1/ω2C2)I2

Crossover Frequency and voltage in a capacitor

For low frequencies, where XC>>R, the circuit is primarily capacitive. A load in parallel with the capacitor will get most of the potential difference and therefore the power.

Crossover Frequency and voltage in a capacitor

For high frequencies, where XC<<R, the circuit is primarily resistive. A load in parallel with the capacitor will be blocked from getting any power.

Crossover Frequency and voltage in a capacitor

Crossover frequency is found where VR and VC are equal:

ωc = 1/RC

fc = ωc /2π

Filters and voltage in a capacitor

• Low pass filter (top):

• connect the output in parallel to capacitor

• at frequencies well below ωc the signal is transmitted with little loss.

• at frequencies well above ωc , the signal is attenuated

Filters and voltage in a capacitor

• High pass filter:

• connect the output in parallel to resistor, as in the lower diagram

• at frequencies well below ωc the signal is attenuated

• at frequencies well above ωc , the signal is transmitted with little loss.

Inductor Circuits and voltage in a capacitor

• An inductor is a device that produces a uniform magnetic field when a current passes through it. A solenoid is an inductor.

|∆VL| = L dI/dt

Inductor circuits (ELI) and voltage in a capacitor

vL = ε = VL cos ωt

Where VL = ε0, and

vL = L diL/dt

For iL,instantaneous current:

diL = (vL/L) dt

di = VL cos ωt dt

L

Integrating, we get:

IL = VL /ωL sin ωt, which we write as:

iL = VL /ωL cos (ωt - π/2)

iL = IL cos (ωt - π/2)

Inductor circuit - ELI and voltage in a capacitor

• The math is similar to that of a capacitor:

iL = IL cos (ωt - π/2) and

iC = IC cos (ωt + π/2)

The difference is that the current in the circuit lags the inductor voltage by 900 or T/4:

ELI

Inductive reactance and voltage in a capacitor

• We can define the inductive reactance to be XL = ωL, then:

IL = VL/XL

(valid for peak values of I, V only)

• Compare to:

XC = 1/ ωC

• both reactances are frequency dependent.

• inductive reactance increases with frequency.

• capacitive reactance decreases with frequency.

The Series RLC circuit and voltage in a capacitor

• This circuit acts as both a low pass and a high pass filter at the same time. It only allows signal to pass from a narrow range of frequencies.

The Series RLC circuit and voltage in a capacitor

• Two important observations:

• The instantaneous current through all 3 elements is the same at a given time:

i = iR + iC + iL

• The sum of the instantaneous voltages add up to the emf at a given time:

ε = vR + vC + vL

Analyzing an RLC circuit and voltage in a capacitor

• Draw the current phasor. All circuit elements in series have the same current at any time. Angle is arbitrary.

• Current is in phase with VR, so draw that phasor in phase with I. Current leads Vc by 900 (ICE), so draw the capacitor voltage phasor 900 behind (i.e. clockwise). Current lags VLby the same amount (ELI) so draw it ahead.

RLC circuit analysis and voltage in a capacitor

• Kirchoff’s loop law says vR + vC + vL = ε, for the instantaneous values. The addition of the peak values is a vector. addition. Therefore ε0 is drawn as the resultant vector as shown.

• VC and VL are in opposite directions and so can be represented as the vector (VL – VC or vice versa).

RLC circuits and voltage in a capacitor

• The length of the emf phasor is the hypotenuse of a right triangle:

ε02 = VR2 + (VL - VC)2

• This relationship is for peak values.

Phase angle between I and and voltage in a capacitorε0

• If VL > VC then the emf leads current:

i = I cos(ωt –φ)

where ωt is the angle between the emf and the horizontal axis.

• If VC > VL then phasor diagram would be below horizontal axis and emf lags current:

i = I cos[ωt – (-φ)] or

i = I cos(ωt +φ)

RLC Circuits and voltage in a capacitor

ε02 = VR2 + (VL - VC)2

• Peak currents are related to peak voltages by:

• VR = IR

• VC = IXC

ε02 = [R2 + (XL- XC )2]I2

Taking the square root of both sides…

Impedance and voltage in a capacitor

The impedance of an RLC circuit is defined as:

Z =

and has units of ohms.

Ohm’s Law for ac circuits can be written as:

I = ε0/Z

This is for peak values only.

Impedance and voltage in a capacitor

The impedance of an RLC circuit is defined as:

Z =

and has units of ohms.

Ohm’s Law for ac circuits can be written as:

I = ε0/Z

This is for peak values only.

Phase angle, revisited and voltage in a capacitor

From the diagram on the right we see that:

tan φ = (VL - VC)/VR

tan φ = (XL - XC)I/IR

φ = tan-1 (XL - XC)/R

φ

Phase angle, revisited and voltage in a capacitor

The phasor diagram at the right shows a case where the current lags emf by:

φ = tan-1 (XL - XC)/R

We can express the peak resistor voltage as:

VR = ε0 cos φ

Resistor voltage oscillates in phase with emf only if φ=0 rads, i.e there are no capacitors or inductors in the circuit.

Resonance Frequency and voltage in a capacitor

Z =

• I = ε0/Z

• In an RLC circuit, current will be limited at low frequency by XC = 1/ωC being large and at high frequency by XL = ωL being large.

• Current will be maximum when impedance, Z is minimized.

• Any ideas what we can do to minimize that term?

Resonance Frequency, and voltage in a capacitorω0

• ω0 = 1/

• this frequency will produce the maximum current in the RLC circuit:

• Imax = ε0/R, as Z = 0 at this frequency.

• at ω0, energy is transferred back and forth between inductor and capacitor.

Peak current as and voltage in a capacitorω is varied.

• ω0 can be considered the natural frequency of the circuit, the frequency at which it would “like” to oscillate.

• When the emf (acting as the driving force) oscillates at the same frequency, there will be a large response in terms of output.

• Note that decreasing the resistance decreases the damping, and narrows the frequency range of large response.

Graphs of emf and current for frequencies below, at, above and voltage in a capacitorω0

• When ω < ω0, current leads emf, φ<0

• When ω > ω0, current lags emf, φ>0

• When ω = ω0, circuit is purely resistive φ=0.

Numerical Problem and voltage in a capacitor

What is the phase angle when the emf frequency is

• 14 kHz

• 18 kHz

• What is ω0 for this circuit