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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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## Ch – 35 AC Circuits

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**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. b. capacitive reactance. c. capacitive impedance. d. capacitive inductance.**Learning Objectives – Ch 35**• To use a phasor analysis to analyze AC circuits. • To understand RC filter circuits. • To understand the series RLC circuit and resonance.**Alternating current - Circuits powered by a sinusoidal emf**are • 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 = 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.**Phasor Diagram**• 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.**Phasor Diagram**• 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.**Stop to Think**The magnitude of the instantaneous value of the emf represented by this phasor is: • Increasing • Decreasing • Constant • Need to know t**Stop to Think**The magnitude of the instantaneous value of the emf represented by this phasor is: • Increasing, since it is traveling ccw**Resistor Circuits**• 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.**Resistor Circuits**• 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.**Capacitor Circuits**• 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)**Capacitor Circuits - ICE**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**Capacitor Circuits - 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**• 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**• 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**• 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**• 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**• 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**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**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**Crossover frequency is found where VR and VC are equal: ωc = 1/RC fc = ωc /2π**Filters**• 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**• 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**• 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)**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**• 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**• 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**• 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**• 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**• 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**• 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**• 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 andε0**• 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**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**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**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**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**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**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, ω0**• 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 ω 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**ω0 • 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**What is the phase angle when the emf frequency is • 14 kHz • 18 kHz • What is ω0 for this circuit

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