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Alternating Current Circuits

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Presentation on theme: "Alternating Current Circuits"— Presentation transcript:

1 Alternating Current Circuits
Chapter 33 Alternating Current Circuits

2 AC Circuit Δv = ΔVmax sin ωt Δv: instantaneous voltage
An AC circuit consists of a combination of circuit elements and an AC generator or source The output of an AC power source is sinusoidal and varies with time according to the following equation Δv = ΔVmax sin ωt Δv: instantaneous voltage ΔVmax is the maximum voltage (amplitude) of the generator ω is the angular frequency of the AC voltage

3 Resistors in an AC Circuit
Consider a circuit consisting of an AC source and a resistor ΔvR = ΔVmax sin ωt ΔvR is the instantaneous voltage across the resistor The instantaneous current in the resistor is The instantaneous voltage across the resistor is also given as ΔvR = ImaxR sin ωt

4 Resistors in an AC Circuit
The graph shows the current through and the voltage across the resistor The current and the voltage reach their maximum values at the same time The current and the voltage are said to be in phase The direction of the current has no effect on the behavior of the resistor

5 Resistors in an AC Circuit
The rate at which electrical energy is dissipated in the circuit is given by i: instantaneous current The heating effect produced by an AC current with a maximum value of Imax is not the same as that of a DC current of the same value The maximum current occurs for a small amount of time

6 rms Current and Voltage
The rms current is the direct current that would dissipate the same amount of energy in a resistor as is actually dissipated by the AC current Alternating voltages can also be discussed in terms of rms values The average power dissipated in resistor in an AC circuit carrying a current I is

7 Ohm’s Law in an AC Circuit
rms values will be used when discussing AC currents and voltages AC ammeters and voltmeters are designed to read rms values Many of the equations will be in the same form as in DC circuits Ohm’s Law for a resistor, R, in an AC circuit ΔVR,rms = Irms R The same formula applies to the maximum values of v and i

8 Capacitors in an AC Circuit
Consider a circuit containing a capacitor and an AC source Kirchhoff’s loop rule gives: ΔvC: instantaneous voltage across the capacitor

9 Capacitors in an AC Circuit
The voltage across the capacitor lags behind the current by 90° The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance (measured in ohms):

10 Inductors in an AC Circuit
Consider an AC circuit with a source and an inductor Kirchhoff’s loop rule gives: ΔvL: instantaneous voltage across the inductor

11 Inductors in an AC Circuit
The voltage across the inductor always leads the current by 90° The effective resistance of a coil in an AC circuit is called its inductive reactance (measured in ohms):

12 Chapter 33 Problem 11 Determine the maximum magnetic flux through an inductor connected to a standard electrical outlet (ΔVrms= 120 V, f = 60.0 Hz).

13 The RLC Series Circuit The resistor, inductor, and capacitor can be combined in a circuit The current in the circuit is the same at any time and varies sinusoidally with time

14 The RLC Series Circuit The instantaneous voltage across the resistor is in phase with the current The instantaneous voltage across the inductor leads the current by 90° The instantaneous voltage across the capacitor lags the current by 90°

15 Phasor Diagrams Because of the different phase
relationships with the current, the voltages cannot be added directly To simplify the analysis of AC circuits, a graphical constructor called a phasor diagram can be used A phasor is a vector rotating CCW; its length is proportional to the maximum value of the variable it represents The vector rotates at an angular speed equal to the angular frequency associated with the variable, and the projection of the phasor onto the vertical axis represents the instantaneous value of the quantity

16 Phasor Diagrams The voltage across the resistor is in phase with the current The voltage across the inductor leads the current by 90° The voltage across the capacitor lags behind the current by 90°

17 Phasor Diagrams The phasors are added as vectors to account for the phase differences in the voltages ΔVL and ΔVC are on the same line and so the net y component is ΔVL - ΔVC

18 Phasor Diagrams The voltages are not in phase, so they cannot simply be added to get the voltage across the combination of the elements or the voltage source  is the phase angle between the current and the maximum voltage The equations also apply to rms values

19 Phasor Diagrams ΔVR = Imax R ΔVL = Imax XL ΔVC = Imax XC

20 Impedance of a Circuit ΔVmax = Imax Z
The impedance, Z, can also be represented in a phasor diagram φ: phase angle Ohm’s Law can be applied to the impedance ΔVmax = Imax Z This can be regarded as a generalized form of Ohm’s Law applied to a series AC circuit

21 Summary of Circuit Elements, Impedance and Phase Angles

22 Problem Solving for AC Circuits
Calculate as many unknown quantities as possible (e.g., find XL and XC) Be careful with units – use F, H, Ω Apply Ohm’s Law to the portion of the circuit that is of interest Determine all the unknowns asked for in the problem

23 Chapter 33 Problem 24 An AC source with Vmax = 150 V and f = 50.0 Hz is connected between points a and d in the figure. Calculate the maximum voltages between (a) points a and b, (b) points b and c, (c) points c and d, and (d) points b and d.

24 Power in an AC Circuit No power losses are associated with pure capacitors and pure inductors in an AC circuit In a capacitor, during 1/2 of a cycle energy is stored and during the other half the energy is returned to the circuit In an inductor, the source does work against the back emf of the inductor and energy is stored in the inductor, but when the current begins to decrease in the circuit, the energy is returned to the circuit

25 Power in an AC Circuit Pav = Irms ΔVR,rms ΔVR, rms = ΔVrms cos 
The average power delivered by the generator is converted to internal energy in the resistor Pav = Irms ΔVR,rms ΔVR, rms = ΔVrms cos  Pav = Irms ΔVrms cos  cos  is called the power factor of the circuit Phase shifts can be used to maximize power outputs

26 Chapter 33 Problem 28 A series RLC circuit has a resistance of 45.0 Ω and an impedance of 75.0 Ω. What average power is delivered to this circuit when ΔVrms = 210 V?

27 Resonance in an AC Circuit
Resonance occurs at the frequency, ω0, where the current has its maximum value To achieve maximum current, the impedance must have a minimum value This occurs when XL = XC and

28 Resonance in an AC Circuit
Theoretically, if R = 0 the current would be infinite at resonance Real circuits always have some resistance Tuning a radio: a varying capacitor changes the resonance frequency of the tuning circuit in your radio to match the station to be received

29 Transformers An AC transformer consists of two coils of wire wound around a core of soft iron The side connected to the input AC voltage source is called the primary and has N1 turns The other side, called the secondary, is connected to a resistor and has N2 turns The core is used to increase the magnetic flux and to provide a medium for the flux to pass from one coil to the other

30 Transformers The rate of change of the flux is the same for both coils, so the voltages are related by When N2 > N1, the transformer is referred to as a step up transformer and when N2 < N1, the transformer is referred to as a step down transformer The power input into the primary equals the power output at the secondary

31 Chapter 33 Problem 43 A transmission line that has a resistance per unit length of 4.50 × 10-4 Ω/m is to be used to transmit 5.00 MW across 400 miles (6.44 × 105 m). The output voltage of the generator is 4.50 kV. (a) What is the line loss if a transformer is used to step up the voltage to 500 kV? (b) What fraction of the input power is lost to the line under these circumstances?

32 Answers to Even Numbered Problems
Chapter 33: Problem 4 25.3 rad/s 0.114 s

33 Answers to Even Numbered Problems
Chapter 33: Problem 10 3.80 J


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