Alternating Current. Symbol for A.C. source. A.C.

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1 Alternating Current Kirchoff s rules for loops and junctions may be used to analyze complicated circuits such as the one below, powered by an alternating current (A.C.) source. But the analysis can quickly become much more complicated. In this chapter we consider the behavior of basic, but useful, circuits driven by A.C. sources. Symbol for A.C. source. A.C.

2 Demo Simple alternating current (A.C.) generator In a basic A.C. generator, a permanent magnet provides a reasonably uniform magnetic field. As the generator loop turns in this field at frequency ω (rad/s), the flux through the loop changes sinusoidally with time. This causes the output voltage (induced emf) to also change sinusoidally, 90 degrees out of phase with the flux. We can see why this happens by applying Faraday s aw of Induction to the flux: et: Φ B = Φ cos( ω ) 0 t Then: emf = = ω d dt Φ [ Φ 0 Motional emf 0 cos( ωt)] sin( ωt) = V sin( ωt)

3 Resistor connected to A.C. source V sin( ωt) V = RI Use Kirchoff s loop equation: So the current in the resistor is: V sin(ωt) = VR = RI R VR V I R = = sin( ω t) = I R R Note that the resistor voltage and current are in phase. (See plot.) Instantaneous power dissipation: P = RI = RI sin ( ω t ) R sin( ωt) To find the average power (root mean square, RMS ), integrate over one cycle: So that: π / ω 1 P RMS = RI sin( ωt) dt = RI 0 = RI I I I RMS = =. 707 I = I RMS RMS

4 Inductor connected to A.C. source V sin( ωt) V = Use Kirchoff s loop equation: I di dt Integrate to solve for I : di dt = V sin( ωt) V V V π = sin( ωt) dt = cos( ωt) = sin( ωt ) ω ω From either the equations or the graph we see that, for A.C. sources, the current in an inductor lags behind its voltage by 90 o

5 Inductive Reactance, X Notice that whenever cos(ωt) is +1 or -1, the magnitude of the current in the inductor is at the maximum. We can express this as follows: I V = ω I V = ω cos( ωt) This equation has the same structure as Ohm s aw, and we can identify the factor in the denominator as setting the ratio between V and I. We call this factor the inductive reactance, and define it as follows: X = ω And since RMS voltage and current are one-half their values at maximum, X can be used in both equations: V = X I V RMS = X I RMS It s clear from these equations that X must have units of ohms. Unlike resistance, inductive reactance changes with frequency! Example

6 Capacitor connected to A.C. source V sin( ωt) V C = Q C Kirchoff s loop equation: Differentiate to find I C : I C Q C = V sin( ω t) Q = CV sin( ωt) dq π = = ω CV cos( ωt) = ωcv sin( ωt + ) dt From either the equations or the graph we see that, for A.C. sources, the current in a capacitor leads its voltage by 90 o

7 Capacitive Reactance, X C Again, whenever cos(ωt) is +1 or -1, the magnitude of the current in the capacitor is at the maximum. We can express this as follows: I C = ωcv cos( ωt) I = ω CV This equation has the same structure as Ohm s aw, and we can identify the factor multiplying V as setting the ratio between V and I. We define the capacitive reactance as the inverse of this factor: 1 X C = ωc And since RMS voltage and current are one-half their values at maximum, X C can be used in both equations: V = X I V RMS = X C I RMS As with X, X C must have units of ohms. And as before, this capacitive reactance changes with frequency! C Example

8 Frequency dependence of R, X, and X C X = ω V = RI 1 X C = ωc

9 The Driven RC Circuit We have already studied the damped oscillations of an RC circuit which has been energized with an initial charge or current. But now we consider the response of this circuit when it is driven by an AC source, which feeds energy into the circuit. We will see how this reaches an equilibrium, with the power from the source being dissipated by the resistor.

10 Analyzing the Driven RC Circuit Since this circuit consists of one loop, the same current, I, passes through every element in the circuit. Once again, we can use Kirchoff s loop equation for voltages. Using V for the source voltage: V = V + V + V C R Now we insert all the equations from previous pages. With this more complicated circuit the phase of the source will no longer be the same as that for the resistor. So we let the source voltage be displaced by φ. V For t = 0: For ωt = π/: sin( ωt ϕ) = Square and add: RI sin( ωt) + X I C cos( ωt) X V sin( ϕ) = X I X I = ( X X ) I V sin( π / ϕ) = V cos( ϕ) = V Define impedance, Z: = I [ ] R + ( X X ) C C RI Z = R + ( X X ) V = C C I ZI cos( ωt) We can think of impedance, Z, as total resistance at a given frequency ω.

11 Measuring voltages in an RC Circuit.

12 Voltages as a function of time in a typical RC Circuit.

13 Frequency dependence of reactances and impedance in RC Circuit At the frequency where X = X C, the impedance of the circuit is at a minimum, and Z = R. Since Z is minimum at the point where X = X C the average current is maximum. The power dissipation is also maximum at this point, meaning that the power delivered to the circuit by the AC source is maximum. At what frequency does this occur? X 1 1 = X C ω = ω = ω = ωc C 1 C But this is the frequency of an C circuit when there is no resistor present! At this frequency, the circuit is in resonance, with the driving frequency, ω, equal to the natural frequency of the C oscillator, ω ο.

14 Current as a function of ω in a driven RC circuit ω 0 = 1 C The plot at right shows current resonance curves for several values of R. The greater the resistance, the lower the maximum current. We can calculate the shapes of these resonance curves as follows: I RMS = V Z RMS = R + V ( X RMS X C ) = R + V ω RMS ( ω 1 C ) = R + V ω RMS ( ω ω ) o

15 Resonance curves for a damped, driven harmonic oscillator: mass + spring + damping.

16 Collapse of Tacoma Narrows bridge in 1940 driven into resonance by wind. Too little damping!

17 Coils sharing the same magnetic flux, BA These can be used as transformers. In air With iron yoke

18 Raising and owering A.C. Voltage: Transformers Faraday s aw of Induction tells us the relationship between the magnitudes of the voltage at the terminals and the rate of change of magnetic flux through each coil: dφ1 V1 = N1 dt dφ V = N dt But the flux through each coil is the same: d Φ1 dφ Φ1 = Φ = dt dt V 1 = So the output (secondary) voltage can be raised or lowered compared to the input (primary) voltage, by the ratio of turns. A transformer that raises (lowers) voltage is called a step-up ( step-down ) transformer. Transformers lose very little power: N 1 V N V = N V1 N1 1 1 = V1I 1 VI 1 I1 V N P = V I = I = As the voltage is increased, the current is lowered by the same factor. N

19 Transformers are not perfect, but they dissipate little power.

Nikola Tesla (Serbian Cyrillic: Никола Тесла; 10 July 1856 7 January 1943) was a Serbian-American inventor, mechanical engineer, and electrical engineer.

late 19th and early 0th centuries building on the theories of electromagnetic technology discovered by Michael Faraday and used in direct current (DC) applications.

20 Nikola Tesla (Serbian Cyrillic: Никола Тесла; 10 July January 1943) was a Serbian-American inventor, mechanical engineer, and electrical engineer. He was an important contributor to the birth of commercial electricity, and is best known for his many revolutionary developments in the field of electromagnetism in the late 19th and early 0th centuries building on the theories of electromagnetic technology discovered by Michael Faraday and used in direct current (DC) applications. Tesla's patents and theoretical work formed the basis of modern alternating current (AC) electric power systems, including the polyphase system of electrical distribution and the AC motor.

21 Tesla Coil Schematic

22 Tesla Never Thought Small

23 Tesla s A.C. Dynamo (generator) used to generate alternating currrent electricity, which became the technology of choice for electrification across the U.S. (power grids) and ultimately, the whole planet. U.S. Patent 39071

24 Wardenclyffe Tower facility ~ 1915 Mark Twain in Tesla s ab.

A.C. power networks Transmission lines: 155,000-765,000 V ocal substation: 7,00 V House: 110/0 V For a given length of power line, P=RI.

25 A.C. power networks Transmission lines: 155, ,000 V ocal substation: 7,00 V House: 110/0 V For a given length of power line, P=RI. If the power line voltage is 550,000 V, the current is reduced by a factor of 5000 compared to that at 110 V, and the power loss is reduced by a factor of 5 million.

Chapter 33. Alternating Current Circuits

Chapter 33. Alternating Current Circuits Chapter 33 Alternating Current Circuits 1 Capacitor Resistor + Q = C V = I R R I + + Inductance d I Vab = L dt AC power source The AC power source provides an alternative voltage, Notation - Lower case