1. Copyright © 2009 Pearson Education, Inc. Lecture 10 – AC Circuits

2. Copyright © 2009 Pearson Education, Inc. Current from a battery flows steadily in one direction (direct current, DC). Current from a power plant varies sinusoidally (alternating current, AC). 25-7 Alternating Current (Chapter 25)

3. Copyright © 2009 Pearson Education, Inc. The voltage varies sinusoidally with time: as does the current: 25-7 Alternating Current,,

4. Copyright © 2009 Pearson Education, Inc. Multiplying the current and the voltage gives the power: 25-7 Alternating Current

5. Copyright © 2009 Pearson Education, Inc. Usually we are interested in the average power: 25-7 Alternating Current.

6. Copyright © 2009 Pearson Education, Inc. The current and voltage both have average values of zero, so we square them, take the average, then take the square root, yielding the root-mean-square (rms) value: 25-7 Alternating Current

7. Copyright © 2009 Pearson Education, Inc. 25-7 Alternating Current Example 25-13: Hair dryer. (a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain?

8. Copyright © 2009 Pearson Education, Inc. Chapter 30 Inductance, Electromagnetic Oscillations, and AC Circuits

9. Copyright © 2009 Pearson Education, Inc. Energy Stored in a Magnetic Field LR Circuits LC Circuits with Resistance ( LRC Circuits) AC Circuits with AC Source LRC Series AC Circuit Resonance in AC Circuits Units of Chapter 30

10. Copyright © 2009 Pearson Education, Inc. Energy Stored in a Magnetic Field Energy U stored in an inductor of inductance L when it is carrying a current I U = ½ LI 2 The energy is stored in the magnetic field of the inductor. Similarly, energy stored in a capacitor C when potential difference across it is V U = ½ CV 2 The energy is stored in the electric field of the capacitor.

11. Copyright © 2009 Pearson Education, Inc. A circuit consisting of an inductor and a resistor will begin with most of the voltage drop across the inductor, as the current is changing rapidly. With time, the current will increase less and less, until all the voltage is across the resistor. 30-4 LR Circuits

12. Copyright © 2009 Pearson Education, Inc. 30-4 LR Circuits The sum of potential differences around the loop gives Integrating gives the current as a function of time: The time constant of an LR circuit is...

13. Copyright © 2009 Pearson Education, Inc. If the circuit is then shorted across the battery, the current will gradually decay away: 30-4 LR Circuits.

14. Copyright © 2009 Pearson Education, Inc. 30-4 LR Circuits Example 30-6: An LR circuit. At t = 0, a 12.0-V battery is connected in series with a 220-mH inductor and a total of 30-Ω resistance, as shown. (a) What is the current at t = 0? (b) What is the time constant? (c) What is the maximum current? (d) How long will it take the current to reach half its maximum possible value? (e) At this instant, at what rate is energy being delivered by the battery, and (f) at what rate is energy being stored in the inductor’s magnetic field?

15. Copyright © 2009 Pearson Education, Inc. Resistors, capacitors, and inductors have different phase relationships between current and voltage when placed in an ac circuit. The current through a resistor is in phase with the voltage. 30-7 AC Circuits with AC Source

16. Copyright © 2009 Pearson Education, Inc. Therefore, the current through an inductor lags the voltage by 90°. 30-7 AC Circuits with AC Source The voltage across the inductor is given by or.

17. Copyright © 2009 Pearson Education, Inc. 30-7 AC Circuits with AC Source The voltage across the inductor is related to the current through it: The quantity X L is called the inductive reactance, and has units of ohms:.

18. Copyright © 2009 Pearson Education, Inc. 30-7 AC Circuits with AC Source Example 30-9: Reactance of a coil. A coil has a resistance R = 1.00 Ω and an inductance of 0.300 H. Determine the current in the coil if (a) 120-V dc is applied to it, and (b) 120-V ac (rms) at 60.0 Hz is applied.

19. Copyright © 2009 Pearson Education, Inc. Therefore, in a capacitor, the current leads the voltage by 90°. 30-7 AC Circuits with AC Source The voltage across the capacitor is given by.

20. Copyright © 2009 Pearson Education, Inc. 30-7 AC Circuits with AC Source The voltage across the capacitor is related to the current through it: The quantity X C is called the capacitive reactance, and (just like the inductive reactance) has units of ohms:.

21. Copyright © 2009 Pearson Education, Inc. 30-7 AC Circuits with AC Source Example 30-10: Capacitor reactance. What is the rms current in the circuit shown if C = 1.0 μF and V rms = 120 V? Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 10 5 Hz.

22. Copyright © 2009 Pearson Education, Inc. Analyzing the LRC series AC circuit is complicated, as the voltages are not in phase – this means we cannot simply add them. Furthermore, the reactances depend on the frequency. 30-8 LRC Series AC Circuit

23. Copyright © 2009 Pearson Education, Inc. We calculate the voltage (and current) using what are called phasors – these are vectors representing the individual voltages. Here, at t = 0, the current and voltage are both at a maximum. As time goes on, the phasors will rotate counterclockwise. 30-8 LRC Series AC Circuit

24. Copyright © 2009 Pearson Education, Inc. Some time t later, the phasors have rotated. 30-8 LRC Series AC Circuit

25. Copyright © 2009 Pearson Education, Inc. The voltages across each device are given by the x -component of each, and the current by its x -component. The current is the same throughout the circuit. 30-8 LRC Series AC Circuit

26. Copyright © 2009 Pearson Education, Inc. We find from the ratio of voltage to current that the effective resistance, called the impedance, of the circuit is given by 30-8 LRC Series AC Circuit

27. Copyright © 2009 Pearson Education, Inc. 30-8 LRC Series AC Circuit The phase angle between the voltage and the current is given by The factor cos φ is called the power factor of the circuit. or

28. Copyright © 2009 Pearson Education, Inc. Power in Inductor, Capacitor, Resistor and AC Circuits Power is dissipated only by a resistance; none is by inductance & capacitance.

29. Copyright © 2009 Pearson Education, Inc. 30-8 LRC Series AC Circuit Example 30-11: LRC circuit. Suppose R = 25.0 Ω, L = 30.0 mH, and C = 12.0 μF, and they are connected in series to a 90.0-V ac (rms) 500-Hz source. Calculate (a) the current in the circuit, (b) the voltmeter readings (rms) across each element, (c) the phase angle , and (d) the power dissipated in the circuit.

30. Copyright © 2009 Pearson Education, Inc. The rms current in an ac circuit is Clearly, I rms depends on the frequency. 30-9 Resonance in AC Circuits

31. Copyright © 2009 Pearson Education, Inc. We see that I rms will be a maximum when X C = X L ; the frequency at which this occurs is f 0 = ω 0 /2π is called the resonant frequency. 30-9 Resonance in AC Circuits

32. Copyright © 2009 Pearson Education, Inc. Summary of Chapter 30 LR circuit: Inductive reactance: Capacitive reactance:..

33. Copyright © 2009 Pearson Education, Inc. Summary of Chapter 30 LRC series circuit: Resonance in LRC series circuit:.