Chapter 32A AC Circuits. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

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1 Chapter 32A AC Circuits A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

2 Objectives: After completing this module, you should be able to: Describe the sinusoidal variation in ac current and voltage,, and calculate their effective values. Write and apply equations for calculating the inductive and capacitive reactances for inductors and capacitors in an ac circuit. Describe, with diagrams and equations, the phase relationships for circuits containing resistance, capacitance, and inductance.

3 Objectives (Cont.) Write and apply equations for calculating the impedance,, the phase angle,, the effective current,, the average power,, and the resonant frequency for a series ac circuit. Describe the basic operation of a step- up and a step-down transformer. Write and apply the transformer equation and determine the efficiency of a transformer.

4 Alternating Currents An alternating current such as that produced by a generator has no direction in in the sense that direct current has. The magnitudes vary sinusoidally with time as given by: AC-voltage and current E = E max sin i = i max sin E max i max time, t

5 Rotating Vector Description The coordinate of of the emf at at any instant is is the value of of E max max sin Observe for incremental angles in in steps of of Same is is true for i. i. E E = E max sin Radius R = = max E max

6 Effective AC Current The average current in a cycle is zero half + and half -. i max I = i max But energy is expended, regardless of direction. So the root-mean- square value is useful. The rms value I rms is sometimes called the effective current I eff : 2 I I Irms The effective ac current: i eff = i max

7 AC Definitions One effective ampere is that ac current for which the power is the same as for one ampere of dc current. Effective current: i eff i eff = i max i max One effective volt is that ac voltage that gives an effective ampere through a resistance of one ohm. Effective voltage: V eff eff = V max max

8 Example 1: For a particular device, the house ac voltage is 120-V and the ac current is 10 A. A What are their maximum values? eff i eff = i max V eff eff = V max i eff 10 A imax max V eff 120V V i max = A V max = 170 V The ac voltage actually varies from +170 V to to -170 V and the current from 14.1 A to to 14.1 A. A

9 Pure Resistance in AC Circuits R V max Voltage A V i max Current a.c. Source Voltage and current are in in phase, and Ohm s law applies for effective currents and voltages. Ohm s law: V eff = i eff R

in i current which then peaks as the emf goes

10 AC and Inductors I 0.63I i Inductor Current Rise I 0.37I i Inductor Current Decay Time, t Time, t The voltage V peaks first, causing rapid rise in i current which then peaks as the emf goes to zero. Voltage leads (peaks before) ) the current by Voltage and current are out of phase.

11 A Pure Inductor in AC Circuit A L V V max Voltage i max Current a.c. The voltage peaks 90 0 before the current peaks. One builds as the other falls and vice versa. The reactance may be defined as the nonresistive opposition to the flow of ac current.

12 Inductive Reactance The back emf induced by a changing current provides opposition to current, called inductive reactance X L. A L V a.c. Such losses are temporary,, however, since the current changes direction,, periodically re-supplying energy so that no net power is lost in one cycle. Inductive reactance X L is a function of both the inductance and the frequency of the ac current.

13 Calculating Inductive Reactance A L V a.c. Inductive Reactance: XL 2 fl Unit is the Ohm's law: VL ix L The voltage reading V in the above circuit at the instant the ac current is i can be found from the inductance in H and the frequency in Hz. V i(2 fl) L Ohm s law: V L = i eff X L

14 Example 2: A coil having an inductance of 0.6 H is connected to a 120-V, 60 Hz ac source. Neglecting resistance, what is the effective current through the coil? Reactance: X L = 2fL L = 0.6 H X L = 2(60 Hz)(0.6 H) A V X L = V, 60 Hz i eff V X eff L 120V 226 i eff i eff = A Show that the peak current is I max = A

15 AC and Capacitance Q max q Capacitor I i Capacitor 0.63 I Rise in Charge 0.37 I Current Decay Time, t Time, t The voltage V peaks ¼ of a cycle after the current i reaches its maximum. The voltage lags the current. Current i and V out of phase.

16 A Pure Capacitor in AC Circuit C V max Voltage A V i max Current a.c. The voltage peaks 90 0 after the current peaks. One builds as the other falls and vice versa. The diminishing current ii builds charge on C which increases the back emf of of V C. C.

17 Capacitive Reactance Energy gains and losses are also temporary for capacitors due to the constantly changing ac current. A C V a.c. No net power is lost in a complete cycle, even though the capacitor does provide nonresistive opposition (reactance( reactance) ) to the flow of ac current. Capacitive reactance X C is affected by both the capacitance and the frequency of the ac current.

18 Calculating Inductive Reactance A C V a.c. Capacitive Reactance: X C 1 2 fc Ohm's law: V C Unit is the ix C The voltage reading V in the above circuit at the instant the ac current is i can be found from the inductance in F and the frequency in Hz. V L 2 i fl Ohm s law: V C = i eff X C

19 Example 3: A 2-F 2 F capacitor is connected to a 120-V, 60 Hz ac source. Neglecting resistance, what is the effective current through the coil? 1 C = 2 F Reactance: X C 2 fc X C i eff X C = 1330 V X eff C (60 Hz)(2 x 10 F) 120V 1330 i eff i eff = 90.5 ma Show that the peak current is i max = 128 ma A V 120 V, 60 Hz

20 Memory Aid for AC Elements An old,, but very effective, way to remember the phase differences for inductors and capacitors is : E L i the I C E man E L I I the i C E Man Emf E is is before current ii in in inductors L; L; Emf E is is after current ii in in capacitors C.

21 Frequency and AC Circuits Resistance R is constant and not affected by f. Inductive reactance X L varies directly with frequency as expected since E i/ i/t. XL 2 fl R, X X C X C X L 1 2 fc Capacitive reactance X C varies inversely with f since rapid ac allows little time for charge to build up on capacitors. R f

22 Series LRC Circuits V T a.c. L Series ac circuit A R C V L V R V C Consider an inductor L, L, a capacitor C, C, and a resistor R all connected in in series with an ac source.. The instantaneous current and voltages can be measured with meters.

23 Phase in a Series AC Circuit The voltage leads current in an inductor and lags current in a capacitor. In phase for resistance R. V V = V max sin V L V C V R Rotating phasor diagram generates voltage waves for each element R, L,, and C showing phase relations. Current i is always in phase with V R.

24 Phasors and Voltage At time t = 0, suppose we read V L, V R and V C for an ac series circuit. What is the source voltage V T? V L V C Phasor Diagram V R V L - V C Source voltage V T V R We handle phase differences by finding the vector sum of these readings. V T = V i. The angle is the phase angle for the ac circuit.

25 Calculating Total Source Voltage Source voltage Treating as vectors, we find: V L - V C V T V R V V ( V V ) 2 2 T R L C VL VC tan V R Now recall that: V R = ir; V L = ix L ; and V C = iv C Substitution into the above voltage equation gives: V i R ( X X ) 2 2 T L C

26 Impedance in an AC Circuit X L - X C Impedance 2 2 Z R V i R ( X X ) T L C Impedance Z is defined: Z R ( X X ) 2 2 L C Ohm s s law for ac current and impedance: V iz or i T VT Z The impedance is is the combined opposition to to ac current consisting of of both resistance and reactance.

27 Example 3: A 60- resistor, a 0.5 H inductor, and an 8-F capacitor are connected in series with a 120-V, 60 Hz ac source. Calculate the impedance for this circuit. X 2 fl and X X L L X C C 1 2 fc 2 (60Hz)(0.6 H) = (60Hz)(8 x 10 F) 0.5 H A 8 F 120 V 60 Hz 60 Z R ( X X ) (60 ) ( ) L C Thus, the impedance is: Z = 122

28 Example 4: Find the effective current and the phase angle for the previous example. X L = 226 X C = 332 R = 60 Z Z = 122 VT 120 V ieff Z H i eff i eff = A Next we find the phase angle: X L - X C Impedance Z R A 120 V 60 Hz 8 F 60 X L X C = = -106 X R X L C R = 60 tan Continued...

29 Example 4 (Cont.): Find the phase angle for the previous example Z X L X C = = -106 X R X L C R = 60 tan tan = The negative phase angle means that the ac voltage lags the current by This is is known as a capacitive circuit.

30 Resonant Frequency Because inductance causes the voltage to to lead the current and capacitance causes it it to to lag the current, they tend to to cancel each other out. X L X C X L = X C R Resonance (Maximum Power) occurs when X L = X C Z R ( X X ) R 2 2 L C Resonant f r X L = X C 2 fl 1 2 fc f r 1 2 LC

31 Example 5: Find the resonant frequency for the previous circuit example: L =.5 H, C = 8 F f f r 1 2 LC (0.5H)(8 x 10 F Resonance X L = X C 0.5 H A 8 F 120 V Resonant f r f r = 79.6 Hz? Hz 60 At resonant frequency, there is is zero reactance (only( resistance) ) and the circuit has a phase angle of of zero.

32 Power in an AC Circuit No power is is consumed by inductance or or capacitance. Thus power is is a function of of the component of of the impedance along resistance: X L - X C Impedance Z R P lost in R only In terms of ac voltage: P = iv ivcos In terms of the resistance R: P = i 2 i 2 R The fraction Cos is known as the power factor.

33 Example 6: What is the average power loss for the previous example: V = 120 V, = , i = 90.5 A, and R = 60. P = i 2 R = ( A) 2 (60 Average P = W The power factor is: Cos Resonance X L = X C A 120 V 0.5 H 8 F Cos = or or 49.2%? Hz 60 The higher the power factor, the more efficient is is the circuit in in its use of of ac power.

34 The Transformer A transformer is a device that uses induction and ac current to step voltages up or down. An ac source of of emf E p is is connected to to primary coil with N p turns. Secondary has N s turns and emf of of E s. Transformer a.c. R N p N s Induced emf s are: t EP NP ES NS t

35 Transformers (Continued): Transformer a.c. E P N P t N p N s R E S N S t Recognizing that /t is the same in each coil, we divide first relation by second and obtain: The transformer equation: E E P S N N P S

36 Example 7: A generator produces 10 A at 600 V. The primary coil in a transformer has 20 turns. How many secondary turns are needed to step up the voltage to 2400 V? Applying the transformer equation: V V P S N N P S I = 10 A; V p = 600 V a.c. 20 turns N p N s R N S NV V (20)(2400 V) 600V P S N S = 80 turns P N S = 80 turns This is is a step-up transformer; ; reversing coils will make it it a step-down transformer.

37 Transformer Efficiency There is no power gain in stepping up the voltage since voltage is increased by reducing current. In an ideal transformer with no internal losses: Ideal Transformer a.c. An ideal transformer: N p N s R E i P P E i S S or i i P s E E S P The above equation assumes no internal energy losses due to to heat or flux changes. Actual efficiencies are usually between 90 and 100%.

38 Example 7: The transformer in Ex. 6 is connected to a power line whose resistance is 12.. How much of the power is lost in the transmission line? V S = 2400 V I = 10 A; V p = 600 V 12 PiP a.c. E i E i i P P S S S is (600V)(10A) 2400V E E S 2.50 A 20 turns N p N s P lost = i 2 R = (2.50 A) 2 (12 ) P lost = 75.0 W P in = (600 V)(10 A) = 6000 W %Power Lost = (75 W/6000 W)(100%) = 1.25% R

39 Summary Effective current: i eff i eff = i max i max Effective voltage: V eff eff = V max max Inductive Reactance: XL 2 fl Unit is the Ohm's law: VL ix L Capacitive Reactance: X C 1 2 fc Ohm's law: V C Unit is the ix C

40 Summary (Cont.) L C V V ( V V ) tan 2 2 T R L C V V V R Z R ( X X ) 2 2 L C XL XC tan R V iz or i T VT Z f r 1 2 LC

41 In terms of ac voltage: Summary (Cont.) Power in AC Circuits: In terms of the resistance R: P = iv ivcos P = i 2 i 2 R Transformers: E E P S N N P S E i P P E i S S

42 CONCLUSION: Chapter 32A AC Circuits

Alternating Current Circuits

Alternating Current Circuits AC Circuit An AC circuit consists of a combination of circuit elements and an AC generator or source. The output of an AC generator is sinusoidal and varies with time according