Unit 9. Alternating current

Transcription

1 nit 9 Alternating current 9. ntroduction 9. Features of an A.C. 9.3 Behaviour of basic dipoles facing an A.C. 9.4 RLC series circuit. pedance and phase lag. 9.5 Power on A.C. 9.6 Questions and probles Objectives To know the ain features of alternating current (A.C.) and their effect on resistors, capacitors and inductors. nderstand the phase lag between drop of potential and intensity of current on A.C. circuits. Copute the rate between drop of potential and intensity of current in RLC series dipoles. Define the ipedance of a circuit. Analyze a RLC series circuit fro the point of view of energy. To know the eaning of power factor on A.C. 9. ntroduction Along 9 th century, the electricity was developed around of direct current (D.C.). This type of electricity was useful for a lot of applications, on electric engines, heating, lighting, telegraph, telephone, etc One of the ost iportant businessen was Thoas Alva Edison, owner of General Electric Copany. But when Edison tried to send the electric energy assively to a large distance, alost all the energy was lost by Joule heating along the transission line. n order to understand the proble of sending electric energy with direct current, let s iagine we want to send, for exaple, a power of W at a voltage of V (an usual voltage), to a little distance, for exaple 3 K; the current along the line will be / A. The resistance of a copper wire having a cross section of 5 is around,4 Ω/K; as the current ust go and return, we need 6 K of wire, with a total resistance of,4*6,4 Ω. The total power lost by Joule heating on transission line will be: Lost Power *,44 W 4% of sent power 9-

2 To solve this proble, the consuers of electricity should be placed near the generating factory and a big project to build a lot of factories in Niagara Falls (to profit its water juping), was presented. But they were soe scientists, engineers and businessen supporters of alternating current (A.C.) because of its easy generation (on before unit we saw that a loop turning inside a agnetic field produces sinusoidal alternating current) and the low quantity of energy lost on the transission line. We know that the transforer doesn t work with direct current, but it works on alternating current, and a voltage of V can be easily converted in, for exaple, V. The lost power sending W at V (intensity of current /, A) to a distance of 3 K with the sae wire we have used for direct current will be: Lost Power(,) *,4,4 W only a,4% of sent power Nikola Tesla, a Croatian engineer, and George Westinghouse were two of the ost active supporters of A.C., and their disagreeent with the wide use of D.C. produced the War of currents on last decade of 9 th century. They won this war and Niagara Falls was preserved. Fro then, the electric energy can be sent at very large distances at very high voltages, even V, and only a few generating factories of electricity are needed. Of course, this voltage is not useful at hoe (it would be very dangerous) and it ust be reduced with a second transforer; this is the reason why we can found transforers along any town. Besides its easy transportation, Fourier s theore states that any periodic function can be written as an addition of sinusoidal functions, and then the results for alternating current can be extrapolated to any type of periodic function, even digital signals. 9. Features of an a.c. We saw on before unit that a sinusoidal alternating current can be produced by turning a loop inside a unifor agnetic field. The sybol for the A.C. generators on circuits is that on right, and the atheatical expression for the difference of potential on terinals of generator can be written as: cos( ω t + ϕ ) this equation has be written as a cosine, but it could be also written as a sine, only adding 9º (/ rad) to the phase: u cos( ω t + ϕu) sin( ωt + ϕu + ) 3 The ain features of this sinusoidal function are: The aplitude,, is the axiu value reached by the function ( Figure 9-). The units of aplitude are the sae that the agnitude we are representing; in this case, volts. The period, T is the tie taken for a whole cycle of function ( Figure 9-). ts unit in the.s. is the second. D.d.p. (V) T tie (s) Figure 9-. Aplitude and period of an a.c. 9-

3 The frequency, f, is the nuber of cycles by second. t s the inverse of period: is Hertz (Hz): Hz s -. f. ts unit T The angular frequency, angular speed or pulsatance, ω, is the angle (in radians) turned on a second. ts unit is rad/s. We ust reeber that the radian is diensionless, and then the diensions of angular frequency are T -. As a whole cycle equals radians and the period, T, is the tie taken by a cycle: ω f T The phase ω t + ϕu, expressed in radians. The initial phase ϕ u, is the agnitude of phase on tie t. For an easier reading it is usual to give the agnitude u in degrees instead radians, but when it s operated, it ust be converted to radians. The initial phase, really, locates the sinusoidal wave along the X axis. On right (Figure 9-) the initial phase of several sinusoidal functions can be seen. t s iportant to note that this figure represents the function cos( ω t + ϕu) ; if the sinus function is used, as 9º ust be added, the initial phases would be different. Figure 9-. nitial phase of several sinusoidal functions (expressed as cosinus) Phase lag is defined for two sinusoidal functions having the sae angular speed. Phase lag is the difference between phases of both functions at the sae tie; as the angular frequency is the sae for both functions, the difference between phases equals the difference between initial phases. Let s consider, for exaple, a voltage cos( ω t + ϕ ) and an intensity of current i(t) cos( ω t + ϕi). The phase lag between voltage and intensity is: u Figure 9-3 Positive phase lag between voltage and intensity; voltage goes ahead intensity (or intensity goes behind voltage) ϕ ui ωt + ϕu) ( ωt + ϕi) ( ϕ ϕ u i For this exaple, if phase lag is positive (Figure 9-3), voltage goes ahead intensity (or intensity goes behind voltage). But if phase lag is negative (Figure 9-4), voltage goes behind intensity (or intensity goes Figure 9-4 Negative phase lag between voltage and intensity; voltage goes behind intensity (or intensity goes ahead voltage) 9-3

4 ahead voltage). f phase lag is zero, both functions go in phase. f we had defined phase lag between intensity and voltage ( ϕ iu instead ϕ ui ) the sign of phase lag would be opposite. But we always will refer to a phase lag between voltage and intensity: ϕ ϕ u ϕ i Root ean square: Root ean square (rs) of a sinusoidal function is defined as the aplitude divided into. For exaple (for a voltage and for a intensity): rs rs The root ean square of a sinusoidal function is very iportant in alternating current because this is the agnitude easured by the easuring devices (volteters, aeters, ultieters, etc ). ts physical eaning is that of a direct current dissipating the sae power that an alternating current of aplitude on a resistor. The deonstration of his equivalence will be done when we deal with power on A.C., at the end of this unit. The current we can found in the wall sockets at our hoes is a current having a voltage rs V, and then the aplitude of this voltage is * 3 V. 9.3 Behavior of basic dipoles facing an A.C. At this point we ll study the behavior of basic dipoles (resistor, inductor and capacitor) when an alternating current is flowing along the. To do it, we ll study two paraeters: the rate between the aplitude of voltage on terinals of dipole and the intensity of current the phase lag between voltage on terinals and intensity of current. Resistor Let s suppose a resistor (resistance R) flowed by a intensity of current i(t) and having a difference of potential between its terinals. n order to siplify the coputations, we ll take initial phase zero for intensity (ϕ i ): i(t) cos( ωt) At any tie, on a resistor ust be verified that u (t) Ri(t) and then Rcos( ωt) The difference of potential on terinals of resistor is a new cosine function, cos( ω t + ϕu) being R and ϕ u Obviously, if we had taken a initial phase not zero for intensity, the initial phase of voltage would have been the sae of intensity. 9-4

5 On a resistor, then R ϕ ϕ ϕ u i Voltage and intensity go on phase. nductor On an inductor, the ain paraeter is the self inductance coefficient L. f we take initial phase zero for intensity (ϕ i ): i(t) cos( ωt) and then At any tie, on an inductor ust be verified that u (t) di(t) L dt d( L cos( ωt)) Lω dt sin( ωt) Lω cos( ωt + ) The difference of potential on terinals of inductor is a new cosine function, On an inductor, then Lω ϕ ϕu ϕi Voltage goes 9º ahead intensity, or ntensity goes 9º behind voltage. cos( ω t + ϕu) being Lω and ϕ u / The factor Lω is known as nductance or nductive Reactance ( X L Lω); it has diensions of a resistance, and thus is easured in Ohs. Really, the inductance on inductor acts as the resistance on a resistor, but the ost iportant difference is that the inductance depends not only on the self inductance coefficient, but also on the frequency. So, on a circuit, the rate between aplitudes of voltage and intensity on inductor can change by only changing the frequency of current. 3. Capacitor n this case, for an easier coputation, we ll suppose the voltage on terinals of capacitor having initial phase zero (ϕ u ): 9-5

6 cos( ωt) For a capacitor ust be reebered that q(t) C and dq(t) i (t) dt Therefore i(t) dq(t) dt dc dt Cd(cos( ωt)) dt sin( ωt) cos( ωt + ) The intensity of current on a capacitor is a new cosine function, i(t) cos( ω t + ϕi) being and ϕ i/ On a capacitor, then ϕ ϕu ϕi Voltage goes 9º behind intensity, or ntensity goes 9º ahead voltage The factor / is known as Capacitance or Capacitive Reactance ( X C ); it has diensions of a resistance, and thus is easured in Ohs. Really, the capacitance on a capacitor acts as the resistance on a resistor, but the ost iportant difference is that the capacitance not only depends on the capacitance, but also on the frequency. So, in a circuit, the rate between aplitudes of voltage and intensity on capacitor can change by only changing the frequency of current. Exaple 9- A resistor 5 Ω sized, an inductor H sized, and a capacitor 5 µf sized, are connected in series. The voltage on terinals of inductor is u L (t) cos(t + ) V. Copute 4 the intensity of current flowing along three dipoles, the voltage on terinals of resistor and capacitor, and plot all these agnitudes on a graph. Solution: As we know voltage on inductor, it s easy to get the intensity of current on inductor. The angular speed of current is ω rad/s, and then: 3 L XL Lω Ω X L A 9-6

7 Phase lag on inductor is 9º, and then ϕ ϕi ϕi rad As all the dipoles are connected in series, the intensity of current on inductor is the sae for all the dipoles, being: i(t) il(t) ir(t) ic(t) cos(t ) 4 A Fro intensity, voltages on terinals of resistor and capacitor can be coputed: on resistor R R 5 5V and as phase lag on resistor is zero: (t) u R 5 cos(t ) 4 V On capacitor: XC Ω C XC 6 5 V And phase lag ϕ ϕ u 3 ) ϕ rad ( u (t) u C 3 cos(t ) 4 V The graph with voltages and intensity: i(t) R(t) L(t) C(t) ntensity (A) wt (rad) Voltage (V) t can be noted that, obviously, intensity and voltage on resistor go in phase, phase lag between voltages on inductor and capacitor is 8º (or -8º), voltage on inductor goes ahead intensity, and voltage on capacitor goes behind intensity. 9-7

8 9.4 RLC series circuit. pedance and phase lag. A RLC series circuit is a circuit having a resistor, an inductor and a capacitor connected in series (RLC dipole). As we have above seen, the intensity of current is the sae for all the basic dipoles on circuit, and the voltage on each device can be coputed. n order to siplify coputations, we have taken initial phase of intensity zero (ϕ i ). i(t) cos(wt) R L C u R (t) u L (t) u C (t) Figure 9-5. RLC series dipole But the ain proble is now to relate the intensity of current i(t) to the voltage on terinals of RLC circuit,. This voltage could be coputed by only adding the voltages on the devices on circuit: ur(t) + ul(t) + uc(t) Rcos( ωt) + Lωcos( ωt + ) + cos( ωt ) () The addition of sinusoidal functions is a new sinusoidal function, and is a sinusoidal function, but fro this equation is difficult to assess its aplitude and its initial phase. Nevertheless, the aplitude and phase of can be coputed using two new features of a RLC circuit, its pedance and its phase lag. Let s suppose that the voltage on terinals of RLC dipole, is: By expanding () reebering that that sin(/) and cos(/): R cos( ω t + ϕ) () cos( A + B) cos AcosB sin AsinB and taking in account cos( ωt) + Lω(cos( ωt)cos( ) sin( ωt)sin( )) + (cos( ωt)cos( ) + sin( ωt)sin( )) Rcos( ω t) Lωsin( ωt) + sin( ωt) (3) And expanding (): (cos( ωt)cos( ϕ) sin( ωt)sin( ϕ)) (4) dentifying coefficients of sin(ω t) and cos(ω t) between (3) and (4): sin(ω t) : Lω + sin( ϕ) (5) cos(ω t) : R cos( ϕ) (6) By dividing both equations (5)/(6) we get the phase lag between voltage on terinals of RLC dipole and intensity of current: Lw R Cw tgϕ 9-8

9 ϕ is the phase lag between voltage and intensity. Squaring (5) and (6) and adding the, taking in account that cos ( ϕ ) + sin ( ϕ) : (R + (Lω ) ) (cos ( ϕ) + sin ( ϕ)) R + (Lω ) The rate between and is the pedance (Z) of RLC series circuit: Z R + (Lω ) R + (XL XC) Obviously, the ipedance of a resistor equals R, for an inductor equals X L and for a capacitor equals X C. Exaple 9- On circuit of exaple 9-, find the voltage on terinals of circuit. Solution: Fro the ipedance of this circuit: Z Phase lag 5 + ( ),8 Ω Z *,8,8 V tgϕ ϕ 63,4º ϕu ϕ + ϕi 63,4 45 8,4º 5 And voltage on terinals of circuit:,8cos(t 8,4º) V The difference between the inductive reactance (X L ) and the capacitive reactance (X C ) is the reactance (X): X X L X C also easured in Ohs The ain agnitudes of an RLC series circuit can be easily reebered drawing the pedance triangle, a triangle being the horizontal cathetus the resistance of circuit, the vertical cathetus the reactance of circuit, the hypotenuse the ipedance of circuit, and the angle between Z and R, the phase lag of circuit. Depending on agnitudes of L and C, X can be positive (inductive circuit), negative (capacitive circuit), or zero (resonant circuit). ϕ Z R X Figure 9-6. pedance triangle for an inductive circuit 9-9

10 Review about phase lag and pedance for soe dipoles Phase lag and ipedance for basic dipoles (R, L and C) and for dipole RLC (RLC series circuit) can be suarized on next table: Dipole Phase lag φφ u - φ i pedance Z Resistor R nductor 9º X L Lω Capacitor -9º X C Lω ϕ arctg RLC series Z R + Lω R Table 9- Suary of phase lag and ipedance for basic dipoles and RLC series circuit. 9.5 Power on A.C. Let s iagine a dipole (both a resistor, an inductor, a capacitor, or a RLC series dipole) flowed by a intensity of current i(t) sin( ωt) ; we ll use the sine function instead cosine and initial phase zero for intensity in order to siplify the coputations. Voltage on terinals of dipole will be sin( ω t + ϕ). The aplitude is depending on ipedance of dipole, and φ is for a resistor, 9º for an inductor, -9º for a capacitor, and any agnitude between -9º and 9º for a RLC series dipole. At any tie, the consued power by the dipole, nstantaneous power, is the product of intensity and voltage: p(t) i(t) sin( ω t) sin( ωt + ϕ) Reebering that sin( A + B) sin AcosB + cos AsinB and that sin A sinacosa expanding the above equation p(t) i(t) sin( ωt)sin( ωt + ϕ) cosϕ sin ωt + [ sin( ωt)cosϕ + cos( ωt)sinϕ] sin( ωt) sinϕsinωt The instantaneous power is the addition of two ters: the first one is always positive, because cosϕ is always positive (ϕ is in the range -9º to 9º) and sin wt is also always positive. This ter is known as Active power or Real power: P (t) a A cosϕsin ωt i(t) sin ωt RLC sin (ωt + ϕ) Figure 9-7. Voltage and intensity on a RLC series dipole B 9-

11 The second ter can be positive or negative, according the agnitudes of ϕ and t; this ter is known as Reactive power: Pr(t) sinϕsinωt Active power (P a ) + Reactive power (P R ) nstantaneous power (P i ) As an exaple, giving values A, V, ω rad/s, ϕ,6 rad and drawing the functions corresponding to active, reactive and instantaneous power, we can see: Power on A.C. Active power Power on A.C. Reactive power Power on A.C. nst. pow er,8,8,8,6,6,6 Power (w),4,,4 + +, Power (w) Power (w),4, , -, -, -,4 -,4 -,4 wt (rad) wt (rad) wt (rad) Active power, as we have said is always positive, eaning that this power is consued at any tie by the dipole; reactive power is a function with a frequency ties that of current, and soeties is consued by the dipole (when it s positive) and soeties is generated by the dipole (when it s negative). Along a whole cycle, the average value of reactive power is zero ( T sin sin tdt T ϕ ω ), but the average value of active power equals the average value of instant power, not being zero: P iav T cosϕ Pa cosϕsin ωtdt av T T T sin ωtdt cosϕ rs rs cosϕ Along a cycle, the average power (both the instantaneous power as the active power) is always positive, eaning that the power is consued by the dipole. Cosϕ is known as Power factor, because for a given intensity and voltage, it deterines the consued power; it depends on the inductive or capacitive feature of circuit. The axiu consued power is reached for a circuit with ϕ (voltage and intensity on phase). Applying these results to basic dipoles: Resistor: For a resistor ϕ (cosϕ) and R The reactive power is zero at any tie, P r (t) Fro above equation, the average values P i av P a av rs R R On a resistor there isn t reactive power at any oent (cosϕ), and all the consued power is active power, lost by Joule heating. Besides, fro this equation can be easily understood the physical eaning of rs (rando ean square) agnitudes: An alternating current consues on a resistor the sae power than a direct current (D.C.) having the rs agnitudes (intensity and voltage). 9-

12 nductor: For an inductor ϕ9º (cosϕ) and Lω The active power is zero at any tie, P a (t) And the reactive power on a tie t is P(t) r sinω t sinωt Lω An inductor takes energy fro circuit along a half of cycle, storing this energy as agnetic field, and it returns this energy to the circuit in the another half of cycle. But an inductor doesn t consue net energy; it neither produces nor consues power. t happens at a frequency two ties that of electric current (ωt). Capacitor: For a capacitor ϕ-9º (cosϕ) and The active power is zero at any tie, P a (t) And the reactive power on a tie t is P(t) r sinωt sinωt As an inductor, a capacitor takes energy fro circuit along a half of cycle, storing this energy as electric field, and it returns this energy to the circuit in the another half of cycle. But a capacitor doesn t consue net energy; it neither produces nor consues power. t happens at a frequency two ties that of electric current (ωt). t s iportant to note the negative sign of reactive power for a capacitor. t only eans that the power is negative (capacitor consuing energy) when sinωt is positive, and then the reactive power on a capacitor is in opposition of phase (phase lag 8º) against the reactive power on an inductor. t eans that when inductor is taking energy fro circuit, the capacitor is giving it, and the other way round on the other half of cycle. f the aplitudes of reactive power on inductor and capacitor are equal (it happens only when X L X C ), then they are cancelled, and the reactive power on circuit is always zero; but if they aren t cancelled, the reactive power isn t zero and the intensity of current necessary to carry this energy produces Joule heating on wires of circuit, losing energy, even if it can t be taken advantage of the reactive power. PROPOSED PROBLEM: A RLC series circuit ade up by L H, C μf and R Oh is connected to an adjustable frequency generator, giving an aplitude V. f generator is adjusted to 6 Hz: a) Find the axiu intensity of current (aplitude of intensity). b) Find the aplitudes of voltage on resistor, inductor and capacitor. c) Find the phase lag. d) Write instantaneous agnitudes of intensity i(t), voltage on generator, resistor u R (t), inductor u L (t), and capacitor u C (t). e) Copute the power factor f) Find the average power given by generator to the circuit, verifying that this power is that consued on resistor. 9-

13 9.6 Questions and probles. Consider a capacitive circuit. How does the capacitive reactance change if the angular frequency increases two ties?. A capacitor C,5μF is connected to an A.C. generator; the aplitude of voltage on terinals of generator is 3 V Which is the aplitude of intensity of current along the capacitor,, if angular frequency is a) ω rad/s? b) ω rad/s? Sol: a),5 A b),5 A 3. An inductor L 45 H is connected to an A.C. generator; the aplitude of voltage on terinals of generator is 3 V and the inductive reactance of inductor is X L 3 Ω. Which is a) the angular frequency ω and the frequency f of generator? b) The aplitude of intensity,? c) Find that angular frequency that equals the inductive reactance of inductor to the capacitive reactance of proble. Sol: a) ω,9. 4 rad/s, f 854,67 Hz b),3 A c) 8,85 rad/s 4. A circuit is ade up by two basic dipoles in series. The terinals of this circuit are connected to an A.C. generator giving a voltage 5 cos(5 t+ º) V, and flowing along the circuit an intensity of current i(t) 3,4cos(5t - 53,4º) A. Deterine the two basic dipoles and their agnitudes. Sol: R 5 Ω L, H 5. Along a circuit ade up by two basic dipoles in series and a generator giving a voltage sen(t + 5º) V flows an intensity i(t) 4cos(t + 3,º) A. Deterine such dipoles and their agnitudes. Sol: R 9,7 Ω C,4 µf 6. On a RL series circuit, R 5 Ω and L,6 H, drop of potential on terinals of inductor is u L (t) 5cost V. Copute: Sol: a) the intensity of current b) phase lag and ipedance of circuit c) voltage on terinals of circuit a) i(t),5 cos(t - 9º) A; b) ϕ 67,4º Z 3 Ω c) 6,3cos(t -,6º) V 7. Along a RLC series circuit, R Ω, L,6 H and C µf, flows an intensity of current i(t) 3cos(5t - 6º) A. 9-3

14 Copute and draw on a graph the sinusoidal functions corresponding to the drop of potential on each device and the drop of potential on terinals of RLC circuit. Sol: u R (t) 6cos(5t - 6º) V, u L (t) 4cos(5t + 3º) V, u C (t) 3cos(5t - 5º) V, 6() / cos(5t - 5º) V 8. A resistor 5 Ω sized and a capacitor are connected in series. Voltage on resistor is u R (t) 5 cos(t + 3º) V. f voltage on terinals of RC dipole goes 6º behind current, which is the capacitance C of capacitor? Sol: C 57,7 µf 9. The voltage applied on terinals of a RLC series circuit goes 3º ahead the current. Aplitude of voltage on inductor is two ties the aplitude of voltage on capacitor, and u L (t) cost V. f R Ω find values of L and C. Sol: L 3, H, C 86,6 µf. Along circuit on picture flows an intensity of current i(t) () / cos(t+9º) A. f R Ω, L,5 H and C µf A i(t) R a) Copute power factor of RLC dipole. b) Copute the average power along a cycle on each basic dipole. C L Sol: a) cos ϕ, b) P Rav W, P Lav P Cav B. How does the power factor depend on resistance R, on inductance L and on capacitance C in a RLC series circuit? 3. Along a RL series circuit, having L,5 H, flows an intensity of current i(t)() / cos5t A. Voltage easured with a volteter on terinals of resistor is V R 5 V. Deterine: a) Magnitude of R b) on terinals of generator c) if a capacitor is added in series with R and L, copute the capacitance of this capacitor in order to get 3º of phase lag between voltage on generator and intensity of current. d) the new i(t) after the capacitor has been added. Sol: a) R 5 Ω b) v(t) cos(5t + 45º) V, c) C 89 µf, d) (t),45() / cos(5t + 5º) A 9-4

15 GLOSSARY Frequency is the nuber of cycles by second of sinusoidal function. nitial phase is the phase of sinusoidal function on tie t. Phase lag ϕ is the difference of phases between the initial phases of two sinusoidal functions, usually voltaje and intensity ϕ ϕ u ϕ Root ean square (rs): square root of average value of square of sinusoidal function along a cycle: rs rs pedance: Rate between aplitudes of drop of potential and intensity in a RLC dipole Z nductive reactance: Rate between aplitudes of difference of potential and intensity on an inductor. X L Lω Capacitive reactance: Rate between aplitudes of difference of potential and intensity on a capacitor. X C Active power: s the consued power by Joule heating in a RLC dipole P a cosϕ Power factor: cosϕ is the cosinus of phase lag. t only depends on devices of circuit and the angular frequency of current. i 9-5