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Alternating Current Short

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Brief note of the chapter AC

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    ALTERNATING CURRENT (SHORT) 2013-14 ALTERNATING CURRENT (A.c. CIRCUITS) Alternating current PAGE: 1 An electrical current, magnitude of which changes with time and polarity reverses periodically is called alternating current (A.C) The sinusoidal alternating current (a.c) is expressed as I = 1m sinot where 1m is the maximum value or peak value or amplitude of a.c. 211 2Ttf , where f is called frequency. In India f=50Hz. Sinusoidal e.m.f. of an a.c. source is given by e = em sin ot where em is the maximum value or peak value or amplitude of e.m.f. The pictorial symbol used to represent the a.c. source is shown in figure. Mean or average value of AC The average value of ac voltages and current over a complete cycle of AC is zero. Average value of alternating emf and current over a half cycle 21 respectively. ROOT MEAN SQUARE (RMS) OR VIRTUAL OR EFFECTIVE VALUE OF A.C. 28 m and are The average value of alternating current or emf over a cycle is zero. Therefore we take root mean square value Root mean square value of alternating current is the direct current which produces the same heating effect in a given resistor in a given time as is produced by the alternating current. It is denoted by I I or by I eff . rms ' v Relation between the effective value and peak value of a.c. 1 0.7071m 1 rms If not otherwise mentioned, the values of alternating voltages or currents quoted any where are virtual (r.m.s) values only. For example 220V a.c. means vrms -220v01t. An ac of IA means 1 ampere 1 rms Clearly r. m. s value of an alternating current is 70.7% of peak value. ROOT MEAN SQUARE VALUE OF AN ALTERNATING VOLTAGE It is defined as the steady voltage that produces the same heating effect in a given resistance in a given time as is produced by the given alternating emf. It is denoted by v rms or V eff or V v . it is give by, v rms On. The peak value of an a.c. supply is 300V. What is the rms voltage? (b) The rms value of current in an a.c circuit is IOA. What is the peak current? Ans. (a) Here v m — 300V . vrms = 0.707Vm = 0.707X300 - 212.1V 10 A (b) Here I rms 1m 1.414x10=14.14A A.C CIRCUIT CONTAINING ONLY A RESISTOR Consider a circuit containing a resistance 'R' alternating voltage. Let the applied voltage be V = vm sinot (1) If I be the current in the circuit at instant t, then the potential drop across R will be m. According to Kirchhoffs loop rule, connected to an ot (a)Graph of V and I versus ot Vmsin ot 1 ot (b)Phasor diagram
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    ALTERNATING CURRENT (SHORT) 2013-14 or 1= -Esinot 1m sinot or I (2) PAGE:2 where 1m —the maximum or peak value of a.c. From eqn(l) and (2), we can understand that the current and voltage are in same phae A.C. CIRCUIT CONTAINING ONLY AN INDUCTOR Consider a circuit containing an inductor of inductance 'L' connected to an alternating voltage. Let the applied voltage be (1) A back emf —L—is developed across the inductor. According to dt Kirchhoffs loop rule Vmsin(Dt-L O or L dt dt v m sin o t — sin o tdt or dl Integrating, f dl — f Y—. sin ot.dt (2) cosot V= Vrn sin ot sin(ot or I Where 1 — — It/ 2) = 1m cos (D t m = YL=the peak value of a.c. The term oLis called inductive reactance (XL ) . (a) Graph of V and I versus Inductive reactance (XL) Inductive reactance XL — OL — 2TtfL where f is the frequency of a.c. supply. reactance is ohm( Q I For a.c., For d.c., f=O, so XL 0 1 (b)Phasor diagram The SI unit of inductive Thus an inductor allows now of d.c through it easily but opposes the now of a.c. through it. On. A 44mH inductor is connected to 220V, 50Hz a.c. supply. Determine the rms value of current in the circuit. Graph of XL versus f.
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    ALTERNATING CURRENT (SHORT) 2013-14 A.C. CIRCUIT CONTAINING ONLY A CAPACITOR PAGE:3 Consider a circuit containing a capacitor of capacitance 'C' connected to alternating voltage. Let the applied voltage be V v m sin ot (1) At any instant, voltage V across the capacitor is V c According to Kirchhoffs loop rule q v m sin 0) t c or q C V m sin ot .. Current at any instant is dq d(CVm sin (D t) dt (D cvm dt or I = 1m cos at or I 1m + / 2) COS CD t (2) where I acv 1 —the current amplitude. 1/(DC —is called capacitive reactance (Xc ) . The term OC Capacitive reactance 1 Capacitive reactance X c 1 O C 2TtfC The S.I unit of capacitive reactance is ohm(Q l. For a.c., For d.c., f=O 1 f Thus a capacitor blocks d.c. Variation of capacitive reactance with frequency 1 1 Capacitive reactance, Xc O C 2TtfC 1 i.e., xcu- f Thus the capacitive reactance varies inversely with the frequency. As f increases, Xc decreases. Figure shows the variation of Xc with f. On. A 60 PIF capacitor is connected to a 110V, 60Hz a.c. supply. Determine the rms value of current in the circuit. c V= Vm sinot Graph of Xc versus f. On. While performing an experiment in laboratory, a student connected an electric bulb and a capacitor is series to a source of direct current (dc). Then will be the bulb glow steadily? Explain. On. What happens when (a) a capacitor is included in a dc circuit. (b) What type of current easily passes through an inductor? (c) Give an explanation for your answer. On. An electric bulb 'B' and a parallel plate capacitor 'C' are connected in series as shown in figure. The bulb glows with some brightness. How will the glow of the bulb affected on introducing a dielectric slab between the plates of the capacitor? Give reason to support yours answer. A. C. VOLTAGE APPLIED TO A SERIES LCR CIRCUIT Let an alternating voltage V v m sin ot applied to a circuit containing an inductance L, a capacitance C and resistance R connected in series. c V= c Vmsinot
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    ALTERNATING CURRENT (SHORT) 2013-14 According to Kirchhoffs loop rule q dt c The solution to this equation can be obtained by different methods. Phasor-diagram solution The phasor diagram for series LCR circuit to which an AC voltage V = Vrn sin Otis applied is shown below. The current in VLm each element is same. Voltage across the resistor is in phase with current. But the voltage across the inductor L leads the current by —radian and voltage across the capacitor C lags VCfiVLm behind the current by — . Let peak value of current in the circuit be 1m. Let V Rm, VLm and Vcm be the peak value of voltage across the resistor, inductor and capacitor respectively. The vector sum of VRm , Vcm and VLm gives the peak value Vm of the applied voltage. VLm)2 By Pythagorean theorem , V12n = VRm + Vcm Substituting the values of VRm , V cm and VLm , we have vm2 (1mR)2 + (1m Xc -1mxL)2 2] or, 1m 2 PAGE:4 VRm 1 Clearly, R 2 + (Xc — XL) is the effective resistance of the series LCR circuit to the now of ac. It is called impedence. It is denoted by Z and its S.I unit is ohm (Q l. 2 1 Thus, impedence Z = R2 + (Xc -XL)2 = R2 + Phase difference Let (b be the phase difference between the voltage and current, from figure we can write vcm V tan Special cases i. If Xc > XL , then current leads the voltage. ii. If XL < Xc , then the current lags the voltage. iii. If XL = Xc , then current is in phase with the voltage. RESONANCE The current amplitude in an LCR circuit is given by, 1m 2 1 oc 1 As angular frequency o of alternating emf is increased, goes on decreasing and oc 1 oLgoes on increasing. For a particular value of 0(= 00, say) — becomes equal to OL Then the impedence is minimum (Z R +0 = R) and the current amplitude becomes maximum — —l. This phenomenon is known as resonance and the angular frequency 00 is called natural or resonant angular frequency.
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    ALTERNATING CURRENT (SHORT) 2013-14 Determination of resonant frequency At resonant angular frequency 0 0 , c 1 or OOL ooc 1 LC 1 The resonant frequency fo 211 27t LC Characterestics of series resonant circuit PAGE:5 1. The current is in phase with voltage and power factor is unity (cos 4) =1 when 0) 2. The voltage across R is equal to the applied voltage. On. A sinusoidal voltage of peak value 283V and frequency 50Hz is applied to a series LCR circuit in which R=3Q , L=25.48mH, and C=796 PIF . Find the impedence of the circuit, (b) the phase difference between the voltage across the source and the currents, (c) the power dissipated in the circuit, and (d) the power factor. SHARPNESS OF RESONANCE: 9-Factor The sharpness of resonance is measured by a quantity called the quality or g-factor of the circuit. The g-factor of a series resonant circuit is defined as the ratio of the resonant frequency to the difference in two frequencies taken on both sides of the resonant frequency at which the current 1 amplitude becomes — times the value at resonant frequency. Mathematically g-factor can be expressed as Re sonant frequency 02 01 Band width 1 1 1 1 01 1 1 1 1 Bandwidth 1 where 01 and 02 are the frequencies at which the current falls to 1 — times its resonant value, as shown in figure. g-factor is a dimensionless quantity. Thus, g-factor of a series LCR-circuit may also be defined as the ratio of either the inductive reactance or the capacitive reactance at resonance to the resistance of the circuit. OOL (I/ooc) Bandwidth of a series resonant circuit R R When R is low, g is high and greater will be the sharpness of resonance. Tuning of a radio receiver The tuning circuit of a radio or TV is an example of LCR resonant circuit. Signals are transmitted by different stations at different frequencies. These frequencies are picked up by the antenna and corresponding to these frequencies, a number of voltages appear across the series LCR-circuit. But maximum current flows through the circuit for that a.c. voltage which has 1 frequency equal to fo . If the g-value of the circuit is LC large, the signals of the other stations will be very weak i.e., circuit will be more selective. By changing the value of the Antenna Series resonant circuit. To receiver adjustable capacitor C, signal from the desired station can be tuned in. AVERAGE POWER IN AN AC CIRCUIT Suppose in an a.c circuit, the voltage and current at any instant are given by V v m sin ot
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    ALTERNATING CURRENT (SHORT) 2013-14 1m sin(ot + d)) and I where is the phase angle by which current I leads the voltage V. The instantaneous power is given by P VI vmlm v m 1m v m 1m 2 sin(ot + (b) sin ot = 2 [cos(b cos(20t + [ 2sin AsinB cos(A B) PAGE:6 COS(A -k B) 2 The average power in the circuit over a complete cycle is 1 av or Pav or Pav f Pdt — cos(20t + v m 1m cos (bdt Icos(20t + (b)dt COS (bf dt 0 V I cos4) rms rms cos(20t + Special cases 1. Pure resistive circuit Here voltage and current are in same phase, i.e., 4) . Pav V .1 COSO-V 1 rms rms rms rms 2. Pure inductive circuit Here voltage leads the current in phase by — , i.e., 4) 2 •.Pav cos— rms rms 2 Thus the average power consumed in an inductive circuit over a complete cycle is zero. 3. Pure capacitive circuit Here voltage lags behind the current in phase by — ,i.e., 4) 2 •.pav cos rms rms 2 Thus the average power consumed in a capacitive circuit over a complete cycle is also zero. POWER FACTOR The average power of an a.c. circuit is given by coscb rms rms The product V I does not give the actual power and is called apparent power. The factor rms rms cos4) is called power factor of an a.c. circuit. . True power=Apparent power x Power factor Thus power factor may be defined as the ratio of true power to the apparent power of an a.c. circuit. Wattless current (Idle current) If the average power consumed in an a.c. circuit is zero, then the current in a.c. circuit is said to be wattless. This happens in the case of a pure inductor or capacitor. The current is called wattles because the current in the circuit does not do any work. AVERAGE POWER ASSOCIATED WITH A RESISTOR V Vmsinot and 1 1m sin ot Instantaneous power P VI — v ml m sin ot = (l cos 20t) 2
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    ALTERNATING CURRENT (SHORT) 2013-14 PAGE:7 Hence the average power consumed by the resistor over a complete cycle (i.e., from t=O to t=T) is 1 av v mlm vmlm 2 f Pdt — cos20 t)dt cos 20tdt = 0 v m 1m rms rms vmlm vmlm f (1 cos20t)dt rms o AVERAGE POWER ASSOCIATED WITH A PURE INDUCTOR V v m sin ot 1m sin ot and I 2 Instantaneous power, 1m cos ot P= VI = Vmlm sin ot cosot v m Im 2 sin ot cos ot 2 v mlm 2 sin 20t Average power over a complete cycle, 1 av v m 1m sin 20tdt 2 0 T 0 sin 20tdt 0 sin 20 tdt 0 =o Thus average power dissipated per cycle in an inductor is zero. AVERAGE POWER ASSOCIATED WITH A CAPACITOR cos 20 tdt o During the first of each quarter current cycle, as the current increases, the magnetic flux through the inductor builds up and energ is stored in the inductor from the external source. In the next quarter of cycle, as the current decreases, the flux decreases and the stored energy is returned to the source. Thus, in half cycle, no net power is consumed by the inductor. V Vm sin ot 1m sin ojt + and I 2 Instantaneous power P sin 20t 1m cos ot VI v m Im sin ot cos ot 2 Average 1 av power 1 Tvmlm over a complete cycle T 0 sin 20tdt 2 When the capacitor is connected across an a.c. source, it absorbs energy from the source for a quarter cycle as it is charged. It returns energy to the source in the next quarter cycle as it is discharged. Thus in a half cycle, no net power is consumed by the capacitor. sin 20 tdt sin 20tdt =O 0 Choke coil Choke coil is simply an inductor. Choke coil offers a reactance XL — 27tfLto the now of alternating current. Average power consumed by a choke coil is zero. Thus, a choke coil reduces current in an a.c. circuit without consuming any power. When an ohmic resistance is used, current reduces but energy loss occur due to heating. So choke is preferred. For d.c, f=O, so XL = O i.e., choke coil cannot limit direct current. On. A choke coil and a bulb are connected in series to an a.c. source. The bulb shines brightly. How does its brighmess change when an iron core is inserted in the choke coil? Ans. When the iron core is inserted in the choke coil, the self-inductance L increases. Consequently, the inducdve reactance, XL = COL increases. 'Ihis decreases the current in the circuit and bulb glows dimmer.
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    ALTERNATING CURRENT (SHORT) 2013-14 LC-OSCILLATIONS When a charged capacitor is allowed to discharge through a non-resisdve inductor, electrical oscillaüons of constant amplitude and frequency are produced. These oscillations are called LC-osci11ations. 1 The frequency of oscillaüon of charge or current f 211 LC However, the LC oscillations are usually damped due to the resisüve losses in the inductor and dielectric losses in the capacitor. TRANSFORMER: Transformer is a device used to convert low alternadng voltage at higher current into high alternating voltage at low current and vice-versa. in other words, a transformer is an electrical device used to increase or decrease altemaüng voltage. Construction It consists of two separate coils of insulated wire wound on same iron core. One of the coils connected to a.c. input is called primary(P) and the other winding giving output is called secondary (S). Theory PAGE:8 P s Load Core When an alternaüng source of emf e is connected to the primary coil, an alternaüng current flows through it. Due to the now of alternating current magneüc flux linked with the primary and secondary changes. This produces an emfin the secondary. Let NP and NS be the number of turns of primary and secondary coils repecüvely. The iron core is capable of coupling the whole of the magneüc flux (b produced by the turns of the primary coil with the secondary coil. According to Faraday's law of electromagnedc inducüon, the induced emfin the primary coil, (1) dt The induced emfin the secondary coil, (2) dt Dividing (2) by (1), we get In a If output voltage is less than the input voltage, the transformer is called step down transformer. step down transformer N s < N and e s < e If the output voltage is greater than the input voltage, the transformer is called step up transformer. In a step up transformer NS > NP and es > e For an ideal transformer(in which there are no energy losses), Output power=lnput power Output power For a transformer, emciency, Input power 81 For an ideal transformer, emciency, n is IOWo. But in a real transformer, the emciency varies from 90-996. This indicates that there are some energ losses in the transformer. Energy losses in transformers 1 . Copper loss. It is energy loss due to heating of the copper windings due their resistance. 2. Eddy current loss. It is the energy loss due to heating of the core by the eddy current. Ihis loss can be reduced by using laminated iron core. 3. Hysteresis loss. It is the energy loss due to the headng of the core due to the application of cyclic magneüä.ng field. It can be minimized by using core material having narrow hysteresis loop. 4. Flux leakage. The magnetic flux produced by the primary may not fully pass through the secondary. This loss can be minimized by winding the primary and secondary coils over one another.


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