Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

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1 AENAING CUEN PAC 7. Introduction : Q. What is direct current? Solution : Direct current does not change direction with tie. Q. What is alternating current? Solution : Alternating currents and voltages change its direction with tie. Q. What is the ain reason for preferring use of AC voltage over DC voltage? Solution : he ain reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted fro one voltage to the other by eans of transforers. Further, electrical energy can also be transitted econoically over long distances. 7. AC Voltage Applied to a esistor : Q. An ac voltage is applied to a resistor. Derive the expression for the current in the circuit. O Prove that the voltage and current are in the phase with each other in pure resistive circuit. Also plot the variation of current and voltage as a function of tie in this circuit. Solution : Figure shows a resistor connected to a source of ac voltage. et this potential difference, also called ac voltage, be given by v = v sin t, where v is the aplitude of the oscillating potential difference and is its angular frequency. o find the value of current through the resistor, we apply Kirchhoff s loop rule ( t), to the circuit shown in figure to get v sin t = i or v i sin t. Since is a constant, we can write this equation as i = i sin t, where the current aplitude i is given by v i. he voltage across a pure resistor and the current through it are plotted as a function of tie as shown in figure. which clearly indicate that the voltage and current are in phase with each other. Q. Find the average value of AC current over one coplete cycle? O Prove atheatically that the average value of a.c. over one coplete cycle is zero. Solution : he alternating current can be expressed as I = I sin t. Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

2 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 hen the average value of a.c. cover one coplete cycle will be I av I PAC I av I cost I sin t I [cos cos ] I I [cos cos] [ ] Q. Find the average value of AC current over one-half cycle. Solution : he alternating current can be expressed as I = I sin t. hen the average value of a.c. cover one-half of cycle will be I av / I / I / / av I cost I sint I [cos/ cos ] I [cos cos ] I [ ] I Q. As the average current over one coplete cycle is zero, does it ean that the average power consued is zero and there is no dissipation of electrical energy? Solution : he fact that the average current is zero, however, does not ean that the average power consued is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i and depends on i (which is always positive whether i is positive or negative) and not on i. hus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. Q. Find the average value of power in the resistor? Solution : he instantaneous power dissipated in the resistor is p = i = i sin t he average value of p over a cycle is p i i sin t where the bar over a letter (here, p) denotes its average value and <...> denotes taking average of the quantity inside the bracket. Since, i and are constants, p i sin we have, sin t. hus, p i. Q. Define the roo ean square current or effective current for ac? Also find the value of rs current? Solution : o express ac power in the sae for as dc power (P = I ), a special value of current is defined and used. It is called, root ean square (rs) or effective current and is denoted by I rs or I. t It is defined by I i i = i sin t ( i = i sint) sin t (by integration) Hence, I i i =.77 i

3 PAC 3 he rs current is the equivalent dc current that would produce the sae average power loss as the alternating current. Q. What is the physical eaning of rs current? Solution : he rs current is the equivalent dc current that would produce the sae average power loss as the alternating current. Q. Define the rs voltage or effective voltage for ac? Also find the value of rs voltage? Solution : he rs voltage or effective voltage is given by V v.77 v It is defined by V v v = v sin t ( v = v sint) sin t Hence, V v v =.77 v Q. What is the rs voltage if peak or axiu voltage equals to V for an ac? Solution : v = V = (.44)( V) = 3 V Q. A light bulb is rated at W for a V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rs current through the bulb. [NCE Solved Exaple 7.]. Solution : (a) 484 (b) 3 V (c).45 A 7.3 epresentation of AC Current and Voltage by otating Vectors - Phasors : Q. What is the use of phasor diagra? Solution : he analysis of an ac circuit is facilitated by the use of a phasor diagra. Q. What is phasor? Solution : A phasor is a vector which rotates about the origin with angular speed as shown in figure. he vertical coponents of phasors V and I represent the sinusoidally varying quantities v and i. he agnitudes of phasors V and I represent the aplitudes or the peak values v and i of these oscillating quantities. Figure shows the voltage and current phasors and their relationship at tie t for the case of an ac source connected to a resistor. he projection of voltage and current phasors on vertical axis are v sin t and i sint respectively represent the value of voltage and current at that instant. Q. Draw the phasor diagra for pure resistive ac circuit. Solution : Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

4 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 Q. What is the phase angle between the voltage and the current in pure resistive ac circuit? Solution : he phase angle between the voltage and the current in pure resistive ac circuit is zero. 7.4 AC Voltage Applied to an Inductor : PAC 4 Q. An ac voltage is applied to an ideal inductor (inductor with zero resistance). Derive the expression for the current in the circuit. Hence define inductive reactance. O Prove that the voltage across the inductor will lead by / w.r.t. current passing through it i.e., current lags the voltage by /. Draw the phasor diagra for this circuit. Also plot the variation of current and voltage as a function of tie. Solution : Figure shows an ac source connected to an inductor with negligible resistance. hus, the circuit is a purely inductive ac circuit. et the voltage across the source be v = v sin t. Using the Kirchhoff s loop rule, ( t), di v di v v, where is the self-inductance of the inductor. We have, sin t. v o obtain the current, we integrate : di sin( v t). and get, i cos( t) constant. he integration constant has the diension of current and is tie-independent. Since the source has an ef which oscillates syetrically about zero, the current it sustains also oscillates syetrically about zero, so that no constant or tie-independent coponent of the current exists. herefore, the integration constant is zero. Using v cos( t) sin t, we have i i sin t, where i is the aplitude of the current. he quantitiy is analogous to the resistance and is called inductive reactance, denoted by v X which is given by X =. he aplitude of the current is, then i. X Fro the coparison of v = v sin t and i i sin t the source voltage and the current in an inductor shows that the current lags the voltage by / or one-quarter (/4) cycle. Figure (a) shows the voltage and the current phasors in the present case at instant t. he current phasor I is / behind the voltage phasor V. Figure (b) shows the graph for the voltage and current versus t.

5 Q. What is the unit and diension of inductive reactance? PAC 5 Solution : he unit and diension of inductive reactance is the sae as that of resistance and its SI unit is oh. he diension of X is [M 3 A ]. Q. What is physical eaning of inductive reactance? Solution : he inductive reactance liits the current in a purely inductive circuit in the sae way as the resistance liits the current in a purely resistive circuit. Q. How does the inductive reactance depend on the inductance and frequency of current? Solution : he inductive reactance is directly proportional to the inductance and to the frequency of the current. Q. Plot a graph of inductive reactance versus angular frequency of ac for the given inductor (pure or ideal). Which quantity can be deterined fro the slope of this graph? Solution : he graph of inductive reactance versus angular frequency of ac is straight line passing through the origin. he slope (tan ) of this graph will give the value of inductance. Q. Plot a graph of inductive reactance versus self inductance () for the given ac source. Which quantity can be deterined fro the slope of this graph? he inductor is ideal. Solution : he graph of inductive reactance versus self inductance () for the given ac source is straight line passing through the origin. he slope (tan ) of this graph will give the value of angular frequency of ac source. Q. he graph of inductive reactance versus frequency of ac for the given inductor (pure or ideal) is as shown in figure. What is the value of self inductance? Solution : Fro X = = ()f, the slope tan equals to where = 45 fro the graph. Hence H. Q. Prove that the average power supplied to an inductor over one coplete cycle is zero. Solution : he instantaneous power supplied to the inductor is P i iv i sin t v sin( t) = i v cos(t)sin(t) v sin(t) Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

6 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 PAC 6 iv i So, the average power over a coplete cycle is P sin(t) = v sin(t) since the average of sin (t) over a coplete cycle is zero. Hence the average power supplied to an inductor over one coplete cycle is zero. Q. A pure inductor of 5. H is connected to a source of V. Find the inductive reactance and rs current in the circuit if the frequency of the source is 5 Hz. [NCE Solved Exaple 7.]. Solution : 7.85, 8A 7.5 AC Voltage Applied to a Capacitor : Q. An ac voltage is applied to a capacitor. Derive the expression for the current in the circuit. Hence define capacitive reactance. O Prove that the voltage across the capacitor will lag by / w.r.t. current passing through it i.e., current leads w.r.t. the voltage by / (the current is / ahead of voltage). Draw the phasor diagra for this circuit. Also plot the variation of current and voltage as a function of tie. Solution : Figure shows an ac source generating ac voltage v = v sin t connected to a capacitor only, a purely capacitive ac circuit. et q be the charge on the capacitor at any tie t. he instantaneous voltage v across the capacitor is equal, sin v q v. Fro the Kirchhoff s loop rule, the voltage across the source and the capacitor are C t q C. o find the current, we use the relation dq i. d i (v C sin t) = Cv cos(t). Using the relation, cos( t) sin t. We have i i sin t, where the aplitude of the oscillating current is i = Cv. We can rewrite it as i v. Coparing it to i = v / for a purely resistive circuit, we find that (/C) plays the role (/ C) of resistance. It is called capacitive reactance and is denoted by X c, X c = /C, so that the aplitude of the v current is i. X C Fro the coparison of v = v sin t and i i sin t the source voltage and the current in an capacitor shows that the voltage lags the voltage by / or one-quarter (/4) cycle i.e., the current is / ahead of voltage. Figure (a) shows the voltage and the current phasors in the present case at instant t. he current phasor I is / ahead the voltage phasor V. Figure (b) shows the graph for the voltage and current versus t.

7 PAC 7 Q. What is the unit and diension of capacitive reactance? Solution : he unit and diension of capacitive reactance is the sae as that of resistance and its SI unit is oh.he diensions of X C is [M 3 A ] Q. What is physical eaning of capacitive reactance? Solution : he capacitive reactance liits the current in a purely capacitive circuit in the sae way as the resistance liits the current in a purely resistive circuit. Q. How does the capacitive reactance depend on the capacitance and frequency of current? Solution : he capacitive reactance is inversely proportional to the capacitance and to the frequency of the current. Q. Plot a graph of capacitive reactance versus angular frequency of ac for the given capacitance. Solution : he graph of capacitive reactance versus angular frequency of ac is hyperbolic. Q. Plot a graph of capacitive reactance versus capacitance (C) for the given ac source. Solution : he graph of capacitive reactance versus capacitance for the given ac source is hyperbolic. Q. Plot a graph of capacitive reactance versus inverse of angular frequency of ac for the given capacitor. Which quantity can be deterined fro the slope of this graph? Solution : he graph of capacitive reactance versus inverse of angular frequency of ac is straight line passing through the origin. he slope (tan ) of this graph will give the value of inverse of capacitance C. Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

8 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 PAC 8 Q. Plot a graph of capacitive reactance versus inverse of capacitance for given ac source. Which quantity can be deterined fro the slope of this graph? Solution : he graph of capacitive reactance versus inverse of capacitance for the given ac source is straight line passing through the origin. he slope (tan ) of this graph will give the value of inverse of angular frequency of ac source. Q. he graph of capacitive reactance versus inverse of frequency of ac for the given capacitor is as shown in figure. Q. What is the value of capacitance? Solution : Fro X C = C, the slope tan equals to where = 3 fro the graph. Hence C C 3 unit. Q. Prove that the average power supplied to a capacitor over one coplete cycle is zero. Solution : he instantaneous power supplied to the capacitor is P i iv i sin t v sin( t) = i v cos(t)sin(t) v sin( t) i v i So, the average power over a coplete cycle is P sin( t) = v sin( t) since the average of sin (t) over a coplete cycle is zero. Hence the average power supplied to a capacitor over one coplete cycle is zero. Q. Capacitors block dc but allow ac to pass. Why. Solution : When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short tie required to charge the capacitor. As charge accuilates on the capacitor plates, the voltage across the increase, opposing the current. hat is, a capacitor in a dc circuit will liit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero. When the capacitor is connected to an ac source, it liits or regulates the current, but does not copletely prevent the flow of charge. he capacitor is alternately charged and discharged as the current reverses each half cycle. Q. A lap is connected in series with a capacitor. Predict you observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? [NCE Solved Exaple 7.3] Solution : When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lap will not glow. here will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (/C) and the current flows in the circuit. Consequently, the lap will shine. educing C will increase reactance and the lap will shine less brightly than before.

9 PAC 9 Q. A 5. µf capacitor is connected to a V, 5 Hz source. Find the capacitive reactance and the current (rs and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? [NCE Solved Exaple 7.4]. Solution :,.4A,.47 A Q. A light bulb and an open coil inductor are connected to an ac source through a key as shown in figure. he switch is closed and after soetie, an iron rod is inserted into the interior of the inductor. he glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons. [NCE Solved Exaple 7.5] Solution : As the iron is inserted, the agnetic field inside the coil agnetizes the iron increasing the agnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. herefore, the glow of the light bulb decreases. 7.6 AC Voltage Applied to a Series C Circuit : Q. Discuss an ac C series circuit and find the value of current flowing in the circuit using the phasor diagra. Define the ipedence. Also find the phase relationship of current to the applied alternating voltage. Discuss the condition for the circuit to be predoinantly capacitive and predoinantly inductive. Solution : Figure shows a series C circuit connected to an ac source. We take the voltage of the source to be v = v sin t. If q is the charge on the capacitor and i the current, at tie t, we have, fro Kirchhoff s loop di q rule : i v. We want to deterine the instantaneous current i and its phase relationship to the C applied alternating voltage v. Fro the circuit shown in figure, we see that the resistor, inductor and capacitor are in series. herefore, the ac current in each eleent is the sae at any tie, having the sae aplitude and phase. et it be i = i sin(t + ), where is the phase difference between the voltage across the source and the current in the circuit. We shall construct a phasor diagra for the present case. Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

10 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 PAC et I be the phasor representing the current in the circuit as given by equation i = i sin(t + ). Further, let V, V, V C and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. he length of these phasors or the aplitude of V, V C and V are v = i, V C = i X C, v = i X respectively. Since V C and V are always along the sae line and in opposite directions, they can be cobined which has a agnitude v C v. Since V is represented as the hypotenuse of a right-triangle whose sides are v and v C v, the pythagorean theore gives : v v (vc v ). Substituting the values of v, v C, and v we have, v = (i ) + (i X C i X ) = i [ + (X C X ) ] or, i v (X C X ). By analogy to the resistance in a circuit, we introduce the ipendance Z in v an ac circuit : i Z, where Z (XC X ). Since phasor I is always parallel to phasor V, the phase angle is the angle between V and V and can be deterine fro figure. vc v XC X tan tan. v If X C > X, is positive and the circuit is predoinantly capacitive. Consequently, the current in the circuit leads the source voltage. If X C < X, is negative and the circuit is predoinantly inductive. Consequently, the current in the circuit lags the source voltage. Q. What is ipedence diagra? Solution : Ipedence diagra is a right triangle with ipedence Z as its hypotenuse and it is shown in the figure. Q. Draw the phasor diagra and variation of voltage and current with t for series C circuit when X c > X. Solution :

11 PAC Q. Find the expression for current flowing in the series C circuit using the analytical solution. Also find the phase relationship of current with the voltage. Solution : he voltage equation for the circuit is di q i v = v C sin t We know that i = dq/. herefore, di/ = d q/. hus, in ters of q, the voltage equation becoes d q dq q C v sin t...(i) his is like the equation for a forced, daped oscillator. et us assue a solution q = q sin (t + ) dq d q so that qcos( t ) and q sin( t ) Substituting these values in equation (i), we get q sin( t ) q sin( t ) q cos( t ) v C By using the relation X c = /C, X =, we get q [cos(t + ) + (X C X ) sin (t + )] = v sin t sin t...(ii) Multiplying and dividing the equation (ii) by Z (X c X ), we have q (X Z cos( t ) Z C X Z ) sin( t ) v sin t...(iii) Now, let cos Z and ( XC X ) sin so that tan Z XC X Substituting this in equation (iii) and siplifying, we get q Zcos(t + ) = v sin t Coparing the two sides of this equation, we see that v = q Z = i Z where i = q and or. herefore, the current in the circuit is dq i qcos( t ) i cos( t ) or i i sin( t ) where i v v and tan Z (X X ) C XC X Q. Discuss the phenoena of resonance in C series AC circuit. O Define resonant frequency and derive the expression for resonance frequency in C series AC circuit. Solution : In the phenoena of resonance in the series C circuit, the current aplitude will have a axiu value at a certain frequency of AC source. his frequency is known as natural frequency or resonance frequency. For an C circuit driven with voltage of aplitude v and frequency, we found that the current aplitude is given by i v v with X Z C = /C and X =. So if is varied, then at a (X X ) C particular frequency, X C = X, and the ipedance is iniu (Z ). his frequency is. called the resonant frequency and it is given by : X or or C C. Xc Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

12 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 At resonant frequency, the current aplitude is axiu; i = v /. PAC Q. Plot the graph for the variation of current aplitude (i ) with angular frequency of ac source in C series ac circuit. Also ention the point on the graph at which resonant frequency will occur. Solution : he graph for the variation of current aplitude (i ) with angular frequency of ac source in C series ac circuit as shown in figure. he point on the graph at which resonant frequency will occur is given by at which the current aplitude will have axiu value v /. Q. Consider the C series ac circuit with =. H, C =. nf for two values of (i) = and =. he source peak voltage is V. (i) Find the resonant freqency and axiu value of current aplitude in each case. (ii) Plot the graph for the variation of current aplitude (i ) with angular frequency of ac source in C series ac circuit. Solution : (i) he resonant frequency is given by and the axiu value of current aplitude C equals to v /. For the given values of, C, and v in the proble, the value of =. 6 rad/s for both values of. he axiu value of current aplitude equals to.5a for = and.a for =. (ii) Graph (i) is for resistance and graph (ii) is for resistance. Q. Draw the phasor diagra for series C ac circuit at resonance. What is the phase difference between current and voltage for series C ac circuit at resonance? Solution : he phasor diagra for series C ac circuit at resonance is as shown in figure. he phase difference between current and voltage for series C ac circuit at resonance is zero i.e., current and voltage are in the sae phase Q. What is the application of resonant C circuits for ac source? O Explain the application of resonant circuit in the tuning echanis of a radio or a V set?

13 PAC 3 Solution : esonant circuits have a variety of applications, for exaple, in the tuning echanis of a radio or a V set. he antenna of a radio accepts signals fro any broadcasting stations. he signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at any frequencies. But to heat one particular radio syste, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant fequency of the circuit becoes nearly equal to the frequency of the radio signal received. When this happens, the aplitude of the current with the frequency of the signal of the particular radio station in the circuit is axiu. Q. Why cannot have resonance in a or C circuit? O Why resonance phenoenon is exhibited by a circuit only if both and C are present in the circuit? Solution : he resonance phenoenon is exhibited by a circuit only if both and C are present in the circuit. Only then do the voltages across and C cancel each other (both being out of phase) and the current aplitude is v /, the total source voltage appearing across. his eans that we cannot have resonance in a or C circuit. Q. What is the power dissipated by series C circuit if the current aplitude is / ties the axiu value of current aplitude? et the power dissipation at the axiu value of current aplitude is P. Solution : he power dissipated by the circuit becoes P/ if the current aplitude is / ties the axiu value of current aplitude. Q. (a) What is bandwih of the circuit and how it is related with the sharpness of resonance? What is sharpness of resonance? O (b) Derive the expression of bandwih of the circuit and the sharpness of resonance and hence define the quality factor? (c) What is the physical eaning of quality factor? Solution : (a) For values of other than, the aplitude of the current is less than the axiu value. Suppose we choose a value of for which the current aplitude is / ties its axiu value. Fro the curve in figure, we see that there are two such values of say, and, one greater and the other saller than and syetrical about. We ay write = + = he difference = is often called the bandwih of the circuit. he quantity ( / ) is regarded as a easure of the sharpness of resonance. he saller the, the sharper or narrower is the resonance. (b) o get an expression for, we note that the current aplitude i is ax ( / )i (the axiu value of i equals to v /) for = +. herefore, at, i v = C i ax v Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

14 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 PAC 4 C C C which ay be written as, ( ) ( )C C Using C in the second ter on the left hand side, we get We can approxiate as since. herefore, Hence the bandwih of the circuit equals to. he sharpness of resonance is given by, he ratio is also called the quality factor, Q of the circuit. (c) Fro equation, we see that. So, larger the value of Q, the saller is the value of or the Q bandwih and sharper is the resonance. Using = /C, equation can be equivalently expressed as Q = / C. If the resonance is less sharp, not only is the axiu current less, the circuit is close to resonance for a larger range of frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit or vice versa. If quality factor is large, i.e., is low or is large, the circuit is ore selective. Q. A resistor of and a capacitor of 5. µf are connected in series to a V, 5 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rs) across the resistor and the capacitor. Is the algebraic su of these voltages ore than the source voltage? If yes, resolve the paradox. [NCE Solved Exaple 7.6] Solution : (a).775 A (b) 5 V, 6.3 V 7.7 Power in AC Circuit : he Power Factor : Q. Find the average power over a cycle in ac circuit? Solution : he voltage v = v sin t is applied to ac circuit drives a current in the circuit given by i = i sin (t + ). herefore, the instantaneous power p supplied by the source is v p = vi = (v sin t) [i sin(t + )] i [cos cos(t )]. [ sinasinb = cos (A B) cos (A + B)] he average power over a cycle is given by the average of the two ters in.h.s. of above equation. It is only the second ter which is tie-dependent. Its average is zero (the positive half of the cosine cancels the v i v i negative half). herefore, P cos cos = V I cos.

15 his can also be written as, P = I Z cos. Q. Define the power factor? PAC 5 Solution : he average power in an ac circuit is given by V I cos. he average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle between the. he quantity cos is called the power factor. Q. Find the power factor for the following circuit (i) resistive circuit (ii) purely inductive or purely capacitive circuit (iii) C series circuit (iv) C series circuit at resonance? Solution : (i) esistive circuit : If the circuit contains only pure, it is called resistive. In that case =, cos =. here is axiu power dissipation. (ii) Purely inductive or capacitance circuit : If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage ande current is /. herefore, cos =, and no power is dissipated even through a current is flowing in the circuit. his current is soeties referred to as wattless current. (iii) C series circuit : In an C series circuit, power dissipated is given by I Z cos, where = tan (X C X )/. So, ay be non-zero in a or C or C circuit. Even in such cases, power is dissipated only in the resistor. (iv) Power dissipated at resonance in C circuit : At resonance X C X =, and =. herefore, cos = and P = I Z = I. hat is, axiu power is dissipated in a circuit (through ) at resonance. Q. How can you obtain wattless current in an AC circuit? O What is wattless current? Solution : If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage ande current is /. herefore, cos =, and no power is dissipated even through a current is flowing in the circuit. his current is soeties referred to as wattless current. Q. Nae device through which power consued an AC circuit is zero. Solution : Pure inductor or pure capacitive circuit. Q. What is the power dissipation in an a.c. circuit in which voltage and current are given by : V = 3 sin (t + /) and I = 5 sin t? Solution : Zero Q. (a) For circuits used for transporting electric power, a low power factor iplies large power loss in transission. Explain. (b) Power factor can often be iproved by the use of a capacitor of appropriate capacitance in the circuit. Explain. [NCE Solved Exaple 7.7] Solution : (a) We know that P = I V cos where cos is the power factor. o supply a given power at a given voltage. If cos is sall, we have to increase current accordingly. But this will lead to large power loss (I r) in transission. (b) Suppose in a circuit, current I lags the voltage by an angle. hen power factor cos = /Z. We can iprove the power factor (tending to ) by aking Z tend to. et us understand, with the help of a phasor diagra, how this can be achieved. et us resolve I into two coponents. I p along the applied voltage V and I q perpendicular to the applied voltage. I q as you have learnt in Section 7.7, is called the wattless coponent since corresponding to this coponent of current, there is no power loss. I p is known as the power coponent because it is in phase with the voltage and corresponds to power loss in the circuit. It s clear fro this analysis that if we want to iprove power factor, we ust copletely neutralize the lagging wattless current I q by an equal leading wattless current I q. his can be done by connecting a Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

16 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 PAC 6 capacitor of appropriate value in parallel so that I q and I q cancel each other and P is effectively I p V. Q. A sinusoidal voltage of peak value 83 V and frequency 5 Hz is applied to a series C circuit in which = 3, = 5.48 H, and C = 796 µf. Find (a) the ipedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor [NCE Solved Exaple 7.8]. Solution : (a) 5 (b) 53.. Since is negative, the current in the circuit lags the voltage across the source. (c) 48 W (d).6 Q. Suppose the frequeny of the source in the previous exaple can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the ipendence, the current, and the power dissipated at the resonant condition. [NCE Solved Exaple 7.9] Solution : (a) 35.4 Hz (b) 3, 66.7 A, 3.35 kw Q. At an airport, a person is ade to walk through the doorway of a etal detector, for security reasons. If she/he is carrying anything ade of etal, the etal detector eits a sound. On what principle does this detector work? [NCE Solved Exaple 7.] Solution : he etal detector works on the principle of resonance in ac circuits. When you walk through a etal detector, you are, in fact, walking through a coil of any turns. he coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with etal in your pocket, the ipendance of the circuit changes resulting in significant change in current in the circuit. his change in current is detected and the electronic circuitry causes a sound to be eitted as an alar. 7.8 C Oscillations : Q. (a) What is C oscillations? (b) Derive the expression for the frequency of C oscillations. Also find the charge on the capacitor at any tie t and current in the C circuit? Solution : (a) We know that a capacitor and an inductor can store electrical and agnetic energy, respectively. When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenoenon of electrical oscillations siilar to oscillations in echanical systes.his is known as C oscillation. (b) et a capacitor be charged q (at t = ) and connected an inductor as shown in figure. he oent the circuit is copleted, the charge on the capacitor starts decreasing, giving rise to current in the circuit. et q and i be the charge and current in the circuit at tie t. According to Kirchhoff s loop rule, q di. C i = (dq/) in the present case (as q decreases, i increases). herefore the equation becoes : d q q. C d x his equation has the for x for a siple haronic oscillator. he charge, therefore, oscillates with a natural frequency C and varies sinusoidally with tie as q = q cos ( t + )

17 PAC 7 where q is the axiu value of q and is a phase constant. Since q = q at t =, we have cos = or =. herefore, in the present case, q = q cos( t). dq he current i is given by i = i sin ( t) where i = q. Q. Show that in the free oscillations of an C circuit, the su of energies stored in the capacitor and the inductor is constant in tie. [NCE Solved Exaple 7.] q Solution :, his su is constant in tie as q C and C, both are tie-independent. Note that it is equal to the initial energy of the capacitor. 7.9 ransforers : Q. (a) What is transforers? On which principle does it work? (b) Explain with the help of a labelled diagra, the construction and working of a transforer (step of transforer or step down transforer). Also write down the assuptions. (c) If the transforer is assued to be % afficient then find the relation of priary and secondary currents with nuber of ters in priary coil and secondary coil. Solution : (a) For any purposes, it is necessary to change (or transfor) an alternating voltage fro one to another of greater or saller value. his is done with a device called transforer using the principle of utual induction. (b) A transforer consists of two sets of coils, insulated fro each other. hey are wound on a soft-iron core, either one on top of the other as in figure (a) or on separate libs of the core as in figure (b). One of the coils called the priary coil has N p turns. he other coil is called the secondary coil; it has N s turns. Often the priary coil is the input coil and the secondary coil is the output coil of the transforer. When an alternating voltage is applied to the priary, the resulting current produces an alternating agnetic flux which links the secondary and induced an ef in it. he value of this ef depends on the nuber of turns in the secondary. We consider an ideal transforer in which the priary has negligible resistance and all the flux in the core links both priary and secondary windings. et be the flux in each turn in the core at tie t due to current in the priary when a voltage v p is applied to it. hen the induced ef or voltage s, in the secondary with N s turns is d s N s. d he alternating flux also induces an ef, called back ef in the priary. his is p N p. But p = v p. If this were not so, the priary current would be inifinite since the priary has zero resistance (as assued). If the secondary is an open circuit or the current taken fro it is sall, then to a good approxiation s = v s, where v s is the voltage across the secondary. herefore, equations can be written as v s d Ns, v N p p d Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

18 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 Fro above equations, we have v v s p N N s p PAC 8 Note that the above relation has been obtained using three assuptions : (i) the priary resistance and current are sall; (ii) the sae flux links both the priary and the secondary as very little flux escapes fro the core, and (iii) the secondary current is sall. (c) If the transforer is assued to be % efficient (no energy losses), the power input is equal to the power output, and since p = iv, i p v p = i s v s Cobining equations we have i i p s v N s s. vp Np Q. What is transforer ratio and hence define step up transforer and step down transforer? What is the value of transforer ratio of step up transforer and step down transforer? Solution : We have, Ns Vs Np V p and I s N N N s /N p is defined as the transforer ratio. p s I p If the secondary coil has a greater nuber of turns than the priary (N s > N p ), the voltage is stepped up (V s > V p ). his type of arrangeent is called a step-up transforer. However, in this arrangeent, there is less current in the secondary than in the priary (N p /N s < and I s < I p ). he value of transforer ratio is greater than one for step up transforer. If the secondary coil has less turns than priary (N s < N p ), we have a step-down tranforer. In this case, V s < V p and I s > I p. hat is, the voltage is stepped down, or reduced, and the current is increased. he value of transforer ratio is less than one for step up transforer. Q. What are different energy losses in actual transforer and how are they iniised? Solution : he different energy losses in actual transforers are following : (i) Flux eakage : here is always soe flux leakage; that is, not all of the flux due to priary passes through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the priary and secondary coils one over the other. (ii) esistance of the windings : he wire used for the windings has soe resistance and so, energy is lost due to heat produced in the wire (I ). In high current, low voltage windings, these are iniised by using thick wire. (iii) Eddy currents : he alternating agnetic flux induces eddy currents in the iron core and causes heating. he effect is reduced by having a lainated core. (iv) Hysteresis : he agnetisation of the core is repeatedly reversed by the alternating agnetic field. he resulting expenditure of energy in the core appears as heat and is kept to a iniu by using a agnetic aterial which has a low hysteresis loss. Q. How the transforer is used for the transission and distribution of electrical energy? he large scale transission and distribution of electrical energy over long distances is done with the use of transforers. he voltage output of the generator is stepped-up (so that current is reduced and consequently, the I loss is cut down). It then transitted over long distances to an area sub-station near the consuers. here the voltage is stepped down. It is further stepped down at distributing sub-stations and unility poles before a power supply of 4 V reaches our hoes.

19 PAC 9 NCE EXECISE 7. A resistor is connected to a V, 5 Hz a.c. supply. (a) What is the rs value of current in the circuit? (b) What is the net power consued over a full cycle? 7. (a) he peak voltage of an a.c. supply is 3 V. What is the rs voltage? (b) he rs value of current in an a.c. circuit is A. What is the peak current? 7.3 A 44 H inductor is connected to V, 5 Hz a.c. supply. Deterine the rs value of the current in the circuit. 7.4 A 6 µf capacitor is connected to a V, 6 Hz a.c. supply. Deterine the rs value of the current in the circuit. 7.5 In exercise 7.3 and 7.4, what is the net power absorbed by each circuit over a coplete cycle. Explain your answer. 7.6 Obtain the resonant frequency r of a series C circuit with =. H, C = 3 µf and =. What is the Q-value of this circuit? 7.7 A charged 3 µf capacitor is connected to a 7 H inductor. What is the angular frequency of free oscillations of the circuit? 7.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 C. What is the total energy stored in the circuit initially? What is the total energy at later tie? 7.9 A sereis C circuit with =, =.5 H and C = 35 µf is connected to a variable-frequency V a.c. supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one coplete cycle? 7. A radio can tune over the frequency range of a portion of MW broadcast band : (8 khz to khz). If its C circuit has an effective inductance of µh, what ust be the range of its variable capacitor? 7. Figure shows a series C circuit connected to a variable frequency 3 V source. = 5. H, C = 8µF, = 4. (a) (b) (c) Deterine the source frequency which drives the circuit in resonance. Obtain the ipendance of the circuit and the aplitude of current at the resonating frequency. Deterine the rs potential drops across the three eleents of the circuit. Show that the potential drop across the C cobination is zero at the resonating frequency. 7. An C circuit contains a H inductor and a 5 µf capacitor with an initial charge of C. he resistance of the circuit is negligible. et the instant, the circuit is closed be t =. (a) What is the total energy stored initially? Is it conserved during C oscillations? (b) What is the natural frequency of the circuit? (c) At what tie is the energy stored (i) copletely electrical (i.e., stored in the capacitor)? (ii) copletely agnetic (i.e., stored in the inductor)? (d) At what ties is the total energy shared equally between the inductor and the capacitor? (e) If a resistor is inserted in the circuit, how any energy is eventually dissipated as heat? Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

20 7.3 A coil of inductance.5 H and resistance is connected to a 4 V, 5 Hz a.c. supply. (a) What is the axiu current in the coil? (b) What is the tie lag between the voltage axiu and the current axiu? PAC 7.4 Obtain the answers (a) to (b) exercise 7.3 if the circuit is connected to a high frequency supply (4 V, khz). Hence, explain the stateent that a very high frequency, an inductor in a circuit nearly aounts to an open circuit. How does an inductor bechave in a d.c. circuit after the steady state? 7.5 A µf capacitor in series with a 4 resistance is connected to a V, 6 Hz supply. (a) What is the axiu current in the circuit? (b) What is the tie lag between the current axiu and the voltage axiu? 7.6 Obtain the answers to (a) and (b) above in the proble if the circuit is connected to a V, khz supply? Hence. explain the stateent that a capacitor is a conductor at very high frequencies. Copare this behaviour with that of capacitor in a d.c. circuit after the steady state. 7.7 Keeping the source frequency equal to the resonating frequency of the series C circuit, if the three eleents, C and are arranged in parallel, show that the total current in the parallel C circuit is iniu at this frequency. Obtain the current rs value in each branch of the circuit for the eleents and source specified in Exercise 7. for this frequency. 7.8 A circuit containing a 8 H inductor and a 6 µf capacitor in series is connected to a 3 V, 5 Hz supply. he resistance of the circuit is negligible. (a) (b) Obtain the current aplitude and rs values. Obtain the rs values of potential drops across each eleent. (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? 7.9 Suppose the circuit in exercise 7.8 has a resistance of 5. Obtain the average power transferred to each eleent of the circuit and the total power absorbed. [Answers : (7.) (a). A (b) 484 W (7.8) (a).64 A (b) V (c) (d) / (e) (7.9) 79.6 W] 7. A series C circuit with =. H, C = 48 nf, = 3 is connected to a 3 V variable frequency supply. (a) (b) (c) What is the source frequency for which current aplitude is axiu? Obtain this axiu value. What is the source frequency for which average power absorbed by the circuit is axiu? Obtain the value of this axiu power. For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current aplitude of these frequencies? (d) What is the Q-factor of the given circuit? 7. Obtain the resonant frequency and Q-factor of a series C circuit with = 3. H, C = 7 µf and = 7.4. It is desired to iprove the sharpness of the resonance of the circuit by reducing its full wih at half axiu by a factor of. Suggest a suitable way. 7. Answer the following questions : (a) (b) (c) (d) In any ac circuit, is the applied instantaneous voltage equal to the angebraic su of the instantaneous voltages across the series eleents of the circuit? Is the sae true for rs voltage? A capacitor is used in the priary instantaneous circuit of an induction coil. An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. he circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across. A choke coil in series with a lap is connected to a dc line. he lap is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lap s brightness. Predict the corresponding observations if the connection is to an ac line. Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

21 (e) PAC Why is choke coil needed in the use of fluorescent tubes with ac ains? Why can we not use an ordinary resistor instead of the choke coil? 7.3 A power transission line feeds input power at 3 V to a step-down transforer with its priary windings having 4 turns. What should be the nuber of turns in the secondary in order to get output power at 3 V? 7.4 At a hydroelectric power plant, the water pressure head is at a height of 3 and the water flow available is 3 s. If the turbine generator effieiency is 6%, estiate the electric power available fro the plant (g = 9.8 s ). 7.5 A sall town with a deand of 8 kw of electric power at V is situated 5 k away fro an electric plant generating power at 44 V. he resistance of the two wire line carrying power is.5 per k. he town gets power fro the line through a 4 - V step down transforer at a sub-station in the town. (i) (ii) (iii) Estiate the line power loss in the for of heat. How uch power ust the plant supply, assuing there is negligible power loss due to leakage? Characterise the step up transforer at the plant. 7.6 Do the sae exercise as above with the replaceent of the earlier transforer by a 4,- V step down transforer (Neglect, as before, leakage losses through this ay not be a good assuption any longer because of the very high voltage transission involved). Hence, explain, why high voltage transission is preferred? Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857

22 Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857 PAC ANSWES 7. (a). A (b) 484 W 7. (a). V (b) 4. A A A 7.5 Zero in each case s J, sae at later ties 7.9 kw 7. he variable capacitor should have a range of about 88 pf to 98 pf. 7. (a) 5 rad s (b) 4, 8. A (c) 7. (a). J. Yes, su of the energies stored in and C is conserved if = (b) = 3 rad s, v = 59 Hz (c) q = q cos t 3 (i) Energy stored is copletely electrical at t,,,, (ii) Energy stored is copletely agnetic (i.e., electrical energy is zero) at t,,... where s. v 3 5 (d) At t,,,..., because q q cos q cos 8 4 q. herefore, electrical energy q = q which is half the total energy. C C (e) daps out the C oscillations eventually. he whole of the initial energy (=. J) is eventually dissipiated as heat. 7.3 (a).8 A (b) 3.94 s s 7.5 (a) 3.4 A (b).55 3 s 7.6 (a) 3.89 A 7.7 In branch, I rs = 5.75 A In branch, I rs =.9 A In C branch, I Crs =.9 A 7.8 (a) I =.6 A, I rs = 8.4 A (b) V rs = 7 V, V Crs = 437 V (c) zero (d) zero (e) zero 7.9 I rs = 7.6 A Average power to = I rs = 79 W Average power to = Average power to C = otal power absorbed = 79 W 7. (a) 4.4 A (b) 3 W (c) A (d).7 7. = rad s ; Q = 45 o double Q without changing, reduce to 3.7.

23 PAC 3 7. (a) Yes. he sae is not true for rs voltage, because voltages across different eleent ay not be in phase. See, for exaple, answer to Exercise (b) he high induced voltage, when the circuit is broken, is used to charge the capacitor, thus avoiding sparks, etc. (c) For dc, ipedance of is negligible and of C very high (infinite), so the dc signal appears across C. For high frequency ac, ipendence of is high and that of C is low. So, the ac signal sppears across. (d) For a steady state dc, has no effect, even if it is increased by an iron core. For ac, the lap will shine dily because of additional ipedance of the choke. It will di further when the iron core is inserted which increases the choke s ipendance. (e) A choke coil reduces voltage across the tube without wasting power. A resistor would waste power as heat MW 7.5 (i) 6 kw (ii) 4 kw (iii) 7 V 7.6 (i) 6 kw (ii) 86 kw (iii) 4,3 V Einstein Classes, Unit No., 3, Vardhan ing oad Plaza, Vikas Puri Extn., Outer ing oad New Delhi 8, Ph. : , 857