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SS —, maa, , ample l. The ., Examp han of alternating current is 5¥2 ampere. Calculate the root mean square value, , Solution. Given : Ig= 5¥2 A, , , , Root mean square value of current 1, = 10. 572, , sr = TS =5A,, rms “Ty 5, Example 2. The reading of a.c. voltmeter is 200 volt. fe, , . What will be the peak value of potential difference ?, Solution. The a.c. voltmeter Measures the root mean square value, therefore Vyms = 200 volt (given)., But Vo, Vins = Re, , oF _ Peak value Vp = Ving V2 = 200 /2 = 282-8 volt,, Example 3. Jn an alternating circuit, the instantaneous current is 1=6 sin 314 t. Calculate : (i) peak value, , . ( or amplitude) of current, (ii) frequency of current and (iii) root mean square value of current., Solution. Given: 1=6 sin 3141, , Comparing with eqn. 1 = Ty sin wt, we get Ig = 6, o = 314, (i) Peak value (or amplitude) of current = Ip=6A., , (ii) Frequency of current f= Da = oS =50 Hz., , (ij) rm.s. value of current Inns = 4 = * =4-242 A., , Example 4. Find the instantaneous value of alternating voltage of 220 volt —50 hertz., Solution. Given : Vn = 220 volt, f= 50 Hz, , Peak value of voltage Vp = Ving x ¥2 =220x /2 =311 volt, -. Instantaneous value of voltage V = Vo sin 2nft, or V=311 sin2 x 3-14 x S0¢ or V=311 sin 3144, Example 5. For a 100 W, 220 V calculate : (i) The resistance of filament, (ii) the peak voltage of AC, source, (iii) the rm.s. value of current flowing through bulb., Solution. Given : P= 100 W, V=220V. ¥, , 4, , , , Vv? _ (220)?, @ The resistance of filament of bulb R = P= 007 = 484.9., (ii) Peak voltage of AC source V) = V 72 =220 72 =311V., , P 100, (iii) r.m.s. value of current through the bulb I= V7 20" 0-45 A., , E ; 6. Resistance, Reactance and Impedance, , Resistance—The resistance offered by a conductor in the flow of current in an alternating circuit, is, called the resistance of that conductor. Its unit is ohm (Q). It is equal to the ohmic resistance of conductor, (ie, the resistance in d.c. circuit). oo oe, , Reactance—The effective resistance of the alternating circuit containing a single component (such as, only the inductive coil L, or only the condenser C), is called the reactance. It is denoted by the letter X. Its, , Unit is ohm (Q)., , , , , , , , , , , , Scanned with CamScanner

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Alternating Current, , , , or _ No ‘ 4), I oLsin (or 5, 2 1=lIpsin (o-4) (7.18), , where Ip (peak value of current) = Vo/wL or Vo/Iy = wL, Lo eels oem ale » iti in circuit i lied potential, difference in phase by 1/2. (7.18), it is clear that the current in circuit lags behind the applied p', , Fig. 7.12 represents the variat, , : : ion in potential difference V and current I with time ¢. Fig 7.13 represents, the phasor diagram which shows th, , e phase difference between the voltage V and current I., , , , , , Fig. 7.12. Variation in voltage and current with, , Fig. 7.13. Phase difference between V and I, time in an a. c. circuit with inductance alone, , in a. c. circuit with inductance alone, The quantity V/I or Vo/Ip is called inductive reactance and it is represented by the letter X,. Thus, , Inductive reactance X, = % = @L = 2nfL -(7.19), Its unit is ohm (symbol Q). 9, , Thus an inductance in an a.c. circuit behaves like an ohmic resistance X, where X;, depends on, frequency of a.c. sources. For direct current since frequency f= 0, or angular frequency « = 0, the inductive, reactance X, is zero, i.e., an inductance allows the d.c. to pass through it easily., , , , , , , , , , , , Solved Examples :, , Example 1. Calculate the reactance of a condenser of capacity 5 uF for a.c. of frequency 108 Hz., Solution. Given : f= 106 Hz, «. @ = 2nf=2 « 3-14 x 106 = 6-28 x 106 rd s|,C=5pF=5x 10°F, , 1 1, Reactance of condenser X¢ oc (628% 105) (5% 10°} 0-0318 ohm., Example 2. The self inductance of a coil is 14 mH. An alternating source of 220 V and of frequency, 100 Hz is joined across it. Calculate the reactance of the coil and current t flowing in it., Solution. Given: L= 14 mH= 14 x 103 H, f= 100 Hz «. @ = 2nf=2 x 3-14 x 100=628 rds,V=220V, Reactance of coil X, = wL = 628 x (14 x 10-3) = 8-8 ohm, Current in the coil I = = = 2 A=25A., , Example 3. A condenser of capacity 15-0 F is joined to an alternating source of 220 volt and 50 Hz,, (a) Calculate : (i) capacitive reactance, (ii) peak and root mean Square va, in circuit. (b) If the frequency of alternating source in the circuit is do, reactance and current affected ?, , Solution. Given : C = 15-0 pF = 15-0 * 10-6 F, V,n5 = 220 volt, f= 50 Hz, oO = 2nf=2 x 3-14 x 50=314 rd st,, , ue of current, ubled, how will the, , 1 1, A Capacitive reactanceXp=—~=——__|, (a) (i) Pp’ c- OG 314*(15-0x10 212-3 ohm., , Scanned with CamScanner

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Shivalal Physics : Class 12, , , , Vins . 220. 1.04, , X. 2123, , (ii) Root mean square value of current In. = Ke, , Peak value of current Iy = 72 Sins = y2 108 =1-47A. nic, (b) If the frequency of alternating source in doubled, the capacitive reactance will reduce to, , half (= aes = 106-15 Q) and current in circuit will be doubled, /.2., now Ims = 2% 1-044, = 2-08 A and Ip =2 x 1-47 =2:94 A., ed to an ac. source of frequency, , Example 4. 4 condenser of capacity 1 picofarad is connect, 50 Hzand 220 volt. Calculate ; (i) reactance of capacitance, (11) cu, and (iii) the phase difference between the current and potential difference., , Solution. Given : V = 220 volt, f = 50 Hz, C= 1 pF = 10-1? F, 1 1, , (i) Reactance of capacitance X¢ = oC 2nfC, , (ii) current flowing in the circuit,, , , , ——___1 ___, =3:18« 10°., 2% 3-14 x50x(1x10-*), , Vv 220, ii ing i ircuit | = —-=————y = 6-92 x 108A., (ii) Current flowing in the circuit I Xe 318x10°, , (iii) The potential difference V lags behind the current I by a phase angle 71/2., (a) What will be the inductive reactance of a coil of negligible resistance and of inductance, 5 a, = mH connected with an a.c. of frequency 50 cycle per second ? (b) /f current flowing in the, circuit is 0-2 A, what will be (i) the potential difference across the ends of the coil and (ii) the, phase difference between the current and the potential difference ?, , Example 5., , 5 5, Solution. Given : f= 50 cycle/second, L = i mH = ee 103H,1=0-2A, (a) Inductive reactance X, = OL = 2nfL, =2x 0x 50x () x 103=0-59., , (b) (i) Potential difference across the ends of coil, , V, =1X, = 0-2 x 0-5 = 0-1 volt., (ii) In the a.c. circuit containing inductance of negligible resistance, the potential difference, V leads ahead the current I in phase by an angle 7/2., , 7.10. Alternating Current Circuit Containing both Resistance, and Inductance (R-L Circuit), , , , In Fig. 7.14, an inductance L and a resistance R are, connected in series with an alternating e.m.f, V = Vo sin wt. Let at, any instant 7, the current in the circuit be I due to which potential, difference across the resistance Ris, , , , , , , , , , , , Ve =IR .(7.20), and the potential difference across the inductance L is, V=Vo sin ot VL -=IX,=IoL (7.21), , At any instant, the sum of the potential difference Vp, , , , Fig. 7.14. R-L a. ¢. circuit, , Scanned with CamScanner

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Alternating Current Ess a, , , , ry) ACE aC Uli, , 1, In all alternating circuits, impedance, Vv, , 2. In pure ohmic circuit, Z=R and phase difference between the current and voltage is zero., . In pure inductive circuit, Z = X, = wL and current lags behind the applied voltage in phase by 7/2., , Rs ey 1, 4.In i a em Zz : Xc= 5G and current leads ahead the applied voltage in phase by 7/2., 5. In R-L circuit, Z= YR? +X? and current lags behind the applied voltage by phase angle 6, where, , 6. InR-C circuit, Z= /R?+ X@ and current leads ahead the applied voltage by phase angle $, where, Xe, tan = ==, , . . R, 7. InL-C circuit, Z = X, — X¢ and current lags behind the applied voltage by phase angle 1/2., , 8. In L-C-R circuit, impedance Z= /R? + (X,-X¢)? and current lags behind the applied voltage, . by phase angle >, where, , , , , , magi X_~Xc, , _ wa, , Solved Examples., , Se, , , , , , , , Example 1. A condenser of capacity 100 uF and a resistance of 40 Q are connected in series to an a.c., , source of frequency 60 Hz, 110 volt. What will be the maximum current in the circuit ?, Solution. Given : C = 100 uF = 100 x 10 F = 104F, R=40Q,V=110 volt,, : S= 60 Hz, w = 2nf= 2 x 3-14 x 60 = 376-8 rad/s, , is Current in the circuit I = ; wv, =, VR?+ (1/02), 110 110, or Ds tg = 229A, coy?+( 1 (40? + (26-54), 3768x104, , -. Peak value of current Ip =I Y2 =2:29 x /2 =3-24A., , Example 2. A resistance 200 Q and a condenser of capacity 15 uF are connected in series with an, , , , , , , , alternating source of 220 volt, 50 hertz. Calculate : (i) impedance of circuit, (ii) current, in circuit, (iii) phase difference between the current and applied potential, (iv) potential, , difference across the resistance and condenser., Solution. Given : R= 200 Q, C = 15 pF = 15 x 10 F, V= 220 volt, f= 50 hertz,, *. O = 2nf=2 «3-14 x 50=314 rds"!, , I 1, ircuit Z = ,/R?+— = 4/(2 (ot, (i) Impedance of circuit Z R we (200) 314x15%10%, , = (200)? + (212-3)? = 291-68 ohm., , el, , Scanned with CamScanner

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Shivalal Physics : Class 12, , V __220_ _ 9.75 ampere., (i) Current in circuit I = Z 7219-68 0-75 _ se alliety, (iii) Current will lead the applied potential in phase by angle >,, , 6:, tang = HOC _ (/B14x15%10) _ 1.969 org = 46-7", ng=—2 =, R 200, , 5 = IR=0-75 x 200 = 150 volt., (iv) Potential difference across the resistance Vp = IR = 0:75, , =[x —5 =0°75 x 212:5 = 159 volt., Potential difference across the condenser Vc =I x oC 0, , ‘i . Vp and Ve are not, Note : We can see that Vp + Vc is not equal to the applied potential V because Vp Cc, , in same phase, but V = Vp" + Ve holds. 7 ; a, A i joii in series with an, Example 3. 4 coil L of negligible resistance and an ohmic resistance R are joined in, , Solution., , Example 4,, , Solution., , iT ii to be 160 volt, alternating voltage source. The ‘potential difference across the coil i oa r difference, and across the resistance 120 volt. Calculate the virtual value of applied pi ,, , If the virtual current in the circuit is 1-0 A, find the total impedance of the circuit., , Given : V, = 160 volt, Vz = 120 volt andI=1-0A ., fi Vt across the coil L (i.e., inductance) leads ahead of current I in the, , The potential difference V, ., circuit in phase by 90° and the potential difference Vp across the resistance R and the current, Tare in same phase, Then V, and Vp are at a phase difference of 90° from each other. If the, applied potential difference is V, then, , v2 = vi? + Vp?, or V? = (160)? + (120)? = (200)?, or V = 200 volt, , Thus the virtual value of applied potential difference = 200 volt, , : Seah lt Potential difference, d it, Z =, Total impedance of circui Cureat, , V_ 200, or Z=7=To =200 ohm., A coil of self inductance 0-525 H when connected to. adic. Source of emf. 120 vol, , : RC. mf, it a, , current 0-5 A flows through it. How much current will ‘low through it, j it is, 120 volt a.c. source of frequency 60 Hz ? Bath Pitts aaa, Here R = 240 Q, f= 60 Hz, w = 2nf'=2 x 3-14 x 60 = 3768 rad s-, 1 = 9.595 H, When coil is connected to a d.c. source (@=0), its Teactance X, = C = c, resistance is effective, Hence ohmic Tesistance of coil is Lal Oand only the aaa, , ~ dc. voltage 120, Re" curent ~~ Gy =240 ohm, , Reactance of coil in the a.c. circuit Z = VRet+@t2, , Z = ¥(240)? + (376-8 x 0-525)? = 31 ohm, , or, in coil] =< © _ 120, Current in coil I = z= 311 ~ 9°386 A,, , Example 5. A coil of negligible resistance and Self inductance 0-05/r he, ite: my and a resistance of, , 12 ohm are connected in series with an alternating Sou, , i , ree, 50 hertz. Calculate : (i) reactance of circuit, (ii) impedance of mn f 260 volt and frequency, (iv) phase difference between the current and aPplied emf, (y) it (iti) current in circuit,, the resistance and inductance, * \Y) Potential diffore fide across, , Scanned with CamScanner