An alternating current i is given by i = 141.4 sin314¢. Find @ the maximum value (ji) frequency (iii) time period and (iv) the instantaneous value when f is 3 milliseconds. Solution : An alternating current / is given by the standard equation i= Im sin ot The given equation is i = 141.4 sin 3141 Comparing the two equations, we have (Jj) = 141.4 Amps (ii) Frequency, ¢ = 2 = 24 = 50 Hz Qn (iii) Time period, T = = 0.02 second . 3 ; (iv) When f= 3x10™ seconds, the instantaneous value of the current is 141.4 sin[314 x 3 x 109 i 114.35 Amps Scanned with CamScannerProblem 3.4 : An alternating voltage has an amplitude of 100 V. Find its (i) R.M.S. value J Average value Solution : E, (@) RMS. value E = = = 0.707 £, v2 ™ Given £,, =100V E = 0.707 x 100 = 70.7 volts (ii) Average value £,, = = 0.637 Em (1/2) Eqy = 0.637 x 100 = 63.7 volts Scanned with CamScannerFig. 3.33 Is applied to a certain circuit. The current taken is 2.4 sin (418 ¢- 1.37). Find ie . i) Frequency if) The phase angle between voltage and current fii) The resistance of the circuit. (Nov/Dec 84, B.U.) Solution : : Peak value of applied voltage, V,, = 400 V Peak value of circuit current, J, = 2.4 A Vn. V, Impedance of the circuit, Z = 7° = we i)" Frequency, Coefficient of time in expression of voltage or current f Qn = 418 _ 66.5 Hz an Scanned with CamScannerii) Phase angle between voltage and current, = 1.37 radians (lag) = 137 199°=78.5° (lag) © iii) The resistance e circuit, R=Z cos 66.6 cos 78.5° 66.6 x 0.1994 = 33.229 Scanned with CamScanner/_A pure inductive coil allows a current of 10 amperes to flow from a 230 volts, 50 Hz supply. Find (i) inductive reactance (ii) inductance of the coil. Also write down the equations for voltage and current. Solution : @ Now, 1-4 x, 104 | V _ 230 or X, =— = = 23 ot ay eee 230 V, 50 Hz Fig. 3.37 Thus, inductive reactance, X, = 23 ohms Scanned with CamScanner(iii) Vn = V2 x V = V2 x 230 = 325.27 volts Im =N2 x1 = V2 x10 = 14.14 Amps @ = 2nf = 2n x50 = 314 rad/sec v= Vp sin wt v = 325.27 sin 314¢ volts i = Im sin ( T =14.14 sin (214-4) Amps Scanned with CamScannerA 318 pF capacitor is connected across a 230 volts, 50 Hz system. Determine (i) the capacitive reactance (ii} R.M.S value of current and (iii) equations for voltage and current. Solution : 318 pF (i) Capactive reactance, 1 1 Xo = —— ule 230 V, 50 Hz a Fig. 3.41 a CT on x 50 x (318 x 10°) = 10 ohms Scanned with CamScannerGi) RMS value of current, f= Xe foo es 10 Gil) V,, = 230 x V2 = 325.27 volts Im = 23x V2 = 32.53 Amps @ = 2nf = 2n x50 = 314 rad/sec v= Vm sinat or v =325.27 sin 314¢ volts Im sin (o+2) 2 = 32.53 sin (a1st-2) Amps 2) & “Problem (it The current drawn by a pure capacitor of 20 microfarads is 1.382 Amps from a 220 volts A.C. supply. What is the supply frequency ? 20 WF 41382A i 220 V A.C. supply Fig. 3.42 Solution : We have p= x or Scanned with CamScannerama | “pacitor of 80 uF takes a current of 1.0 A when the alternati C=8uF Fig. 10.24 Circuit diagram for Example 10.5 voltage applied acrons it is 230 V.( alculate: (a) the frequency of the applied voltage; , (b) the resistance to be connected in series with the capacitor reduce the current in the circuit 100.5 A at the same frequen’ (c) the phase angle of the resulting circuit v @) Xe 1-20 ol. 2nfC s i 1 f=t es WOK. Ie XBX10* x 29 = 86-5 Hz (b) When a resistance is - ot A ven in ego ,onnected in series with the capacitor, cuit is now as given in Fig. 10.24, 7 05 = (B+ X2F but Xe=230Q hence §=R=398Q ak © $= 008 Zoo OS 430° 06 30° ead Scanned with CamScannerBESET 8 coil having a resistance woo Fig. 10.27 Circuit diagram for Example 10.6 Fig. 10.28 Phasor diagram for Example 10.6 of 12Q and Hz supply. in inductance of 6.11; culate connected across a 100 V, 5 (a) the reactance and the impedance of the coil; (b) the current; (c) the phase difference between the current and the apple voltage. When solving problems of this kind, students should first of all dnw a circuit diagram (Fig. 10.27) and insert all the known quantities. ‘The should then proceed with the phasor diagram (Fig. 10.28). It is not essentil to draw the phasor diagram to exact seal imately correctly since itis then easy to values, le, but it is helpful to draw it appro- make a rough check of the calculated (a) Reactance = X, = 2a1, = 22x 500.1 =314Q Impedance = Z = V(R? + 2) =V(I2+ 31.4) = 33.6.9 v 100 () Current= 7 100 336 2974 ©) tan ¢= pm) 314 = 7 = 2.617 Scanned with CamScanneroe erene ok) supply. Caleulate: (a) the impedance; (b) the current; (c) the voltages across R, L and C; (d) the phase difference between the current and the supply voltage. The circuit diagram is the same as that of Fig. 10.25. (a) From equation [10.21], 2 % 2 -—__" _. z= | + (2xausasoxo1s alesse = V{144 + (47.1 - 31.85)}} = 19.4.9 0 i94 S154 ‘4 (b) Current = Z* Scanned with CamScanner[ee Fig. 10.31 Phasor diagram for Example 10.8 Voltage across R = Vp = 12 x 5.15 = 61.8 V Voltage across L = V, = 47.1 x 5.15 = 242.5 V Voltage across C= Vi; = 31.85 x 5.15 = 164.0 V © and ‘These voltages and current are represented by the respective Phasors i, Fig. 10.31 (d) Phase difference between current and supply voltage is Y, 61.8 = cos! 2 = cog! 9 = 510597 9 = cos"! Ft = cos = 51°50 Or, alternatively, from equation [10.22], 471-31, $= tan E1385 = tan 1.271 = 51°48” Scanned with CamScannera purely resistive circuit V=IR y inductive circuit For a pure V=IX,=2nfLI = oll fly sin(2mft + 1/2) a purely capacitive circuit V=IX¢=I/2nfC = 1/@C Af CV, sin(2ft + 1/2) ‘or Rand Lin series VaIz Z=(R+o'L)? For R and C in series Re XP! 1 z-(e+ Sa) =(R?+ X23) or R, L and Cin series 2) 1 1 2a {rs(on- 2) | = (R'+(X,- Xo)! Scanned with CamScanner (103 (107) (106) (10.16) [10.14] (109) (10.10) fot pio2tlA current i= sin (31¢-10°) produces a potential drop v = 220 sin (31 ¢+20°) in a circuit. Find the values of circuit parameters, assuming a series combination. (May/June 86, B.U,) Solution : ee We notice that the voltage leads by 20° and the current lags by 10°, with regard to the reference quantity. The phase difference between the voltage and current is 20° - (-10°) = 30°, with the current lagging. The angular frequency is @ = 31 rad/sec. As current lags, the circuit is inductive. Vin = 220 Vo and I,=1A z=¥n 200 _so99 Im 1 R = Zcoso = 220 cos 30° = 220 x 0.866 = 190.5 2 X, =Zsin 9 = 220 sin 30° = 220 x 0.5 = 110.0 @ = 2nf 31 or 31=2nf » fo 2 aot on X,=2n/L = 1109 pelo nx 4.9 =3.57H Scanned with CamScannerPropiem 3.24 ‘A 100V, 50Hz inductive circuit takes a current of 10 Amps, lagging the ‘oltage by 30°. Calculate the resistance and inductance of the circuit, Draw the waveforms of current and voltage. (B.U. Feb/Mar 83) Solution : The current lags behind the applied voltage by 30°, hence the circuit is inductive The p.f., cos 30° = 0.866 (lagging). Resistance R= Z cos > = 10 x 0.866 = 8,662 Inductive reactance X,=Z sino =10x0.5=5Q Fig. 3.51 2nfL=5 L= = 0.016 H 2nx Scanned with CamScanner‘A series circuit with R = 100, L = 50 mH and C = 100 uF is supplied with tye V, 50 Hz. Find (1) the impedance, (ii) current, (ili) power and (iv) power factor. (May/June 86, B.U.) Solution : Circuit resistance, R = 109 Inductive reactance of the circuit, X, = 2n/L = 2m x 50 x 50 x 107° =15.719 Capacitive reactance of the circuit, X¢ = 2nfC 1 2nx50%100x10-% = 3189 Scanned with CamScannerImpedance of the circuit Z = |R?+(X,~X¢)* = fio? +(15.71-31.8)" = (100 +258.8 = 18.949 Circuit current, / = v = -200_ =10.55A Z 18.94 - Circuit power factor, cos ¢ = & = 10 3 2 Os a6 Z wae 18 94 = 0.947 (leading). (Q Xe > X,) Power consumed, P = V/ cos > = 200 x 10.55 x 0.947 = 1998 Watts Scanned with CamScanner