Unit-3. Question Bank
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1 Unit- Question Bank Q.1 A delta connected load draw a current of 15A at lagging P.F. of.85 from 400, -hase, 50Hz suly. Find & of each hase. Given P = = I = 15A Ans. 4.98, 5.7mH So I P = 15 =8.66A so I = 8.66 Cos 1.85 (lagging load) = A Z P = I P P j = P +jx P P = 4.98 Ans. = X P f mH Ans. Q. A balanced -hase star connected load of 100 KW takes leading current of 80A, when connected across -hase, 1100, 50Hz suly. Find comonent of load er hase. Given P = 100KW I = 80A Ans. = 5., C = 5µF In case of star connected load I P = I I P = I = 80A & = 1100 = f = 50Hz I = I P = P Cos 1100Cos
2 Cos = = 49, Sin =.754 load imedance/hase = P /I P = 1100 / =Z Cos = = 5. Ans. X C = Z Sin = =5.986 Xc = 1 fc (As leading current is drawn) C = 1 fx C = = 5µF Ans.
3 Q. Each hase of star connected load consists of non-inductive resistance of 50 in arallel with a caacitance of 6.6µF. Calculate the line current, total ower absorbed, total ka and the ower factor when this load is connected to a 81 (line voltage), -hase, 50Hz suly. Ans. 6.A,.9kW, 4.1kA,.707 = 81 = 81 0 =0 0 = (0+j0). Admittance of each hase is, Y = 1 +jωc= j = (0.0+j0.0) S Hence I = Y = 0 (0.0+j0.0)=4.4+j4.4=6. 45 A Pf=cos = cos45 =.707 Ans I =I = 6.A Ans Power factor = cos45 = leading = (0+j0), I = (4.4+j4.4)A Hence comlex ower/hase S = I* S = (0+j0) (4.4 j4.4)=968 j968 = A Total aarent ower in the hase = 169 A = ka Ans. Total ower absorbed in the hase = 968W =.904kW Ans. Q. 4 Calculate the active and reactive current comonents in each hase of a star connected, 10,000, hase alternator sulying 5000 kw at a ower factor of 0.8. If the total current remains the same when the load ower factor is raised to 0.9, find the new outut.
4 Ans. 89A, 16A, 565kW I = P = I Cos P A cos 10, The active current = I cos = = 89 A Ans. The reactive current = I sin = = 16 A Ans. The new ower at 0.9 P.F. = = 565kW Ans. Q. 5 A balanced -hase star connected load of 150kW takes a leading current of 100A with a line voltage of 1100, 50Hz. Find the circuit constants of the load er hase. 816µF Ans. = 5 X C =.9, C = Phase imedance Z = = Phase oltage Phase current The Power er hase = I = W 5010 = Ans. X = Z.9 The reactive art is caacitive (leading current)
5 1 C.9 Ans. C = = F 816µF Ans. Q. 6 Three equal star connected inductors take 8kW at ower factor 0.8 when connected to 460, hase, wire suly. Find the line currents if one inductor is short circuited. P = I Cos Ans A, A, A I = A cos The hase imedance = I P (I =I for star connection) cos = 0.8; = 7 Since Z is shorted and N are at the same otential. The three line voltages as hasors will be 1 = = = Since and N are at the same otential I 1 Z
6 = A Ans. =.64 j1.5 A I Z = A Ans. = j8.48 A I I I 1 (As at N I 1 + I + I = 0) = j( ) A =.61 + j9.98 = A Ans. (i.e.) the three line currents are 1.7A, 1.7A and 7.6 A Q. 7 The circuit shown in which = 00, r = 100 and = 0.55H is connected to a symmetrical hase, 400, 50 Hz system at oints 1, and. Find the.d. between A, B and C, when the hase sequence of the suly is (a) 1,, (b) 1,, Ans. (a) 0, (b) j = 100+j17. = (+r)+j = (00+100) + j17. =
7 Phase 1 1 = 400 0, = , 1 = AB = (r+j )I 1 +I = = = o 0 Ans. Phase 1 I = A Hence, AB = = = 00+j115.5+j1 = 00+j46.5 = Ans. Q. 8 A wire hase suly feeds a load consisting of three equal resistors. By how much is the load reduced if one of the resistors be removed (a) when the load is star connected (b) when the load is delta connected? Ans. (a) 50%, (b).% (a) Star connected With all three resistors I = =I The total ower = I = With one resistor removed.(i)
8 I = The ower =I =.(ii) 4 This is half of the above. The reduction in ower = (c) Delta connected % Ans. With all three resistors current in each resistor I = P The Power = I = (i) With one resistor removed
9 Current in each resistor I = The ower =I =. (ii) Hence reduction in ower =.% Ans. Q. 9 A -hase star-connected alternator feeds a 000 h delta-connected induction motor having a P.F. of 0.85 and an efficiency of 0.9. Calculate the current and the active and reactive comonents in (a) each alternator hase (b) each motor hase. The line voltage is 00. Ans. (a) 495A, 41A, 61A (b) 86A, 4A, 151A Motor inut = HP Watts 0.9 If I is the line current we have I cos Inut I = A Ans. Power factor = 0.85 = cos sin = 0.57 (a) Alternator (star connected) Phase current = line current = 495A 41A Ans. Active comonent = I cos = eactive comonent = I sin = 61A Ans. (b) Induction motor (delta connected)
10 Phase current I = A Ans. Active comonent = I cos = 4 A Ans. eactive comonent = I sin = 151 A Ans. Q. 10 Three star-connected imedances Z 1 = 0+j7.7 er hase are connected in arallel with three delta-connected imedances Z = 0 j159. er hase. The line voltage is 98. Find the line current, P.F, Power and reactive volt-ameres taken by the combination. A Ans..6 98A,.985 lagging, 84W, 94 The hase voltage = 98 0 Z 1 = 0+j7.7 = I 1 = Z = 0 j159. The equivalent star imedance =.5 j4.76 A Z = Z 10 j I = j4.19 A 54 The total current I = I 1 +I =.1 j0.57 =.6 9.8A Ans. The P.F. = cos = cos9.8 = lagging Ans. The ower sin = 0.17 = ower er hase 84W Ans.
11 The reactive A = I sin = = = = 94 A Ans. Q. 11 Prove that P = P Y if load er hase same. Ans. P = P Y Sol et ine voltage resistive load er hase Z imedance er hase Then P = / For Y connective P Y = P I P Cos = P Z. Z = P Z Z P Z..(1) For connective P = P P IPCos P = P.. P Z Z Z = () Z By Comarison (1) & ()
12 P = P Y Ans. Q. 1 Three identical coils are connected in star to a three hase 50Hz suly. If the line current is 10A, total ower consumed is 1kW and volt amere inut is 15 KA, find the line voltage, hase voltage, A inut, resistance and inductance of each coil. I =10A; P = 1kW; S = 15 KA cos =.f.= P = I cos kw 1 ka 15 = 0.8. Hence sin = 0.6. Ans. 866, 500, 9000A, 40, 0.095H Hence = P 110 = I cos volt. Ans. 866 = = 500 volt. Ans. Q = I sin = = 9000 A Ans. I =I =10A; Z = I A = Z cos = = =50 40 Ans. X = Z sin = =0 X Inductance/hase is = f = 0 = H Ans. Q. 1 Three 100 non-inductive resistance are connected in (a) star (b) delta across a 400-, 50Hz, -hase mains. Calculate the ower taken from the suly system in each case. In the event of one of the three resistances getting oen-circuited, what would be the value of total ower taken from the mains in each of the two cases? Ans. 1600W, 4800W, 800W 00W
13 (i) Star Connection = 400/ I = A I Z 100 cos = 1 as load is non-inductive resistance P = I cos 1600W = / (1) (ii) Delta Connection = 400 ; = 100 I = 400 / 100 = 4A I = 4 A 4800W P = = () When one of the resistors is disconnected (i) Star Connection The circuit no longer remains a -hase circuit but consists of two 100 resistors in series across a 400- suly. Current in lines A and C is = 400/00=A Power absorbed in both = I = = 800 W...() Comaring () with (1) it is seen that by disconnecting one resistor, the ower consumtion is reduced by half. (ii) Delta Connection In this case, currents in AB and AC remain as usual 10 out of hase with each other.
14 Current in each hase = 400/100=4A Power consumtion in both = = 00W..(4) Comaring (4) with () it is seen that when one resistor is disconnected, the ower consumtion is reduced by one-third. Q. 14 A -hase, 7.kW, 440-, 50Hz induction motor oerates on full load with an efficiency of 89% and at a ower factor of 0.85 lagging. Calculate the total ka rating of caacitors required to raise the full-load ower factor at 0.95 lagging. What will be the caacitance er hase if the caacitors are (a) delta-connected and (b) star-connected? Motor ower inut P = 7./0.89 = kW Power Factor = 0.85 (lag) cos 1 =.85 1 = 1.78 tan 1 =.6197 K 1 = tan 1 = = 5.5 Ans µF, 197.4µF cos ; tan Motor ka = P tan = = 1.54 The difference in the values of ka is due to the caacitors which suly leading ka to artially neutralize the lagging ka of the motor. leading ka sulied by caacitors is =ka 1 ka = = 1 Since caacitors are loss-free, their ka is the same as ka ka/caacitors = 1/ = 4 A/caacitor=4 (a) In -connection, voltage across each caacitor is 440 Current drawn by each caacitor I c = 4000/440 = 9.09 A Now, I c = X 1/ C C c C = Ic / 9.09 / F = 65.79µF Ans. (b) In star connection, voltage across each caacitor is = 440/
15 Current drawn by each caacitor, I c = A 440 / I c = X c 440 C or, C 197.4µF Ans. Or, C = Q.15 Three imedance coils, each having a resistance of 0 and a reactance of 15, are connected in star to a 400-, -, 50Hz suly. Calculate (i) the line current (ii) ower sulied and (iii) the ower factor. If three caacitors, each of the same caacitance, are connected in delta to the same suly so as to form arallel circuit with the above imedance coils, calculate the caacitance of each caacitor to obtain a resultant ower factor of 0.95 lagging. Ans. (i) 9.4A (ii) 5,10W (iii).8, C = 14.µF Z = cos 1 = /Z = 0/5 =0.8 lag; sin 1 = 0.6 lag When caacitors are not connected (i) I = 400 / 5 = 9.4A =I (star connected load) I = 9.4A Ans. (ii) P = I cos = 5,10 W Ans. (iii) Power factor = 0.8 (lag) Ans. Motor A 1 = I sin ,840 When caacitors are connected Power factor cos =0.95, =18. ; tan 18. =0.88 Since caacitors themselves do not absorb any ower, active ower remains the same i.e. 5,10 W even when caacitors are connected, what changes is the A. Now A = P tan = =1684 eading A sulied by the three caacitors is
16 = A 1 A = = 156 A A/Caacitor= 156/ = 719 For delta connection, voltage across each caacitor is 400 Ic = 719/400 = 1.798A Also Ic = 1/ c 6 C C / F = 14.µF Ans. Q.16 A symmetrical -hase, -wire suly with a line voltage of 17 sulies two balanced - hase laods; one Y-connected with each branch imedance equal to (6+j8) ohm and the other -connected with each branch imedance equal to (18+j4) ohm. Calculate (i) The magnitudes of branch current taken by each -hase load (ii) the magnitude of the total line current and (iii) the ower factor of the entire load circuit Draw the hasor diagram of the voltages and currents for the two loads. Ans. (i) 10A (ii) 0A (iii) 0.6 lag The equivalent Y-load of the given -load is = (18+j4)/ = (6+j8). With this, the roblem now reduces to one of solving two equal Y-loads connected in arallel across the -hase suly as shown in (a) Phasor diagram for the combined load for one hase only is given in (b). Combined load imedance = (6+j8)/ = +j4 = Ω = 17/ = 100
17 I A = I Current in each load = A
18 (i) branch current taken by each load is 10 A Ans. (ii) line current is 0 A Ans (iii) combined ower factor = cos 5.1 = 0.6 (lag) Ans. Q.17 Three identical imedance of 0 0 ohms are connected in delta to a -hase, -wire, 08 abc system by conductors which have imedances of (0.8+j0.6) ohm. Find The magnitude of the line voltage at the load end. Ans The equivalent Z Y of the given Z connections become as shown in is 0 0 / = 10 0 = (8.66+j5). Hence, the load Z an = (0.8+j0.6)+(8.66+j5) = 9.46+j5.6 = an = = 08/ = 10 et an = 10 0 I an = 10 0 / A Zaa = 0.8+j0.6 = oltage dro on line conductors is I Z j1.196 aa an aa (10 j0) (10.8 j1.196) an an aa = j Ans. Q.18 A balanced delta-connected load having an imedance Z = (00+j10) ohm in each hase is sulied from 400, -hase suly through a -hase line having an imedance of Z s =
19 (4+j8) ohm in each hase. Find the total ower sulied to the load as well as the current and voltage in each hase of the load. Ans W, 1.78A, 16.9 The equivalent Y-load of the given -load is = (00+j10)/ = (100+j70) Zs = 4+j8 = = Zaa Hence, connections become as shown in Za0 (4 j8) (100 j70) 104 j a / 1 Ia / A Ans. ine dro I Z j7.1 aa ao aa (1 j0) (14. j7.1) = (16.8 j7.1) = ao ao aa Phase voltage at load end, ao = o Ans. Phase current at load end, I a0 = 1.78A Power sulied to load = = W Ans. Incidentally, line voltage at load end ac = 16.9 = 75.7 Q. 19 Two wattmeters are used for measuring the ower inut and the ower factor of an overexcited synchronous motor. If the reading of the meters are (.0kW) and (+7.0 kw) resectively, calculate the inut and ower factor of the motor. Ans. 5kW,.05 ( 1 ) tan W W W W W 1 = kw W = 7kW 1 ( 7) 9 tan tan (.1176) 7. ( lead ) cos cos 7..05( lead ) Ans.
20 Inut W1 W 7 5kW Ans. Q.0 A wattmeter reads 5.54 kw when its current coil is connected in hase and its voltage coil is connected between the neutral and the hase of a symmetrical -hase system sulying a balanced load of 0 A at 400. What will be the reading on the instrument if the connection to the current coil remain unchanged and the voltage coil be connected between B and Y hases? Take hase sequence YB. Draw the corresonding hasor diagram. Ans. 7.kW As seen from W 1 = I cos Or =(400/ ) 0 cos cos =0.8,sin =0.6 In the second case W = YB I cos(90 ) 4000 sin = 7.kW Ans.
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