HAPTE 5 Slutins fr Exercises E5. (a We are given v ( t 50 cs(00π t 30. The angular frequency is the cefficient f t s we have ω 00π radian/s. Then f ω / π 00 Hz T / f 0 ms m / 50 / 06. Furthermre, v(t attains a psitive peak when the argument f the csine functin is zer. Thus keeping in mind that ωt has units f radians, the psitive peak ccurs when π ω tmax 30 tmax 0.8333 ms 80 (b P avg / 5 W (c A plt f v(t is shwn in Figure 5.3 in the bk. E5. We use the trignmetric identity sin( z cs( z 90. Thus 00 sin(300π t + 60 00 cs(300πt 30 E5.3 ω πf 377 radian/s T / f 6.67 ms 55.6 m The perid crrespnds t 360 therefre 5 ms crrespnds t a phase angle f ( 5 /6.67 360 08. Thus the vltage is v ( t 55.6 cs(377t 08 E5.4 (a 0 0 + 0 90 0 j0 4.4 0 cs( ω t + 0 sin( ωt 4.4 cs( ωt 45 45 (b 0 30 + 5 60 8.660 + j 5 +.5 4. 330 j.6 + j 0.670.8 3.44 0 cs( ω t + 30 + 5 sin( ωt + 30.8 cs( ωt + 3.44 (c 0 0 + 5 60 0 + j 0 + 7.5. 99 j 7.5 j.99 30.4 5.8 0 sin( ω t + 90 + 5 cs( ωt 60 30.4 cs( ωt 5.8
E5.5 The phasrs are 0 30 0 + 30 and 3 0 45 v lags v by 60 (r we culd say v leads v by 60 v leads v 3 by 5 (r we culd say v 3 lags v by 5 v leads v 3 by 75 (r we culd say v 3 lags v by 75 E5.6 (a E5.7 (a E5.8 (a Z jω j 50 50 90 / 00 / j 50 Z 00 0 90 (b The phasr diagram is shwn in Figure 5.0a in the bk. Z / jω j 50 50 90 00 0 / 00 /( j 50 90 Z (b The phasr diagram is shwn in Figure 5.0b in the bk. Z 50 50 0 00 0 / 00 /(50 0 (b The phasr diagram is shwn in Figure 5.0c in the bk. E5.9 (a The transfrmed netwrk is: s Z 0 90 50 + j50 8.8 35
i ( t 8.8 cs(500t 35 7.07 35 j ω 7.07 45 (b The phasr diagram is shwn in Figure 5.6b in the bk. (c i(t lags v s (t by 45. E5.0 The transfrmed netwrk is: Z /00 + /( j 50 + /( + j00 55.47 56.3 Z 77.4 56.3 /( j 50 5.547 33.69 A /( j00.387 46.3 /(00.774 56.3 A A Ω E5. The transfrmed netwrk is: We write K equatins fr each f the meshes: j 00 + 00( 00 j 00 + j00 + 00( 0 Simplifying, we have 00 + j 00 00 00 ( 00 + (00 j 00 0.44 45 A and 0 ( t.44 cs(000t 45 A and i ( t cs(000t Slving we find A. Thus we have i. 3
E5. (a Fr a pwer factr f 00%, we have cs( θ, which implies that the current and vltage are in phase and θ 0. Thus, Q P tan( θ 0. Als P /[ cs( θ ] 5000 /[500 cs(0] 0 A. Thus we have 4.4 and 4.4 40. m (b Fr a pwer factr f 0% lagging, we have cs( θ 0., which implies that the current lags the vltage by θ cs (0. 78.46. Thus, Q P tan( θ 4.49 ka. Als, we have P /[ cs( θ ] 50.0 A. Thus we have 70.7 A and 70.7 38.46. m (c The current ratings wuld need t be five times higher fr the lad f part (b than fr that f part (a. Wiring csts wuld be lwer fr the lad f part (a. E5.3 The first lad is a 0 µ F capacitr fr which we have Z /( jω 65.3 90 Ω θ 90 / Z 3.770 A P cs( θ 0 Q sin( θ 3.770 ka The secnd lad absrbs an apparent pwer f 0 kawith a pwer factr f 80% lagging frm which we have cs θ (0.8 36.87. Ntice that we select a psitive angle fr θ because the lad has a lagging pwer factr. Thus we have P cs( θ 8.0 kw andq sin( θ 6 ka. Nw fr the surce we have: P P + P 8 kw Q Q + Q.3 s s ka P + Q 8.305 ka / 8.305 A s s s factr P s /( s s s pwer 00% 96.33% E5.4 First, we zer the surce and cmbine impedances in series and parallel t determine the Thévenin impedance. 4
Z t 50 j5 + 50 j5 + 50 + j 50 /00 + / j00 00 + j5 03. 4.04 Then we analyze the circuit t determine the pen-circuit vltage. t c 00 00 70.7 45 00 + j00 n t t / Z 0.6858 59.04 E5.5 (a Fr a cmplex lad, maximum pwer is transferred fr * Z Z 00 j5 + jx. The Thévenin equivalent with the lad t attached is: The current is given by 70.7 45 0.3536 45 00 + j5 + 00 j5 The lad pwer is P 00(0.3536 / 6.5 W 5
(b Fr a purely resistive lad, maximum pwer is transferred fr Z 00 + 5 03. Ω. The Thévenin equivalent with the t lad attached is: The current is given by 70.7 45 0.3456 37.98 03. + 00 j5 The lad pwer is P 03.(0.3456/ 6.57 W E5.6 The line-t-neutral vltage is 000 / 3 577.4. N phase angle was specified in the prblem statement, s we will assume that the phase f an is zer. Then we have an 577.4 0 bn 577.4 0 cn 577.4 0 The circuit fr the a phase is shwn belw. (We can cnsider a neutral cnnectin t exist in a balanced Y-Y cnnectin even if ne is nt physically present. The a-phase line current is an 577.4 0 aa 00 + j 75.40 Z 4.60 37.0 The currents fr phases b and c are the same except fr phase. 4.60 57.0 4.60 8.98 bb c 577.4 4.60 3 Y P cs( θ 3 cs(37.0 3.88 kw 6
577.4 4.60 3 Y Q sin( θ 3 sin(37.0.404 ka E5.7 The a-phase line-t-neutral vltage is an 000 / 3 0 577.4 The phase impedance f the equivalent Y is Z / 3 50 /3 6.67 Ω. 0 Z Y Thus the line current is an 577.4 0 aa 34.63 0 A Z 6.67 Y Similarly, 34.63 0 A and 34.63 0 A. bb c Finally, the pwer is P 3( aa / y 30.00 kw Answers fr Selected Prblems P5.* phase angle θ 60 t peak f 500 Hz T ms P W 0.3333 ms ω 000π rad/s 7.07 π 3 radians 7
4 P5.6* v ( t 8.8 cs(π 0 t 7 P5.9* 3.536 P5.*.5 P5.5* P5.7* v s.58 ( t 4.4 cs( ωt 45 lags by 90 lags by 45 s leads s by 45 P5.* ( t 0 cs( 400πt + 30 v v ( t 5 cs(400πt + 50 ( t 0 cs( 400πt + 90 v 3 v ( t lagsv ( t by 0 v ( t lagsv3( t by 60 v ( t leadsv ( t by 60 3 P5.3* 5 cs( ω t + 75 3cs( ωt 75 + 4 sin( ωt 3.763cs( ωt + 8.09 P5.8* i Z 00π 90 0 0 ( 0π 90 ( t ( 0π cs( 000πt 90 ( 0π sin( 000πt i ( t lagsv ( t by 90 8
P5.9* i Z 5.9 90 0 0 Z Ω 0.683 90 ( t 0.683 cs( 000πt + 90 0.683 sin( 000πt i ( t leadsv ( t by 90 P5.3* ω 500 : Z 58. 7.57 ω 000 : Z 50 0 ω 000 : Z 58. 7.57 P5.34* 70.7 45 7.07 45 7.07 45 lags s by 45 9
P5.37* 4.47 63.43 4.47 63.43 8.944 6.57 leads s by 63.43 P5.39* 8.944 6.56 89.44 6.56 44.7 63.44 lags s by 6.56 P5.4* 0 0 50 90 50 90 The peak value f i ( t is five times larger than the surce current! P5.45* P5.46* 9.40 9.58 4.986.45.644 80.54.977 74.0 0
P5.53* P5.59* This is a capacitive lad. P.5 kw Q.5 ka pwer factr 89.44% 5. 0.66 P 0 kw Q 3.770 ka Apparent pwer 0.68 ka Pwer factr 93.57% leading P5.6* P kw s Q 3.84 ka s Apparent pwer 6 ka Pwer factr 84.6% lagging P5.64* (a (b 400 75.5 07µF The capacitr must be rated fr at least 387.3 ka. 00 0 (c P5.67* (a The line current is smaller by a factr f 4 with the capacitr in place, reducing lsses in the line by a factr f 6. n.789 6.57 (b (c P 50 W lad P 47. W lad
P5.70* lad lad.5 Ω 06. µ F P5.73* P5.74* 76. 4.67 A P 9.36 kw Z 70.9 6. 05 Ω P5.78* P aa AB AB lad P line 6.9 4.40 363.4.60 67.49 34.40 68.3 kw 0.50 kw