CHAPTER 5. Solutions for Exercises

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1 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 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 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 / Thus the vltage is v ( t 55.6 cs(377t 08 E5.4 (a j cs( ω t + 0 sin( ωt 4.4 cs( ωt (b j j.6 + j cs( ω t sin( ωt cs( ωt (c j j 7.5 j sin( ω t cs( ωt cs( ωt 5.8

2 E5.5 The phasrs are and 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 / 00 / j 50 Z (b The phasr diagram is shwn in Figure 5.0a in the bk. Z / jω j / 00 /( j Z (b The phasr diagram is shwn in Figure 5.0b in the bk. Z / 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 j

3 i ( t 8.8 cs(500t j ω (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 + /( + j Z /( j A /( j /( A A Ω E5. The transfrmed netwrk is: We write K equatins fr each f the meshes: j ( 00 j 00 + j ( 0 Simplifying, we have 00 + j ( 00 + (00 j A and 0 ( t.44 cs(000t 45 A and i ( t cs(000t Slving we find A. Thus we have i. 3

4 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 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 ( Thus, Q P tan( θ 4.49 ka. Als, we have P /[ cs( θ ] 50.0 A. Thus we have 70.7 A and 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 Ω θ 90 / Z A P cs( θ 0 Q sin( θ ka The secnd lad absrbs an apparent pwer f 0 kawith a pwer factr f 80% lagging frm which we have cs θ ( 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 ka / 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

5 Z t 50 j j j 50 /00 + / j j Then we analyze the circuit t determine the pen-circuit vltage. t c j00 n t t / Z 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 j j5 The lad pwer is P 00( / 6.5 W 5

6 (b Fr a purely resistive lad, maximum pwer is transferred fr Z Ω. The Thévenin equivalent with the t lad attached is: The current is given by j5 The lad pwer is P 03.(0.3456/ 6.57 W E5.6 The line-t-neutral vltage is 000 / N phase angle was specified in the prblem statement, s we will assume that the phase f an is zer. Then we have an bn cn 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 aa 00 + j Z The currents fr phases b and c are the same except fr phase bb c Y P cs( θ 3 cs( kw 6

7 Y Q sin( θ 3 sin( ka E5.7 The a-phase line-t-neutral vltage is an 000 / The phase impedance f the equivalent Y is Z / 3 50 / Ω. 0 Z Y Thus the line current is an aa A Z 6.67 Y Similarly, A and A. bb c Finally, the pwer is P 3( aa / y kw Answers fr Selected Prblems P5.* phase angle θ 60 t peak f 500 Hz T ms P W ms ω 000π rad/s 7.07 π 3 radians 7

8 4 P5.6* v ( t 8.8 cs(π 0 t 7 P5.9* 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 cs( ωt sin( ωt 3.763cs( ωt P5.8* i Z 00π ( 0π 90 ( t ( 0π cs( 000πt 90 ( 0π sin( 000πt i ( t lagsv ( t by 90 8

9 P5.9* i Z Z Ω ( t cs( 000πt sin( 000πt i ( t leadsv ( t by 90 P5.3* ω 500 : Z ω 000 : Z 50 0 ω 000 : Z P5.34* lags s by 45 9

10 P5.37* leads s by P5.39* lags s by 6.56 P5.4* The peak value f i ( t is five times larger than the surce current! P5.45* P5.46*

11 P5.53* P5.59* This is a capacitive lad. P.5 kw Q.5 ka pwer factr 89.44% P 0 kw Q 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 µF The capacitr must be rated fr at least ka (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 (b (c P 50 W lad P 47. W lad

12 P5.70* lad lad.5 Ω 06. µ F P5.73* P5.74* A P 9.36 kw Z Ω P5.78* P aa AB AB lad P line kw 0.50 kw

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