Solutions HW#1 - dsafds PDF

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ELE 212 Electric Circuits II Solutions HW #1 1. Given the circuit below, find I and the overall complex power supplied.

200 90˚ V

Consider the network shown below. Io

I2

+

I1

S2

So

Vo

S1

S3



S 2  1.2  j0.8 kVA

S3  4  j

4 sin(cos -1 (0.9))  4  j1.937 kVA 0.9

Let

S 4  S 2  S3  5.2  j1.137 kVA

But

S4  Vo I *2 S4 (5.2  j1.137) 103  5.685  j26  Vo 20090 I 2  5.685  j26 I*2 

Similarly, But

2 2 j sin(cos-1 (0.707))  0.707 S1  Vo I*1

S1 

2 (1  j) kVA

S1 (1.4142  j1.4142) 10 3  - 7.071  j7.071  Vo j200 I1  – 7.071 + j7.071 I*1 

Io  I1  I2  -1.385  j33.07  33.192.4 A S o  Vo I *o So  ( 20090)(33.1 - 92.4 ) VA –2.4˚ kVA So  6.62

2. For the circuit shown below, find Vs.

Consider the circuit below. 0.2 + j0.04 

I

I2

0.3 + j0.15 

I1 Vs

+ 

S 2  15  j But

+

+

V1

V2





15 sin(cos -1 (0.8))  15  j11.25 0.8

S 2  V2 I *2 S2 15  j11.25  120 V2 I 2  0.125  j0.09375 I*2 

V1  V2  I 2 (0.3  j0.15) V1  120  (0.125  j0.09375)(0.3  j0.15) V1  120.02  j0.0469

S1  10  j But

10 sin(cos -1(0.9))  10  j4.843 0.9

S1  V1 I *1

S1 11.111 25.84   V1 120.020.02 I 1  0.093 - 25.82  0.0837  j0.0405 I*1 

I  I 1  I 2  0.2087  j0.053

Vs Vs Vs Vs

 V1  I (0.2  j0.04)  (120.02  j0.0469)  (0.2087  j0.053)(0.2  j0.04)  120.06  j0.0658  120.060.03 V

3. Two loads connected in parallel draw a total of 2.4 kW at 0.8 lagging pf from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW at a 0.707 lagging pf. Determine: (a) The pf of the second load, (b) The parallel element required to correct the pf to 0.9 lagging for the two loads.

(a) P  S cos 1

  S

P 2.4   3.0 kVA cos1 0.8

pf  0.8  cos1   1  36.87o Q  S sin 1  3.0sin36.87 o  1.8 kVAR Hence, S = 2.4 + j1.8 kVA 1.5 P S1  1   2.122 kVA cos  0.707 pf  0.707  cos      45o Q1  P1  1.5 kVAR   S1  1.5  j1.5 kVA Since, S  S 1  S 2   S 2  S S 1  (2.4  j 1.8)  (1.5  j 1.5)  0.9  j 0.3 kVA S2  0.9497 18.43o pf = cos 18.43o = 0.9487   2  25.84o (b) pf  0.9  cos 2 P(tan 1  tan 2 ) 2400(tan 36.87  tan 25.84)   117.5  F C 2 Vrms 2 x 60 x (120) 2

4. A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find: (a) The apparent power, (b) The current drawn from the supply, (c) The kVAR rating and capacitance required to improve the power factor to 0.96 lagging, (d) The current drawn from the supply under the new power-factor conditions.

(a)

S  10  j15  j22  10  j7 kVA S  S  10 2  7 2  12.21 kVA

(b)

S  V I*

  I* 

S 10, 000  j7,000  240 V

I  41.667  j29.167  50.86 - 35 A

(c)

 7 1  tan -1    35 ,  10 

2  cos-1 (0.96)  16.26

Q c  P1 [ tan 1  tan 2 ]  10 [ tan(35) - tan(16.26) ] Qc  4.083 kVAR C (d)

Qc 4083   188.03 F 2  Vrms (2 )(60)(240)2

S 2  P2  jQ 2 ,

P2  P1  10 kW

Q 2  Q 1  Q c  7  4.083  2.917 kVAR S 2  10  j2.917 kVA

But

S 2  V I *2

S2 10,000  j2917  240 V I2  41.667  j12.154  43.4  - 16.26 A I*2 

5. A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in the figure. (a) Find the power factor of the parallel combination. (b) Calculate the value of the capacitance connected in parallel that will raise the power factor to unity.

(a)

1  cos -1 (0.8)  36.87 P1 24   30 kVA S1  cos 1 0.8 Q 1  S1 sin 1  (30)(0.6)  18 kVAR S 1  24  j18 kVA 2  cos -1 (0.95)  18.19 P2 40   42.105 kVA S2  cos2 0.95 Q 2  S 2 sin 2  13.144 kVAR S 2  40  j13.144 kVA S  S 1  S 2  64  j31.144 kVA  31.144    25.95    tan -1   64  pf  cos   0.8992

(b)

 2  25.95 ,

1  0

Q c  P[ tan 2  tan 1 ]  64 [ tan(25.95)  0 ]  31.144 kVAR C

Qc 31,144   5.74 mF 2  Vrms (2)(60)(120) 2...