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(ELEC102)ELEC102 Test 2 2003.pdf
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[b] Q = P tan= 1k tan(cos.1 0.7) = 1k tan(45.58o ) = 1.02kVAR(L)
(35)
1. Load A is connected in parallel to a 200 2 cos(250t)V supply as shown.
(a)
Explain very briefly the physical meaning of apparent power S, reactive power Q and average power P.

(b)
Show that Q of load A is 1.02kVAR.

(c)
If load A is a resistance R in parallel with an element X, find R and X.

(d)
Find the magnitude of current I in Arms. Find also I(t). (35)


I


Load A 1000W supply 0.7lag

[a]
P = real (or average) power dissipated by load S = power supplied by source to load Z Q = maximum reactive power stored in Z
(21)
2. Refer to the circuit in Question 1, a load B is now connected in parallel to load A.
(a)
If P of the combined load is 4kW, find the P of load B.

(b)
If Q of the combined load is 0.02kVAR inductive, find the Q of load B.

(c)
If power factor PF of the combined load is 1, find the element and value of load B. Find also the magnitude of current in load B in Arms.

(21)


V 2 2002
P == = 1kW
[c] RR
2002
R == 40.
1k
V 2 2002
Q == = 1.02kVAR
L 2 (50)L X = L = 124mH
P 1k
[d]
S = == 1.43KVA
cos 0.7
S 1.43k
I == = 7.14A
V 200
I(t) = 7.14
2 cos( 2 50t . 45.6o )A
[b]
If total Q = 0.02kVAR(L), then Q of load B = 1kVAR(C) or C1kVAR
supply

[a]
If total P = 4kW, then P of load B = 3kW
[c]
Q = 1.02kVAR(C) = V 2C Q 1.02k
C = 2 = 2 = 0.082mF
V 200 ( 2 50 )
Q 1.02k
I == = 5.1Arms
V 200
(20)
3. In the circuit, if V1(t) = 4 cos (4k t + 90o) V, use complex method and find V2 (t) . The voltage controlled voltage source is in volt and equal to 3V2 (t) . (20)
2.
V2 0.5mH
3V2

V1
0.25mF
V(t) = 4cos( 4kt + 90o )V .V = 490o = 4 j jL = j(4k)0.5m = 2j. 11
= =. j.
jC j( 0.25m)4k
20A
i(t)


t
[a]
. 20o
i(t) leads v(t) by 20o
I = 2040oA
[b]
V = 4020oV


[c] Capacitive element (C in parallel with R)
V 4020oV
Z == o
I 2040 A = 2. 20o .
2j2.
V2
4j

3V2
. j
4 j .V2 V2 V2 . 3V2
=+
2 j . j 2
4 j .V =.2V . 2V ( j)

2 22
4 j = V2(1. 2 . 2 j) = V2(.1. 2 j) 4 j 4 j
V = =.
2 .1. 2 j 1+ 2 j 490o 4
=.
=.26.6oV 563.4o 5
. 4 o
V2(t) = cos(4kt + 26.6 )V 5
(30)
11
Y =20oS =+ jC
2 R
Y = 0.5( cos20o + j sin 20o) 1
= 0.47 + j0.17 =+ jC
R
1
R == 2.13.
0.47
0.17 0.17
C === 0.017F
10
4. In the following circuit, v(t) = 40 cos (10t + 20o) V, i(t) = 20 cos (10t + 40o) A.
(a)
Sketch v(t) and i(t) together. Show clearly the phase angles and amplitudes. Does v(t) lead i(t) ? Find also the phase angle between v(t) and i(t).

(b)
Plot v and i in a phasor diagram.

(c)
If Z is composed of two elements connected in parallel, find the two elements and the values. (30)


i(t)
Z
v(t)



R1 = 1k. R3 = 2k.
15V

Vo
R6 = 1k.
-15V
V1
R4 = 2k.
R2 = 1k.
R5 = 1k.
[a] V.= V+= v V1 . vvv .VO
=+