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(ELEC102)test2past5.pdf
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(23)


Z = 25 (cos -20o + j sin -20o) .

i(t) v(t) = 250 cos (100t + 20o) V = 23.5 - j8.55 .
= R C j(1/
C)


i(t) = 10 cos (100t + 40o) A
Z
v(t) (4)
R = 23.5 .


(3)
Z =
25
.
20o .

.
20o
1. In the following circuit, v(t) = 250 cos (100t + 20o) V, i(t) = 10 cos
i(t) leads v(t) (2)
(100t + 40o) A.
(a)
Sketch v(t) and i(t) together. Show clearly the phase angles and amplitudes. Does v(t) lead i(t) ?

(b)
If Z is two elements in series, find the two elements and the values.


Capacitive element (C in series with
(c) Sketch impedance Z and phasors V, I in a complex plane. (23)
R)

C =
1 =
1 =
1.17mF (5)
8.55
8.55(100)
I =
10
40oA V =
250
20oV


i(t)
o

V 250
20 V


Z =
=
(3)
I 10
40oA
=
25
.
20o .



(31)

2
When L is connected (PF = 1), S = P = 1174.62VA, hence 10 (3)

P =
VI cos
=
(250)( )cos(20) =
1174.62W
S 1174.62
2


I =
=
=
4.7 2 Arms
(5)
10 V 250/2

Q =
VIsin
=
(250)( )sin(.
20) =
.
427.53VAR (3)
2

S =
VI =
(10)(250) =
1250VA (3)
2 If V is 500V (2x of original V)

o I is 2x of original I



PF =
cos(.
20 ) =
0.94 leading (3)
but R and C are unchanged hence P andQ (
I 2 ) are 4x of original P and Q Hence new L
V 2 / Q = original L = 0.73H
(5)
P = real (or average) power dissipated by load Z
2. Using the same circuit in question (1),
(a) find the apparent power S, reactive power Q, average power P and S = power supplied by source to load Z power factor PF of load Z.
(b) What are the physical meanings of S, P, Q?
(4)
Q = maximum reactive power stored in Z (c) A load B is now connected in parallel to Z to make the total power factor = 1. Find the load B and the value. Find also the current (in rms) supplied by v(t) when PF is 1.
(d) If v(t) is now changed to 500 cos (100t + 20o) V, find the new load
Connect L in parallel to improve PF = 1
B required.
(31)
V2
(250/ 2)2


L =
=

=
0.73H
(5)
OQ 100(427.53)

(37)


Q 35.355

C =
=
=
2.5
F (3)

OR 14142(1k)
max VO =
iR =
1m(1k) =
1Vrms (3) max VO
14142
max iC =
=
IR ( j
OC) =
jIQ
1/j
OC
(rad/s)



O =
14142rad/ s
max iC(t) =
35.355(
2)cos(14142t +
90o)mA (6)

O (3)

fO =

=
2250.8Hz )
(Vrms

2
If R is changed to 2k

BW =
400rad/s (2)

BW (2)



2 =

O +
2 =
14142 +
200 =
14342rad / s
BW (2)



1 =

O .
=
14142 .
200 =
13942rad / s 14142
(5)2
(rad/s)

O 14142


Q =
=
=
35.355 (3)
3. A LCR circuit has the following resonance curve (magnitude of Vo(t)
BW 400

versus frequency ).
(a) Find the resonant frequency (in rad/s and in Hz), bandwidth (BW),
(2)

upper and lower cut-off frequencies (in rad/s), and the Q-factor of the
Q > 10, good resonant circuit