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(ELEC102)exampast9a.pdf
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i(t)

(35) t < 0, S is at a, v = 4V


\i(0) = 2A


i(t)
2 W
v(t)

2A



\v(0) = 4V
2 W
v(t)

2A

At t = 0, S is switched to b


i(t)
\v() =-10V 2 W
t = CRv(t)

0.1 mF
-10V = (0.1m)(2) = 0.2ms
-t /CR -t / 0.2ms

v(t) = v() +[v(0) + v()] e =-10 +[4 + 10]e V
-t / 0.2ms
-10 -v(t) -10 + 10 -14e -t / 0.2ms
\i(t) == =-7e A
R2
t < 0, S is at a, i = 2A

At t = 0, S is switched to b


i(t)
\i() =-5A 2 W t = L/R
v(t)

10 mH
-10V = 10m / 2 = 5ms

-t/ t-t / 0.2ms


\i(t) = i() + [i(0) -i()]e =-5 + [2 + 5]e A
-t / 0.2ms -t / 0.2ms

\v(t) =-10 -i(t)2 =-10 -[-5 + 7 ]2 =-14e V
1. In the circuit, the switch has been at terminal a for a long time.
At t = 0, the switch is switched to terminal b.

(a)
If X is a 0.1mF capacitor, find i(t) for t . 0. Find also the power
supplied or absorbed by the C10V source at t = 0.


(b)
If X is a 10 mH inductor, find v(t) for t . 0.
Given that i(t) = i() + [i(0) C i()] e Ct / t and v(t) = v() + [v(0) C
v()] e Ct / t (35)





11
= =-2j
jwC1


j(1k) m

2 jwL = j(1k)10m = 10j
v(t) = 80cos(1kt)V . V = 80


a
80
2W

i Voc

14W
-2j 10j 4i

b
Q4i = 80 + i(14 + 10j) -80 -8

\i ==
10 + 10j 1 + j
-2j

\Voc = 4i
2 -2j
-j -8 -j 32j

= 4i = 4 == 16j
1-j1+ j1 -j2 \Voc(t) = 16cos(1000t + 90o)V
Hence Voc(t) leads v(t)


a Isc

4i
b

4i
\Isc == 2i
2
-8 11-j



= 2 =-16
1 + j1+ j1-j 1-j -45
=-16 =-8(1 -j) =-8
2
2 -8
\Isc(t) =
cos(1000t -45o)A
2
2. In the circuit, v(t) = 80 cos (1k t) V. Find Voc(t) and Isc(t) at terminals ab. Does Voc(t) lead V(t)? (33)

(33)

3. In the circuit, load A is 50 kW at 0.866 lagging power factor. Load B is 100 kW at 0.5 leading power factor.
(a)
Find the total apparent power S, reactive power Q, average power P and power factor PF of the combined load (load A and B).

(b)
If a load X is connected to terminals ab to make the total power factor = 1, find the element and value of load X. Find also the new load current I in Arms. (33 )


Load A

For load A and B total P = 50k + 100k = 150kW total Q = 28.87k -173.2k = -144.33 kVAR
22
2 2

total S =
P + Q = 150k + 144.33k = 208.16kVA P 150k
total PF = == 0.72leading
S 208.16k
2000o Vrms 50Hz

For load A


P = 50kW Q = Ptan q= 50k tan(cos-1 0.866) = 28.87kVAR (L)
For load B
P = 100kW Q = Ptanq= 100k tan(cos-1 0.5) = 173.2kVAR(C)

Add Load X (inductance L)

V2

\QL =
wL
V2 2002

\L == = 0.88mH
wQL 2p50(144.33k)

New I

S 150k
I == = 750A V 200 rms





i(t) =
2 coswtmA




(30)
Vo(t)
Vo 0.707Vo


1k

120k w (rad/s)
wO = 120krad /s BW = 1krad /s BW
w2 =wO += 120.5krad / s
2 BW
w1 =wO -= 119.5krad / s
2 w 120k
Q = O == 120
BW 1k
Vo is maximum at resonance,
hence network is parallel LCR

R

\Q = wOL
\R = QwOL = 120(120k)0.4m = 5.76kW
11
\C == = 0.174mF
wO2L (12