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(ELEC101)[200X](s)quiz~1270^_10268.pdf
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(26)
1k.
a
Rth = 2k.
20/3 mA
a
20/3 mA
2k.
1k.
Voc
b
VOC = 20V
2k.
20V
VOC 20V
ISC === 10mA
RTH 2k.
V1 + 6V 12V
1k.
4mA
V1 + 6V + 12V + 2V
6V
2mA
V = V . V = 20V
OC ab
1. (a) Show that the Thevenin resistance at terminals ab is 2k. .
(b) If Io = 20
mA
3 , find the Thevenin voltage at terminals ab. (c) If the load is changed from 1k. to 2 k. , find the new Io. (d) Find the short circuit current Isc at terminals ab (from a to b). (26)
12V
1k.
4mA
a
6V
2mA
IO == 5mA
4k.
(16)
12. 2.
VOC
4.
4.
VO = 2V
12. VOC 6.
12V
2. In the circuit, use Thevenins Theorem to find Vo. (15)
12V +12V
VOC = *12..12V
12. 2.
18.
VO
= 4V
4.
12.Rth = 6. //12.= 4.
VOC 2.
4.
(13)
I(t)
10mA
5mA
0
2ms 3ms 4ms
V(t) dI(t)
v(t) = L
30mV
dt
10mA
= 6mH *
0
2ms
-30mV
= 30mV
4ms
2ms 3ms
L*i26mH* (10mA)2
EL == = 300nJ
22
3. In the circuit, if X = 6mH and I(t) is given as shown.
Determine and label clearly the waveform of the voltage v(t). Find also the maximum energy stored in X. (13)
I(t)I(t)
10mA
+
v(t)
X -
4ms
(14)
18V 18V
2m
-12V -30V
Io1
12k.6k.
1m
5m
18V
IO == 1mA
1 18k.
6k. 0V
4(a). Use Superposition Method to show that IO = 5mA. (14)
18V
12k.
IO1 = 6mA * = 4mA
18k.
IO
IO = 5mA
12k.
6k.
(20)
. 30V . VO VO + 12V
= 6mA +
6k. 12k. . 60V . 20VO = 72V + VO + 12V VO =.48V
P = 6mA* 48V = 288mW Power is delivered
. 30V + 48V
I == 3mA
16k.
18V
2m
-12V
-30V
1m
5m
6k.
0V
4(b). Find Vo .
Find also I1 and the power absorbed or delivered by the 6mA source. (20)
18V
6k. 12k.
Vo
IO
6k.
(10)
5 4A 4. 20V 2Io 2Io + 4 0)4*(2Io2204 =.+.+.. V* Io A. VIo VV)Io( 1 58 12 082044 = . = =+. .. + Io 5. In the circuit, find Io . (12) 4A -14V 20V 3A 6V
20V
(21)
I2
4.
V2
(6a)
R
I2
R = 0
R =
8A
16V V2
Frequency when V
measured = 0.707 * V
(b)
measured at low
frequency by meter
(c) 50W
6 (a) Plot I2 versus V2 for all values of R (R = 0.to R = .). .
(b)
Explain briefly the meaning of bandwidth for a meter.
(c)
Two 200V 100W light bulbs are connected in parallel to a 100V supply. Find the total power dissipated by the light bulbs. (21)
(14)
4. 2.
.= 4R N Vo Io
4. 2.
ISC = 4A
Io
Vo
Norton equivalent
RN = 4.
ISC = 4A
4.
I2 = 2A
7. In the circuit, Io and Vo are unknowns. If R = 0. , I1 = 4A .
If R is changed to 4., find the new I1 . (Hint: find Norton equivalent) . (14)
4. 2.
IO
VO
4. 2.
VO
IO
I1
(20)
2.
12.
VO
4.
4.
24V
6.
VO
2mA
16Vo
VO VO
4.
8Vo/3 -2A
VO =.(8VO /3 . 2