l. The superposition thborem can only be applied to circuits that are fil) linear. (3) nonlinear. (5) inverted. (2) bilateral. (4) unilateral. 2. ln the circuit of Fig. 84 find the currenr through R,...

1 answer below »
l. The superposition thborem can only be applied to circuits that are
fil) linear. (3) nonlinear. (5) inverted.
(2) bilateral. (4) unilateral.
2. ln the circuit of Fig. 84 find the currenr through R, by the superposition method. (Hint: Fig. 60)
(l) /r : 4.16 A (3) /r = 0.375 A
(2) I,: 0.416 A (4) It = 3.75 A
Rl
30n
R2
25.n
20v
n
Fig. 84
3. In the circuit of Fig. 84 find the current
tion method.
(1) /3 : 1.16 A
(2) It = 0.116 A
(1) analogs of each other.
* (2) duals of each other.
5. In the circuit shown in
minalsaandbis
{(2) 4.67 mhos. (4)
7. Given the values in Question 6, what
circuit of Fig. 8(bX
(1) 0.12 mho (3) 0.5 mho
dd2) 0.25 mho (4) 3 mho.s.
4 : 0.021 A
/s = 0.883 A
through fts by the superpo$r-
(s) 0.771 A
(3) equivalent to each other.
(4) variations of each other.
Fig. 85, the total conductance between ter2.1 mhos. (5) 21 rnhos.
2.14 mhos. (6) 21.4 mhos.
(3)
(4)
4. Kirchhoffs current law and Kirchhoffs voltage law are
\ t'lh+F'6
Fig. 85
6. The voltage source in Fig. 86 has the parameter values E, = 12 V and
R" : 40. What is the value of the equivalent current source {r?
(1) r.2 A X(3) 3 A (s) 12 A ,.
(2) 2A (4)4A (6)4FA t
is the conductance G, in the
o't+ D'V+
O'N
t'
,6,v
..NIE
)
(5) 4 mhos
(6) 25 mhos
8. If the current source shown in Fig. 8(b) has the value I, : 14 A and if
G, : 0.2 mho, what is the value of the equivalent constant-voltage
source E" in Fig. 8(aX
v1
I
v
lrL
' (o)
l ,
.ry h
I
'*'
tY, o'L"l
""(ru
Fig. 86
80 (1) 0.7 v
(2) 1.4 v
(3) 2.8 v (5) 14 v
(4) 7 v y(6) 70 v
9. Given the values in Question 8, what is the value of R" in Fig. 86(a)?
(l)0.20 (3)20 (5)70
(2) 0.5 0 (4) 5 o (6) 28 o
J 10. In the circuit of Fig. 87 which of the following equations is not
correct? (Hint: check by using 15v battery and 10o resistors)
*5)t(
Sl'(l) Iz: Ir - It ($ f
:
*
,r\ 12 R3 ') E: I,
, E _ I,R
(5) 1: :
u -'t^'t 9 '
r12
(4) I' _
E - /rRr -,
R3
R1 " i0P
\6{ t to)L
Fig. 87
Jtt. Figure 88 shows the equivalent circuit of an FET amplifier. What is
the output voltage Eo of this stage? (Hint: Topic l)
(t) G^EG, 6 %ES Ups*U2 *(rffi
In the circuit of Fig. 88, what is the current through
tance G1?
t2. the load conduce-- EG ^\
V--'Y'S b5 t hL)
!"nL'''
/ ,^

121',) r
Fig. 88
13. In the circuit of Fig. 89(a) furd the current through R3 with the switch in
position 1. Assume a perfect diode (zerc forward resistance and infrnite
back resistance). (Hint: See exarnple 2 in Topic 16. Short out the 60V
supply and find the current through R3, observing the polarity. Next,
short out the 40V supply and again find the current, noting the direction.)
+
(I) Zero
(2) 0.2s mA
(3) 0.7s mA
lokn 30 krI
+60 V -4OV
1r7,.rz!G"
17y GtG^Eo,
\-" Gt * Gos
rJ/ E:ffi
(4\ G tEe
,
\'/ G^ * Gt + Gog
,
f.t, ^rt
tr4ffi1
\J) v2n6r{JDS
(6) G^GrGo, {\Y h
5ko
2 7.5 kfr
o-^AA+-2Ov
I
-40 v
R3
Fis. 89
14. In the circuit of Fig. 89 (b) find the current through the 5-kcl fesis!9r with the
switch in positioni. lchoot".your answer from the choices givenfor
Question 13.)
84
15 mA
d-c
Fig.93 '
Questions 17 to 19 refer to Fig. 93.
17. What value of current flows through R5? (tlint: Topic 16 example 3)
(1) 6 mA (3) 19 mA (s) 33 mA
(2) 12 mA (4) 27 n1A (6) 39 mA
18. What is voltage .E o?
(i) -3.e v (4) {.6 v
(2) -3.3 V (5) +0.6 V
(3) -1.2 V (6) +1.2 v
(7) +1.9 V
(8) +3.3 v
i9. How much power is dissipated on Rr?
(1) 5.8 mW (2) 10.3 mW (3) 28.9 mw (4) 60 mW
Questions 20 to 23 rcfer to Fig. 94. (Hint App. Ex. 11, problems 1 and 2)
20. What is the value of curent flowing through R1?
(1) 0 mA (3) 240 mA (s) 660 mA
(2) 180 mA (+) 300 mA (6) 790 mA
21. What is the direction of curent through R,?
(1) Downward *(2) Upward
Rt =5 n'
6V
R+=2o nRz =lOO
E -
f(2) UPward
Rz =lo n'
Rs=20 n'
Fig.94
22. Whatis the current through ^Rr? (I) 0 mA (3) 180 mA
@ ls mA @) 3rs mA
(s) 6ts mA
(6) 770 mA
23. What is the rtiqection of current through ^Rr? g (t) Downward (t- U;**d
k
Answered Same DayDec 20, 2021

Answer To: l. The superposition thborem can only be applied to circuits that are fil) linear. (3) nonlinear....

David answered on Dec 20 2021
133 Votes
Solution 2:
Taking source voltage 25:
Hence net resistance is:
Current in R1 is




Taking source 15 V
Hence net resistance is:
Current due to source is:




Current in R1 is



(


)
The direction of this current is same as IR11
Taking source 20 V
Hence net resistance is:
Current due to source is:




Current in R1 is



(


)


The direction of this current is opposite to IR11
Hence net current is:




Hence part 3
Solution 3:
Current trough R3 by source 25 V:












Current trough R3 by source 15 V:

...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here