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Solutions to FE Problems
Chapter 2
2FE-1 What is the power generated by the source in the network in Fig. 2PFE-1?
The correct answer is b.
V120
SR
1R 2R
3R
R !R
a
bc
"esistorsR1# R2# and R3are connected in de$ta. % de$ta-wye trans&or'ation can be conducted.
=++
=++
= kkkk
kk
RRR
RRRa 3
121()
*1(+)
321
21
=++
=++
= kkkk
kk
RRR
RRRb )
121()
*12+1(
321
32
=++=++= k
kkkkk
RRRRRRc 2
121()*12+)
321
31
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V120
SR
cR bR
R !R
a
bc
aR
*,))+*2-+3!*,+*-+ ! kkkkkkRRRRRRR bcaseq ++++=++++=
=+
+= kkk
kkkReq 12
12)
*12+)(
WkR
V
P eq
s
Vs 2.112
*120+ 22
===
2FE-2 Find Vabin the circuit in Fig. 2PFE-2.
The correct answer is d.
A
1!
10 10
!
ab
2I1I
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( ) AI !.11!10!10
!101 =
++++
=
( ) AI !.21!10!10
1!102 =
++++=
VV !.12*!.2+!! ==
VV !.22*!.1+1!1! ==
VVab 10!.22!.12 ==
2FE-3 & Req/10.(in the circuit in Fig. 2PFE-3# what is R2?
The correct answer is a.2*(+ 2 ++= RReq
)(
(*+
2
2 ++
=R
RReq
(
()(.10
2
2
+
=R
R
.3(2.3 2 =R
=122R
2FE- Find the euia$ent resistance o& the circuit in Fig. 2PFE- at the ter'ina$s %-.The correct answer is c.
)R
1R
3R!R
R
2R
RABR
[ ][ ]{ } *+*+ 213!) RRRRRRRRAB +++=[ ][ ]{ } *)12+)122 kkkkkkRAB ++=
++
+
+=
kk
kkkk
kk
kkRAB
)12
*)+12)
122
*12+2
kkk
kkkkkRAB
)12
*)+12*)12+ +
+=+=
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= kRAB (
2FE-! The 1004 source is absorbing !0W o& power in the network in Fig. 2PFE-!. What isR?
The correct answer is a.
A!
10
V100
R
V200
101I2I
A
1VIP=
1100!0 I=
AI !.01 =
546 around the right $oop7
20010100 12 +=+ IRI586 at node %7
12! II +=AI !.!!.0!2 =+=
200*!.0+10*!.!+100 +=+ R
= 2.1R
2FE-) Find the power supp$ied by the 04 source in the circuit in Fig. 2PFE-).The correct answer is b.
A3
!0
V100
1I
2I3I I
V099.9
100*2!20+=eqR
=+
=+
= 101001111.11
*100+1111.11100
2!20
*2!+20eqR
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5867 213 III =+
123 III =
5467 I!00100 +=)0!0 =I
AI!
)=
31 =II
AI (.13!
)1 ==
AI (.!(.110
03 =+=
WVIP V 232*(.!+030 ===
2FE- What is the currentIoin the circuit in Fig. 2PFE-?
The correct answer is d.
SV
2R R
1R
!R
3R
)R
R
SI
oI
I
=+= kRRRA 921
=+
++=++= kkk
kkkkRRRRRB 1(
)3
*)+312*+ )!
=+
== kkk
kkRRR BAC )
1(9
*1(+9
mAkkRR
VI
C
s
s 1))
12
3
=+
=+
=
( )s
BA
A IRR
RI
+
=
( ) mAmkk
kI
3
11
1(9
9 =
+=
( )
+=
+= m
kk
kI
RR
RIo
3
1
)3
)
)
mAIo 22.0=
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2FE-( Find the o$tage Voin the network in Fig. 2PFE-(.
The correct answer is c.
SI
3R
1R
2R
R
oI
!R
)R
+
oV
=+= kRRRA 21
=+
== kkkkkRRRB )12*)+12
)!
*+3
s
BA
Ao I
RRRR
RI
+++=
mAmkkkk
kIo )*2+
)2
=
++=
VkmRIV oo 3)*)+) ===
2FE-9 What is the o$tage Voin the circuit in Fig. 2PFE-9?
The correct answer is a.
2R
A21R
3R
oI
+
oV
ARRR
RI
o 3
2*2+
312
2*2+
321
1
=
++=++=
VVo 2*3+3
2==
2FE-10 Find the current : in Fig. 2PFE-10.
The correct answer is d.
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xI
V12
3R1R
2R!R
R
)R
[ ][ ]{ } 123)! *+ RRRRRRReq +++=[ ][ ]{ } 13110*2(+ +++=eqR
=+= 31*3)+eqR
AR
Ieq
s 3
1212===
[ ] =++= )*+; 3)! RRRRR
( ) AIx3
(
9
2
)3
)==
+=
