### Solutions to FE Problems Chapter 2 2FE-1 What is the - KU Leuven

```Solutions to FE Problems
Chapter 2
2FE-1 What is the power generated by the source in the network in Fig. 2PFE-1?
a
RS
R2
R1
120V
c
b
R3
R4
R5
Resistors R1, R2, and R3 are connected in delta. A delta-wye transformation can be conducted.
R1 R2
6k (18k )
Ra =
=
= 3kв„¦
R1 + R2 + R3 6k + 18k + 12k
R2 R3
18k (12k )
=
= 6kв„¦
R1 + R2 + R3 6k + 18k + 12k
R1 R3
6k (12k )
Rc =
=
= 2kв„¦
R1 + R2 + R3 6k + 18k + 12k
Rb =
a
RS
Ra
120V
Rc
Rb
c
b
R4
Req = Rs + Ra + [( Rc + R4 ) ( Rb + R5 )] = 5k + 3k + [(2k + 4k ) (6k + 6k )]
6k (12k )
= 12kв„¦
6k + 12k
(120) 2
=
= 1.2W
12k
Req = 8k +
PVs =
Vs2
Req
2FE-2 Find Vab in the circuit in Fig. 2PFE-2.
I1
10в„¦
b
15в„¦
I2
10в„¦
a
4A
5в„¦
R5
10 + 5
пЈ«
пЈ¶
I1 = пЈ¬
пЈ·( 4 ) = 1.5 A
пЈ­ 10 + 5 + 10 + 15 пЈё
10 + 15
пЈ«
пЈ¶
I2 = пЈ¬
пЈ·( 4 ) = 2.5 A
пЈ­ 10 + 5 + 10 + 15 пЈё
V5в„¦ = 5(2.5) = 12.5V
V15в„¦ = 15(1.5) = 22.5V
Vab = 12.5 в€’ 22.5 = в€’10V
2FE-3 If Req=10.8в„¦ in the circuit in Fig. 2PFE-3, what is R2?
Req = ( R2 8) + 4 + 2
( R2 )8
+6
R2 + 8
8 R2
10.8 в€’ 6 =
R2 + 8
3.2 R2 = 38.4
R2 = 12в„¦
Req =
2FE-4 Find the equivalent resistance of the circuit in Fig. 2PFE-4 at the terminals A-B.
R1
R6
R2
R3
R5
R AB
R7
R4
R AB = { [[( R6 + R7 )
R5 ] + R4 ]
R AB = { [[24k 12k ] + 4k ]
R3 } + ( R1 R2 )
6k } + (12k 6k )
пЈ± пЈ® 24k (12k )
пЈј пЈ« 12k (6k ) пЈ¶
пЈ№
R AB = пЈІ пЈЇ
+ 4k пЈє 6 k пЈЅ + пЈ¬
пЈ·
пЈ»
пЈі пЈ° 24k + 12k
пЈѕ пЈ­ 12k + 6k пЈё
R AB = (12k 6k ) + 4k =
12k (6k )
+ 4k
12k + 6k
R AB = 8kв„¦
2FE-5 The 100V source is absorbing 50W of power in the network in Fig. 2PFE-5. What is R?
A
10в„¦
I1 10в„¦
I2
R
100V
5A
200V
P = VI 1
50 = в€’100 I 1
I 1 = в€’0.5 A
KVL around the right loop:
100 + I 2 R = 10 I 1 + 200
KCL at node A:
5 = I 2 + I1
I 2 = 5 + 0.5 = 5.5 A
100 + (5.5) R = 10(в€’0.5) + 200
R = 17.27в„¦
2FE-6 Find the power supplied by the 40V source in the circuit in Fig. 2PFE-6.
I3
I
I1
I2
40V
9.99в„¦
50в„¦
3A
100V
Req = (20 25) 100
20(25)
11.1111(100)
= 10в„¦
100 =
20 + 25
11.1111 + 100
KCL: I 3 + I 1 = I 2
I 3 = I 2 в€’ I1
KVL: 100 = 40 + 50 I
50 I = 60
6
I= A
5
I1 = I в€’ 3
6
I 1 = в€’ 3 = в€’1.8 A
5
40
I3 =
+ 1.8 = 5.8 A
10
P40V = VI 3 = 40(5.8) = 232W
Req =
2FE-7 What is the current Io in the circuit in Fig. 2PFE-7?
R2
I
R4
R5
VS
IS
R1
R3
R6
R7
Io
R A = R1 + R2 = 9kв„¦
RB = R4 + R5 + ( R6 R7 ) = 4k + 12k +
9k (18k )
= 6kв„¦
9k + 18k
Vs
12
Is =
=
= 1mA
R3 + RC 6k + 6k
RC = R A RB =
пЈ« RA
I = в€’пЈ¬пЈ¬
пЈ­ R A + RB
пЈ¶
пЈ·пЈ·(I s )
пЈё
3k (6k )
= 18kв„¦
3k + 6k
1
пЈ« 9k пЈ¶
I = в€’пЈ¬
пЈ·(1m ) = в€’ mA
3
пЈ­ 9k + 18k пЈё
пЈ« R7 пЈ¶
6k пЈ¶пЈ« 1 пЈ¶
пЈ·пЈ·(I ) = пЈ«пЈ¬
I o = пЈ¬пЈ¬
пЈ·пЈ¬ в€’ m пЈ·
пЈ­ 3k + 6k пЈёпЈ­ 3 пЈё
пЈ­ R6 + R7 пЈё
I o = в€’0.22mA
2FE-8 Find the voltage Vo in the network in Fig. 2PFE-8.
R3
R2
R1
IS
Io
R4
+
Vo
в€’
R5
R6
R A = R1 + R2 = 4kв„¦
12k (6k )
RB = R5 R6 =
= 4kв„¦
12k + 6k
RA
Io =
(I s )
R A + R3 + R4 + RB
4k
Io =
(24m) = 6mA
4k + 2k 6k + 4k
Vo = I o R4 = 6m(6k ) = 36V
2FE-9 What is the voltage Vo in the circuit in Fig. 2PFE-9?
R2
2A
R1
Io
+
R3
Vo
в€’
R1
2 пЈ¶
2
пЈ«
(2) = пЈ¬
пЈ· ( 2) = A
R1 + R2 + R3
3
пЈ­ 2 +1+ 3 пЈё
2
Vo = (3) = 2V
3
Io =
2FE-10 Find the current Ix in Fig. 2PFE-10.
R1
12V
Ix
R2
R3
R4
R6
Req = { [[( R5 + R6 )
R4 ] + R3 ]
Req = { [[(8 + 2) 10] + 1]
3} + 1
Req = (6 3) + 1 = 3в„¦
Is =
12 12
=
= 4A
Req
3
R' = [( R5 + R6 )
R4 ] + R3 = 6в„¦
24 8
пЈ« 6 пЈ¶
Ix = пЈ¬
= A
пЈ·(4) =
9 3
пЈ­3+ 6пЈё
R2 } + R1
R5
Solutions to FE Problems
Chapter 3
3FE-1 Find Vo in the circuit in Fig. 3PFE-1.
A
2в„¦
6в„¦
1в„¦
I1
I
12V
+
2в„¦
Vo
в€’
KCL at node A: I 1 = I + I 2
I = I1 в€’ I 2
KVL around the left loop:
12 = 2 I 1 + 1( I ) + 2 I
12 = 2 I 1 + 1( I 1 в€’ I 2 ) + 2( I 1 в€’ I 2 )
12 = 5 I 1 в€’ 3I 2
KVL around the right loop:
6 = 6 I 2 в€’ 2 I в€’ 1( I )
6 = 6 I 2 в€’ 3( I 1 в€’ I 2 )
6 = 9 I 2 в€’ 3I 1
Two equations and two unknowns:
12 = 5 I 1 в€’ 3I 2
6 = в€’3 I 1 + 9 I 2
Therefore, I 1 = 3.5 A and I 2 = 1.833 A
V0 = 2 I = 2( I 1 в€’ I 2 ) = 2(3.5 в€’ 1.833)
V0 = 3.33V
I2
6V
3FE-2 Determine the power dissipated in the 6-ohm resistor in the network in
Fig. 3PFE-2.
V1
V2
4в„¦
I1
12V
V1 = 12V
KCL at node 2:
V2 в€’ V1 V2 V2
+
+
= 2I1
4
6 12
V в€’ V2
I1 = 1
4
V2 в€’ V1 V2 V2
пЈ« V в€’ V2 пЈ¶
+
+
= 2пЈ¬ 1
пЈ·
4
6 12
пЈ­ 4 пЈё
3V2 в€’ 3V1 + 2V2 + V2 = 6V1 в€’ 6V2
12V2 = 9V1
12V2 = 9(12)
V2 = 9 V
P6 в„¦ =
V22 9 2
=
= 13.5W
6
6
6в„¦
text
2I1
12в„¦
3FE-3 Find the current Ix in the 4-ohm resistor in the circuit in Fig. 3PFE-3.
12V
3в„¦
I1
6в„¦
2A
+
Vx
I2
4в„¦
в€’
I2 = I x + I3
I x = I2 в€’ I3
I1 + 2 = I 2
I 2 в€’ I1 = 2
12 = 6 I 1 + 4 I x
12 = 6 I 1 + 4( I 2 в€’ I 3 )
12 = 6 I 1 + 4 I 2 в€’ 4 I 3
KVL around the right loop:
2V x + 4(в€’ I x ) + 3I 3 = 0
в€’ 2V x = в€’4( I 2 в€’ I 3 ) + 3I 3
в€’ 2(4 I x ) = 4( I 3 в€’ I 2 ) + 3I 3
в€’ 8( I 2 в€’ I 3 ) = 4 I 3 в€’ 4 I 2 + 3I 3
8 I 2 в€’ 8 I 3 + 4 I 3 в€’ 4 I 2 + 3I 3 = 0
4I 2 в€’ I 3 = 0
I 3 = 4I 2
12 = 6 I 1 + 4 I 2 в€’ 4(4 I 2 )
12 = 6 I 1 в€’ 12 I 2
Two equations and two unknowns:
в€’ I1 + I 2 = 2
6 I 1 в€’ 12 I 2 = 12
Therefore, I 1 = в€’6 A and I 2 = в€’4 A .
I 3 = 4(в€’4) = в€’16 A
I x = в€’4 + 16
+
2V x
-
Ix
Vx = 4I x
I3
I x = 12 A
3FE-4 Determine the voltage Vo in the circuit in Fig. 3PFE-4.
12V
4в„¦
I1
4в„¦
2в„¦
Ix
KCL: I 1 = I x + I 2
I x = I1 в€’ I 2
KVL around the left loop:
12 = 4 I 1 + 2 I x
12 = 4 I 1 + 2( I 1 в€’ I 2 )
12 = 6 I 1 в€’ 2 I 2
KVL around the outer loop:
12 = 4 I 1 + 4 I 2 + 4 I 3
I 2 = 2I x + I 3
2I x = I 2 в€’ I 3
2( I 1 в€’ I 2 ) = I 2 в€’ I 3
I 3 = в€’2 I 1 + 3 I 2
12 = 4 I 1 + 4 I 2 + 4(в€’2 I 1 + 3I 2 )
12 = в€’4 I 1 + 16 I 2
Two equations and two unknowns:
12 = 6 I 1 в€’ 2 I 2
12 = в€’4 I 1 + 16 I 2
Therefore, I 1 = 2.45 A and I 2 = 1.36 A .
I 3 = в€’2(2.45) + 3(1.36)
I 3 = в€’0.82 A
V o= 4 I 3 = 4(в€’0.82)
V o= в€’3.28V
I3
I2
text
2I x
+
4в„¦
Vo
в€’
3FE-5 What is the voltage V1 in the circuit in Fig. 3PFE-5?
I1
V1
1в„¦
V2
I2
V3
2в„¦
10V
3в„¦
15V
8A
4A
KCL at the supernode: 4 = I 1 + 8 + I 2
I 1 + I 2 = в€’4
V1 V2 в€’ V3
+
= в€’4
4
2
V1 + 2V2 в€’ 2V3 = в€’16
V3 = 15V
V1 + 2V2 в€’ 2(15) = в€’16
V1 + 2V2 = 14
V2 в€’ V1 = 10
Two equations and two unknowns:
в€’ V1 + V2 = 10
V1 + 2V2 = 14
Therefore, V1 = в€’2V and V2 = 8V .
Solutions to FE Problems
Chapter 4
4FE-1 Given the summing amplifier shown in Fig. 4PFE-1, what value of R2 will
produce an output voltage of вЂ“3V?
пЈ« в€’ R2 пЈ¶
пЈ« R пЈ¶
Vo = пЈ¬
пЈ·(4) в€’ пЈ¬ 2 пЈ·(в€’2)
пЈ­ 4k пЈё
пЈ­ 12k пЈё
Vo = в€’3V
1
в€’ 3 = в€’1m( R2 ) + m( R2 )
6
5
в€’ m ( R2 ) = в€’3
6
R2 = 3.6kв„¦
4FE-2 Determine the output voltage Vo of the summing op-amp circuit shown in Fig.
4PFE-2.
36kв„¦
18kв„¦
-
V1
+
6kв„¦
+
6kв„¦
12 kв„¦
12kв„¦
2V
3V
1V
пЈ« 18k пЈ¶
пЈ« 18k пЈ¶
V1 = в€’2пЈ¬
пЈ·(4) + 1пЈ¬
пЈ· = в€’4.5V
пЈ­ 6k пЈё
пЈ­ 12k пЈё
Vo
пЈ« 36k пЈ¶ пЈ« 36k пЈ¶
V0 = в€’V1 пЈ¬
пЈ· в€’ 3пЈ¬
пЈ·
пЈ­ 6k пЈё пЈ­ 12k пЈё
пЈ« 36k пЈ¶ пЈ« 36k пЈ¶
V0 = 4.5пЈ¬
пЈ· в€’ 3пЈ¬
пЈ·
пЈ­ 6k пЈё пЈ­ 12k пЈё
V0 = 18V
4FE-3 What is the output voltage Vo in Fig. 4PFE-3?
if
2в„¦
i=0
-
Vo
+
3в„¦
4в„¦
6V
2V
V1 V2 6 2
+
= + = 2.5 A
R1 R2 3 4
V0 = в€’i f R f = в€’(2.5)(2)
if =
V0 = в€’5V
4FE-4 What value of Rf in the op-amp circuit of Fig. 4PFE-4 is required to produce a
voltage gain of 50?
The op-amp is a noninverting op-amp.
Rf
A = 1+
R1
R f = ( A в€’ 1) R1
R f = (50 в€’ 1)5k = 245kв„¦
4FE-5 What is the voltage Vo in the circuit in Fig. 4PFE-5?
R1 = 6kв„¦
+
+
-
R2 = 2kв„¦
5V
R3 = 1kв„¦
8kв„¦
Vo
2 kв„¦
в€’
The 8kв„¦ and 2kв„¦ resistors make up a noninverting op-amp.
пЈ« 8k пЈ¶
V1 = пЈ¬1 +
пЈ·5 = 25V
пЈ­ 2k пЈё
Use nodal analysis at node A:
Vo Vo в€’ Vi Vo в€’ 25
+
+
=0
R3
R1
2k
Vo Vo в€’ 5 Vo в€’ 25
+
+
=0
1k
6k
2k
6Vo + Vo в€’ 5 + 3Vo в€’ 75 = 0
10Vo = 80
Vo = 8V
Solutions to FE Problems
Chapter 5
5FE-1 Determine the maximum power that can be delivered to the load RL in the network in Fig.
5PFE-1.
I'
R2
R1
R4
+
Voc'
R3
4mA
в€’
Use superposition.
пЈ«
пЈ¶
R1
пЈ·пЈ·(4m)
I ' = пЈ¬пЈ¬
пЈ­ R1 + R2 + R3 пЈё
1k
пЈ«
пЈ¶
I'= пЈ¬
пЈ·(4m) = 1mA
1
k
+
1
k
+
2
k
пЈ­
пЈё
'
Voc = I ' R3 = (1m)(2k )
Voc' = 2V
R1
12V
R2
R4
+
Voc"
R3
в€’
пЈ«
пЈ¶
R3
пЈ·пЈ·(12)
Voc" = пЈ¬пЈ¬
пЈ­ R1 + R2 + R3 пЈё
2k
пЈ«
пЈ¶
Voc" = пЈ¬
пЈ·(12)
пЈ­ 1k + 1k + 2k пЈё
Voc" = 6V
Voc = Voc' + Voc" = 2 + 6 = 8V
R2
R1
R4
R3
RTH = [(R1 + R2 ) R3 ] + R4
RTH =
2 k ( 2k )
+ 1k = 2kв„¦
2k + 2k
RTH = 2kв„¦
Voc = 8V
RL
PLmax = I L2 R L
RL = RTH for maximum power.
2
PLmax
PLmax
пЈ« V пЈ¶
= пЈ¬пЈ¬ oc пЈ·пЈ· RTH
пЈ­ 2 RTH пЈё
V2
82
= oc =
= 8mW
4 RTH 4(2k )
RTH
5FE-2 Find the value of the load RL in the network in Fig. 5PFE-2 that will achieve maximum
power transfer, and determine the value of the maximum power.
+ Vx в€’
2kв„¦
12V
1kв„¦
I
+
Voc
+
в€’
-
2V x
12 = 2kI + 1kI + 2V x
V x = I ( 2k )
12 = 2kI + 1kI + 2(2kI )
12
I = mA
7
12 = 2kI + Voc
пЈ« 12 пЈ¶
12 = 2k пЈ¬ m пЈ· + Voc
пЈ­7 пЈё
Voc = 8.57V
+ Vx в€’
2kв„¦
12V
1kв„¦
I1
I sc
I2
+
-
пЈ« 12 пЈ¶
I 1 = пЈ¬ пЈ· = 6mA
пЈ­ 2k пЈё
2V x
1kI 2 + 2V x = 0
1kI 2 + 2(2kI 1 ) = 0
I 2 = в€’24mA
I 1 = I 2 + I sc
I sc = 6m в€’ (в€’24m) = 30mA
Voc 8.57
=
= 285.7в„¦
I sc 30m
RL = RTH for maximum power.
RTH =
RTH = 285 .7 в„¦
Voc = 8.57V
PLmax =
RL
Voc2
(8.57) 2
=
= 64.3mW
4 RTH 4(285.7)
5FE-3 Find the value of RL in the network in Fig. 5PFE-3 for maximum power transfer to this
Ix
+
R1
I
12V
R2
2I x
text
Voc
в€’
I = I x + 2I x
I = 3I x
12 = 3I x + 12 I
12 = 3I x + 12(3I x )
4
Ix = A
13
пЈ«4пЈ¶
Voc = 12 I = 12(3I x ) = 12(3)пЈ¬ пЈ· = 11.08V
пЈ­ 13 пЈё
Ix
R1
I2
12V
2I x
R2
I sc
text
I sc = I x + I 2 + 2 I x = 3I x + I 2
I2 = 0A
I sc = 3I x
12 = 3I x
I x = 4A
I sc = 3(4) = 12 A
V
11.08
RTH = oc =
= 0.92в„¦
I sc
12
RTH = 0.92в„¦
Voc = 11.08V
RL = 0.92 + 12 = 12.92в„¦
RL = 12.92в„¦ for maximum power transfer.
12в„¦
RL
5FE-4 What is the current I in Fig. 5PFE-4?
2в„¦
I'
3в„¦
2в„¦
4в„¦
10 A
Use superposition.
пЈ« 2 пЈ¶
I' =пЈ¬
пЈ·(в€’10) = в€’4 A
пЈ­ 2 + 3пЈё
I"
3в„¦
20
= 4A
5
I = I' + I"
I = в€’4 + 4
I = 0A
I" =
2в„¦
4в„¦
20V
2в„¦
5FE-5 What is the open circuit voltage Voc at terminals a and b of the circuit in Fig. 5PFE-5?
Use superposition.
a
4в„¦
I
2в„¦
3в„¦
'
12 A
+
Voc'
в€’
b
пЈ« 4 пЈ¶
I' =пЈ¬
пЈ·(12) = 8 A
пЈ­2+ 4пЈё
Voc' = I ' (2) = 8(2) = 16V
a
3в„¦
4в„¦
12V
2в„¦
+
Voc"
в€’
b
пЈ« 2 пЈ¶
Voc" = пЈ¬
пЈ·(в€’12) = в€’4V
пЈ­2+ 4пЈё
Voc = 16 в€’ 4 = 12V
Solutions to FE Problems
Chapter 6
6FE-1 Given three capacitors with values 2ВµF, 4ВµF, and 6ВµF, can the capacitors be
interconnected so that the combination is an equivalent 3ВµF?
The correct answer is a. Yes. The capacitors should be connected as shown.
6Вµ F
2Вµ F
4Вµ F
C eq =
6 Вµ (6 Вµ )
= 3Вµ F
6Вµ + 6Вµ
Ceq
6FE-2 The current pulse shown in Fig. 6PFE-2 is applied to a 1ВµF capacitor. What is the energy
stored in the electric field of the capacitor?
q(t ) = в€« i (t ) dt
пЈ±0 C , t в‰¤ 0
пЈґ
q(t ) = пЈІ6t C , 0 < t в‰¤ 1Вµ s
пЈґ6 Вµ C , t > 1Вµ s
пЈі
q(t )
C
пЈ±0 V , t в‰¤ 0
пЈґ
v(t ) = пЈІ6 x10 6 t V , 0 < t в‰¤ 1Вµ s
пЈґ6 V , t > 1Вµ s
пЈі
v(t ) =
1
C v 2 (t )
2
пЈ±0 J , t в‰¤ 0
пЈґ
w(t ) = пЈІ18 x10 6 t 2 J , 0 < t в‰¤ 1Вµ s
пЈґ18Вµ J , t > 1Вµ s
пЈі
w(t ) =
6FE-3 The two capacitors shown in Fig. 6FE-3 have been connected for some time and have
reached their present values. Determine the unknown capacitor Cx.
The voltage across the unknown capacitor Cx is (using KVL):
24 = 8 + V x
V x = 16V
q = Cv
The capacitors are connected in series and the charge is the same.
q = 60Вµ (8) = 480Вµ C
q 480Вµ
Cx = =
= 30Вµ F
v
16
6FE-4 What is the equivalent inductance of the network in Fig. 6PFE-4?
2 mH
3mH
12mH
Leq
3mH
Leq = [ {[(3m + 9m) 12m] 6m} 3m] + 2m
Leq = [ {[(12m) 12m] 6m} 3m] + 2m
Leq = [ {6m 6m} 3m] + 2m
Leq = [ 3m 3m] + 2m
Leq = 1.5m + 2m
Leq = 3.5mH
6mH
9mH
6FE-5 The current source in the circuit in Fig. 6PFE-5 has the following operating
characteristics:
пЈ±0 A, t < 0
i(t ) = пЈІ
.
в€’2t
пЈі20te A, t > 0
What is the voltage across the 10 mH inductor expressed as a function of time?
di(t )
v(t ) = L
dt
di(t )
= 20te в€’2t (в€’2) + 20e в€’ 2t = 20e в€’2t в€’ 40te в€’ 2t
dt
[
]
v(t ) = 10m 20e в€’2t в€’ 40te в€’2t
пЈ±0 V , t < 0
v(t ) = пЈІ
в€’2t
в€’2t
пЈі0.2e в€’ 0.4te V , t > 0
Solutions to FE Problems
Chapter 7
7FE-1 In the circuit in Fig. 7PFE-1, the switch, which has been closed for a long time, opens at
t=0. Find the value of the capacitor voltage vc(t) at t=2s.
Find the initial condition:
8kв„¦
+
vc (0в€’)
6kв„¦
12V
R' =
6 kв„¦
в€’
12k (6k )
= 4kв„¦
12k + 6k
8kв„¦
+
12V
пЈ« 4k пЈ¶
v c (0 в€’ ) = пЈ¬
пЈ·(12) = 4V
пЈ­ 4k + 8k пЈё
R ' = 4 kв„¦
v c (0 в€’ )
в€’
6 kв„¦
The t > 0 circuit:
6kв„¦
+
v c (t )
6kв„¦
в€’
v c (t ) + 12kic (t ) = 0
ic (t ) = C
dvc (t )
dt
vc (t ) + 12k (100Вµ )
dvc (t )
=0
dt
dvc (t ) 1
+
vc (t ) = 0
dt
1.2
1
r+
=0
1.2
1
r=в€’
1.2
vc (t ) = Ae
в€’t
1.2
vc (0в€’) = 4V
A=4
в€’t
vc (t ) = 4e 1.2 V , t > 0
vc (2) = 0.756V
7FE-2 In the network in Fig. 7PFE-2, the switch closes at t=0. Find v0(t) at t=1s.
Find the initial condition:
12kв„¦
4kв„¦
+
12 kв„¦
v0 (0в€’)
12V
в€’
vo (0в€’) = 0V
The t > 0 circuit:
R1 = 12kв„¦
R 2 = 12 kв„¦
12V
R3 = 4kв„¦
+
v0 (t )
в€’
R1 = 12kв„¦
R3 = 4kв„¦
+
voc
R 2 = 12 kв„¦
в€’
12V
voc =
R2
12k
(12) =
(12) = 6V
R1 + R2
12k + 12k
R1 = 12kв„¦
R 2 = 12 kв„¦
RTH = ( R1 R2 ) + R3
RTH =
12k (12k )
+ 4k = 10kв„¦
12k + 12k
R3 = 4kв„¦
RTH
RTH = 10kв„¦
100Вµ F
voc = 6V
в€’
10ki(t ) + vo (t ) = 6
dv (t )
ic (t ) = C o
dt
dvc (t )
RTH C
+ vo (t ) = voc
dt
dvo (t )
v
1
+
vo (t ) = oc
dt
RTH C
RTH C
1
r+
=0
RTH C
1
r=в€’
RTH C
The natural solution is:
в€’t
von (t ) = Ae RTH C
в€’t
10 k (100 Вµ )
von (t ) = Ae
= Ae в€’t
The forced solution is:
vo f (t ) = k
dvo f (t )
dt
0+
=0
1
RTH C
k = voc
+
v0 (t )
k=
voc
RTH C
vo f (t ) = 6V
vo (t ) = 6 + Ae в€’t
vo (0в€’) = 0V
0 = 6+ A
A = в€’6
v o (t ) = 6 в€’ 6e в€’ t V , t > 0
vo (1) = 6 в€’ 6e в€’1
vo (1) = 3.79V
7FE-3
Assume that the switch in the network in Fig. 7PFE-3 has been closed for some time.
At t=0 the switch opens. Determine the time required for the capacitor voltage to decay
to one-half of its initially charged value.
Find the initial condition:
12kв„¦
+
v c (0 в€’ )
12V
в€’
пЈ« 6k пЈ¶
v c (0 в€’ ) = пЈ¬
пЈ·(12) = 4V
пЈ­ 6k + 12k пЈё
The t > 0 circuit:
+
vc (t )
6kв„¦
в€’
vc (t ) + 6kic (t ) = 0
ic (t ) = C
dvc (t )
dt
vc (t ) + 6k (100Вµ )
dvc (t )
=0
dt
dvc (t ) 1
+
vc (t ) = 0
dt
0.6
6kв„¦
1
=0
0.6
1
r=в€’
0.6
r+
в€’t
vc (t ) = Ae 0.6
vc (0в€’) = 4V
A=4
в€’t
vc (t ) = 4e 0.6 V , t > 0
в€’t
0.5(4) = 4e 0.6
в€’t
1
= e 0.6
2
1 в€’t
ln =
2 0.6
t = в€’0.6 ln
1
2
t = 0.416s
7FE-4 Find the inductor current iL(t) for t>0 in the circuit in Fig. 7PFE-4.
Find the initial condition:
i L ( 0 в€’)
2в„¦
2в„¦
10V
i L (0 в€’ ) =
10
= 5A
2
+
voc
1в„¦
5A
2в„¦
в€’
1 2=
2
в„¦
3
+
voc
2
в„¦
3
6A
в€’
пЈ«2пЈ¶
voc = 6пЈ¬ пЈ· = 4V
пЈ­3пЈё
2в„¦
2в„¦
RTH = (2 2 ) 2 = 1 2 =
RTH
2
в„¦
3
2в„¦
1A
RTH =
voc = 4V
2
в„¦
3
iL (t )
4H
di L (t ) 2
+ i L (t ) = 4
dt
3
di L (t ) 1
+ i L (t ) = 1
dt
6
1
r+ =0
6
1
r=в€’
6
4
в€’t
i Ln (t ) = Ae 6
i L f (t ) = k
di L f (t )
=0
dt
1
0+ k =1
6
k =6
i L f (t ) = 6
i L (0 в€’ ) = 5 A
в€’t
i L (t ) = 6 + Ae 6
5=6+ A
A = в€’1
в€’t
i L (t ) = 6 в€’ e 6 A, t > 0
7FE-5
Find the inductor current iL(t) for t>0 in the circuit in Fig. 7PFE-5.
Find the initial condition:
i L (0в€’)
3в„¦
1в„¦
12V
i L (0 в€’ ) =
12
= 3A
4
L
iL (t )
R1
R2
Vs
di L (t )
+ (R1 + R2 )i L (t ) = Vs
dt
V
diL (t ) R1 + R2
+
i L (t ) = s
dt
L
L
R1 + R2
r+
=0
L
R + R2
r=в€’ 1
L
L
i Ln (t ) = Ae
i Ln (t ) = Ae
i L f (t ) = k
пЈ« R +R пЈ¶
в€’пЈ¬ 1 2 пЈ· t
пЈ­ L пЈё
пЈ«5пЈ¶
в€’пЈ¬ пЈ·t
пЈ­3пЈё
di L f (t )
=0
dt
5
0+ k = 4
3
12
k=
5
i L (0 в€’ ) = 3 A
5
в€’ t
12
+ Ae 3
5
3 = 2.4 + A
A = 0.6
i L (t ) =
i L (t ) = 2.4 + 0.6e
5
в€’ t
3
A, t > 0
Solutions to FE Problems
Chapter 8
8FE-1
Find V0 in the network in Fig. 8PFE-1.
2в„¦
12в€ 0В°V
j1в„¦
+
4в€ 0В°V
в€’ j1в„¦
1в„¦
V0
-
j1в„¦
+
8в€ 0В° A
2в„¦
в€’ j1в„¦
1в„¦
V0
-
Z =
2(в€’ j1)
= 0.4 в€’ j 0.8в„¦
2 в€’ j1
I0
j1в„¦
8в€ 0В° A
1в„¦
Z
пЈ«
пЈ¶
0.4 в€’ j 0.8
пЈ·пЈ·(8в€ 0В°) = 1.6 в€’ j 4.8 A
I 0 = пЈ¬пЈ¬
пЈ­ 0.4 в€’ j 0.8 + 1 + j1 пЈё
V0 = (1.6 в€’ j 4.8)(1) = 5.06в€ в€’ 71.6В°V
8FE-2
Find V0 in the circuit in Fig. 8PFE-2.
text
2I x
1в„¦
V1
I1
2в„¦
V2
I2
12в€ 0В°V
в€’ j1в„¦
Ix
2в€ 0В° A
V0
V2 = 12в€ 0В°V
V1
в€’ j1
2I x + I x + I1 = 0
Ix =
пЈ« V пЈ¶ V
V в€’ 12в€ 0В°
2пЈ¬пЈ¬ 1 пЈ·пЈ· + 1 + 1
=0
1
пЈ­ в€’ j1 пЈё в€’ j1
2V1 + V1 + в€’ j1V1 + j1(12в€ 0В°) = 0
в€’ j1(12в€ 0В°)
V1 =
= 1.2 в€’ j 3.6V
3 в€’ j1
KCL at node 0:
2 I x + I 2 + 2в€ 0В° = 0
пЈ« V пЈ¶ V в€’ V0
2пЈ¬пЈ¬ 1 пЈ·пЈ· + 2
+ 2в€ 0В° = 0
2
пЈ­ в€’ j1 пЈё
пЈ« 1.2 в€’ j 3.6 пЈ¶ 12в€ 0В° в€’ V0
пЈ·пЈ· +
2пЈ¬пЈ¬
+ 2в€ 0В° = 0
2
пЈ­ в€’ j1 пЈё
V0 = 30.8в€ 8.97В°V
8FE-3
Find V0 in the network in Fig. 8PFE-3.
I
2в„¦
в€’ j1в„¦
j 2в„¦
I1
I2
+
4в„¦
12 в€ 0В°V
+
2V0
-
I1 = I + I 2
I = I1 в€’ I 2
KVL around the left loop:
12в€ 0В° = 2 I 1 + j 2 I + 2V0
V0
в€’
V0 = 4I 2
(2 + j 2) I 1 + (8 в€’ j 2) I 2 = 12в€ 0В°
KVL around the right loop:
2V0 = в€’ j 2 I в€’ j1I 2 + 4 I 2
j 2 I 1 + (в€’4 + j1) I 2 = 0
Two equations and two unknowns:
(2 + j 2) I 1 + (8 в€’ j 2) I 2 = 12в€ 0В°
j 2 I 1 + (в€’4 + j1) I 2 = 0
It follows that:
I 1 = 4.24в€ 45В° A
I 2 = 2.06в€ в€’ 30.96В° A
V0 = 4(2.06в€ в€’ 30.96В°)
V0 = 8.24в€ в€’ 30.96В° V
8FE-4 Determine the midband (where the coupling capacitors can be ignored) gain of the
single-stage transistor amplifier shown in Fig. 8PFE-4.
Z c is small at midband.
Vx
5k
5
=
=
Vs 5k + 1k 6
V0
пЈ« 6k (12k ) пЈ¶
= в€’40 x10 в€’3 пЈ¬
пЈ· = в€’160
Vx
пЈ­ 6k + 12k пЈё
V0 пЈ« V x пЈ¶пЈ« V0
= пЈ¬ пЈ·пЈ¬
Vs пЈ¬пЈ­ Vs пЈ·пЈёпЈ¬пЈ­ V x
пЈ¶ 5
пЈ·пЈ· = (в€’ 160) = в€’133.33
пЈё 6
8FE-5 What is the current I0 in the circuit in Fig. 8PFE-5?
Is
6в€ 0В°V
Zeq
(1 в€’ j1)(1 + j3)
+ 3 = 1.58в€ в€’ 18.43В° + 3 = 4.53в€ в€’ 6.34В° в„¦
1 в€’ j1 + 1 + j 3
6в€ 0В°
Is =
= 1.3245в€ 6.34В° A
4.53в€ в€’ 6.34В°
Z eq =
3в„¦
в€’ j 1в„¦
Io
j 3в„¦
1в„¦
6в€ 0В° V
1в„¦
KVL around the outer loop:
6в€ 0В° = 3(1.3245в€ 6.34В°) + I 0 (1 в€’ j1)
I 0 = 1.48в€ 32.92В° A
Solutions to FE Problems
Chapter 9
Find V0 in the network in Fig. 8PFE-1.
From the power triangle:
Qold = Pold tan Оё old
9FE-1
Qold = 120k tan 45В° = 120k var
S old = (120k + j120k ) VA = 170в€ 45В° kVA
S new = Pold + jQnew
Оё
new
= cos в€’1 (0.95) = 18.19В°
Qnew = Pold tan Оё
new
Qnew = 120k tan 18.19В° = 39.43k var
S new = (120k + j 39.43k )VA
S new = 126.31в€ 18.19В° kVA
S cap = S new в€’ S old = в€’ j80.57 kVA
Qcap = в€’П‰ CV 2
в€’ 80.57k = в€’2ПЂ 60(C )(480 2 )
C = 928Вµ F
9FE-2 Determine the rms value of the following waveform.
T
Vrms
1 2
=
v (t ) dt
T в€«0
Vrms =
Vrms =
Vrms =
1
2
3
4
пЈ№
1пЈ® 2
2
2
1
dt
2
dt
1
dt
0 2 dt пЈє
+
+
пЈЇв€«
в€«
в€«
в€«
4 пЈ°0
1
2
3
пЈ»
[
]
1 1
2
3
t 0 + 4t 1 + t 2 + 0
4
1
[1 + (8 в€’ 4) + (3 в€’ 2)] = 1.22V
4
9FE-3
Find the impedance ZL in the network in Fig. 9PFE-3 for maximum power transfer.
2в„¦
в€’ j1в„¦
ZTH
Z TH = (2 в€’ j1) + j 2 = 0.4 + j1.2в„¦
*
Z L = Z TH
= 0.4 в€’ j1.2в„¦
j 2в„¦
Solutions to FE Problems
Chapter 10
10FE-1 In the network in Fig. 10PFE-1, find the impedance seen by the source.
M
4в„¦
24 в€ 0В°V
I1
в€’ j 5в„¦
j 8в„¦
j 2в„¦
k=0.5
M = k L1 L2 = 1H
24в€ 0В° = I 1 (4 + j 2) в€’ j 2 I 2
в€’ j 2 I 1 + (5 + j3) I 2 = 0
I 1 = 4.92в€ в€’ 19.75В° A
I 2 = 1.69в€ 39.29В° A
24в€ 0В°
Z1 =
= 4.88в€ 19.75В°в„¦
4.92в€ в€’ 19.75В°
I2
5в„¦
N2
N1
to achieve impedance matching for maximum power transfer. Using this value of n,
calculate the power absorbed by the 3в„¦ resistor.
10FE-2 In the circuit in Fig. 10PFE-2, select the value of the transformerвЂ™s turns ratio n =
48 n 2 в„¦
j 32n 2 в„¦
в€’ j 2в„¦
R L = 3в„¦
V2
Z TH = 3в„¦ for maximum power transfer.
Reflecting the primary into the secondary:
Z TH = 48n 2 + j 32n 2 в€’ j 2 = 3
48n 2 + j 32n 2 в€’ j 2 = 3
1
if n = :
4
Z TH =
48
пЈ« 32
пЈ¶
+ j пЈ¬ в€’ 2 пЈ· = 3в„¦
16
пЈ­ 16
пЈё
3в„¦
I
RL = 3в„¦
30в€ 0В°V
30в€ 0В°
= 5в€ 0В° A
6
1
1
PL = I M2 R L = (5) 2 (3) = 37.5W
2
2
I =
10FE-3 In the circuit in Fig. 10PFE-3, select the turns ratio of the ideal transformer that will
match the output of the transistor amplifier to the speaker represented by the 16в„¦ load.
+
1kв„¦
+
5kв„¦
Vs
4V x
100
Vx
10kв„¦
Voc
-
пЈ« 5 пЈ¶пЈ« 4
пЈ¶
Voc = пЈ¬ Vs пЈ·пЈ¬ в€’
x10 4 пЈ·
пЈ­ 6 пЈёпЈ­ 100
пЈё
1kв„¦
+
5kв„¦
Vs
Vx
-
пЈ« 5 пЈ¶пЈ« 4 пЈ¶
I sc = пЈ¬ Vs пЈ·пЈ¬ в€’
пЈ·
пЈ­ 6 пЈёпЈ­ 100 пЈё
Find Rout of the amplifier:
5
V x = Vs
6
Rout =
Voc
I sc
пЈ« 5 пЈ¶пЈ« 4
пЈ¶
x10 4 пЈ·
пЈ¬ Vs пЈ·пЈ¬ в€’
6 пЈёпЈ­ 100
пЈё = 10kв„¦
=пЈ­
пЈ« 5 пЈ¶пЈ« 4 пЈ¶
пЈ¬ Vs пЈ·пЈ¬ в€’
пЈ·
пЈ­ 6 пЈёпЈ­ 100 пЈё
16a 2 = 10k
a = 25
4V x
100
I sc
10 kв„¦
10FE-4 What is the current I2 in the circuit shown in Fig. 10PFE-4?
Z1 =
ZL
= 2 2 (1 в€’ j1) = 4 в€’ j 4в„¦
2
n
120в€ 0В°
= 11.77в€ 11.31В° A
6 + j2 + 4 в€’ j4
I
I2 = 1
n
I 2 = 23.54в€ 11.31В° A
I1 =
10FE-5 What is the current I2 in the circuit in Fig. 10PFE-5?
I1 =
V1
Z1
ZL
n2
N
n= 2 =2
N1
10 + j10
Z1 =
= 2.5 + j 2.5в„¦
22
120в€ 0В°
I1 =
= 24 в€’ j 24 A
2.5 + j 2.5
I
24 в€’ j 24
I2 = 1 =
n
2
I 2 = 12 в€’ j12 A = 16.97в€ в€’ 45В° A
Z1 =
Solutions to FE Problems
Chapter 11
11FE-1 The correct answer is d.
I = 6 A rms
V R = 84.85V rms
V L = 84.85V rms
R=
VR
I
П‰L =
84.85
= 14.14в„¦
6
=
VL
=
I
84.85
= 14.14в„¦
6
Z load = 14.14 + j14.14 в„¦
2
S 3в€’П† = 3 I Z load
2
S 3в€’П† = 3(6) (14.14 + j14.14)
S 3в€’П† = 2.16в€ 45В° kVA
11FE-2 The correct answer is a.
Z в€† = 12 + j12 в„¦
Finding the equivalent per-phase wye:
Z Y = 4 + j 4в„¦ = 5.66в€ 45В°в„¦
V AB = 230V rms
V AN =
I aA =
230
3
V AN
ZY
= 132.79V rms
=
132.79
= 23.5 A rms
5.66
P3в€’П† = 3 V AN I aA cosОё Z = 3(132.79)(23.5) cos 45В° = 6.62kW
11FE-3 The correct answer is b.
24 + j18
ZY1 =
= 8 + j 6в„¦
3
Z Y 2 = 6 + j 4в„¦
V AB = 208V rms
V AN =
208
3
= 120V rms
120в€ 0В°
= 12в€ в€’ 36.87В° A
8 + j6
120в€ 0В°
I aN 2 =
= 16.64в€ в€’ 33.69В° A
6 + j4
I aA = I aN 1 + I aN 2 = 28.63в€ в€’ 35.02В° A rms
I aN 1 =
11FE-4 The correct answer is c.
S 3в€’П† = 24в€ 30В° kVA = 20784.61 + j12000 VA
P3в€’П† = 20.78kW
P1в€’П† = 6.93kW
11FE-5 The correct answer is d.
By use of the power triangle:
S 2 = P2 + Q2
Q = S 2 в€’ P2
Q = (100k ) 2 в€’ (90k ) 2
Q = 43.59k var
Solutions to FE Problems
Chapter 12
12FE-1 The correct answer is a.
1
1
=
= 10,000
П‰0 =
s
LC
1m(10 Вµ )
пЈ«
пЈ¶
пЈ«
пЈ¶
1
1
пЈ® Z ( jП‰ 0 ) пЈ№
пЈ¬
пЈ·пЈ· = 6в€ 0В°пЈ¬пЈ¬
пЈ·пЈ· = 60в€ в€’ 90В°V
V0 ( jП‰ 0 ) = 12в€ 0В°пЈЇ c
=
6
в€ 0
В°
пЈє
пЈ¬
2
П‰
C
в€ 90
В°
10
,
000
(
10
Вµ
)
в€ 90
В°
пЈ°
пЈ»
пЈ­
пЈё
пЈ­ 0
пЈё
12FE-2 The correct answer is c.
П‰0 =
1
=
1
LC
20m(50Вµ )
R
BW = = 200
L
s
R = 200(20m) = 4в„¦
= 1000
s
12FE-3 The correct answer is d.
1
1
V0
ZC
jП‰C
RC
=
=
=
= Gv ( jП‰ )
1
1
VI Z C + R
+ R jП‰ +
jП‰C
RC
at DC, Gv=1=0dB
1
at 3dB down, П‰ =
RC
1
П‰=
= 200
(5k )(1Вµ )
s
П‰ = 2ПЂf
П‰
f =
= 31.83 = 32 Hz
2ПЂ
12FE-4 The correct answer is b.
1
П‰0 =
= 1000
s
LC
1
1
L= 2 =
= 100mH
П‰ 0 C (1000) 2 (10 Вµ )
R
BW = = 100
L
s
R = 100m(100) = 10в„¦
12FE-5 The correct answer is a.
f = 8 Hz , П‰ 1 = 16ПЂ
s
1
ZC =
= в€’ j 2kв„¦
jП‰C
1
1
jП‰C
RC
G v ( jП‰ 1 ) =
=
1
1
+ R jП‰ +
jП‰C
RC
For f=8Hz and П‰1=16ПЂ
V0
ZC
в€’ j 2000
= G v ( jП‰ 1 ) =
=
= 0.707в€ в€’ 45В°
VS
Z C + R 2000 в€’ j 2000
Gv ( j16ПЂ ) = 0.707в€ в€’ 45В°
fC =
1
1
=
= 7.96 = 8Hz
2ПЂRC 2ПЂ (2k )(10Вµ )
Solutions to FE Problems
Chapter 13
13FE-1 The correct answer is c.
12
12
V0 ( s ) =
=
2
s ( s + 3s + 2) s ( s + 2)( s + 1)
A
B
C
V0 ( s ) = +
+
s s + 2 s +1
12
A
B
C
= +
+
s ( s + 2)( s + 1) s s + 2 s + 1
let s=-1
12 = C (в€’1)(1)
C = в€’12
let s=-2
12 = B (в€’2)(в€’1)
B=6
let s=0
12 = 2 A
A=6
6
6
в€’ 12
+
+
s s + 2 s +1
v0 (t ) = 6 + 6e в€’2t в€’ 12e в€’t u (t ) V
V0 ( s ) =
[
]
13FE-2 The correct answer is d.
120
V0 ( s ) =
s ( s + 10)( s + 20)
A
B
C
V0 ( s ) = +
+
s s + 10 s + 20
120
A
B
C
= +
+
s ( s + 10)( s + 20) s s + 10 s + 20
let s=0
120 = A(10)(20)
3
A=
5
let s=-10
120 = B (в€’10)(10)
6
B=в€’
5
let s=-20
120 = C (в€’20)(в€’10)
C=
3
5
3
6
3
в€’
5 + 5
V0 ( s ) = 5 +
s s + 10 s + 20
3
пЈ®3 6
пЈ№
v0 (t ) = пЈЇ в€’ e в€’10t + e в€’ 20t пЈє u (t ) V
5
пЈ°5 5
пЈ»
3
пЈ®3 6
пЈ№
v 0 (100m) = пЈЇ в€’ e в€’10(100 m ) + e в€’ 20(100m ) пЈє
5
пЈ°5 5
пЈ»
v 0 (100m) = 0.24V
13FE-3 The correct answer is b.
12( s + 2)
V0 ( s ) =
s ( s + 1)( s + 3)( s + 4)
lim v 0 (t ) = lim sV0 ( s )
t в†’в€ћ
s в†’0
12( s + 2)
( s + 1)( s + 3)( s + 4)
12(2)
lim v 0 (t ) = lim
= 2V
t в†’в€ћ
s в†’0 (1)(3)( 4)
lim v 0 (t ) = lim
t в†’в€ћ
s в†’0
13FE-4 The correct answer is a.
2s
V0 ( s ) =
( s + 1) 2 ( s + 4)
A
B
C
V0 ( s ) =
+
+
s + 4 s + 1 ( s + 1) 2
2s
A
B
C
=
+
+
2
s + 4 s + 1 ( s + 1) 2
( s + 4)( s + 1)
let s=-1
2(в€’1) = C (в€’1 + 4)
2
C=в€’
3
let s=-4
2(в€’4) = A(в€’4 + 1) 2
8
A=в€’
9
let s=0
0 = A + B(1)(4) + 4C
8
B=
9
8
8
2
в€’
в€’
3
V0 ( s ) = 9 + 9 +
s + 4 s + 1 ( s + 1) 2
8
2 пЈ№
пЈ® 8
v0 (t ) = пЈЇв€’ e в€’ 4t + e в€’t в€’ e в€’t пЈє u (t ) V
9
3 пЈ»
пЈ° 9
13FE-5 The correct answer is c.
2
s 2 X ( s ) + 6[sX ( s )] + 8 X ( s ) =
s+3
2
X ( s) =
( s + 2)( s + 4)( s + 3)
A
B
C
X ( s) =
+
+
s+2 s+4 s+3
2
A
B
C
=
+
+
( s + 2)( s + 3)( s + 4) s + 2 s + 4 s + 3
let s=-3
2 = C (в€’3 + 2)(в€’3 + 4)
C = в€’2
let s=-4
2 = B(в€’4 + 2)(в€’4 + 3)
B =1
let s=-2
2 = A(в€’2 + 3)(в€’2 + 4)
A =1
1
1
в€’2
X ( s) =
+
+
s+2 s+4 s+3
[
]
x(t ) = e в€’2t + e в€’4t в€’ 2e в€’3t u (t )
Solutions to FE Problems
Chapter 14
14FE-1 A single loop, second-order circuit is described by the following differential equation.
2
dv 2 (t )
dv(t )
+4
+ 4v(t ) = 12u (t )
2
dt
dt
t >0
Which is the correct form of the total (natural plus forced) response?
v f (t ) = K 1
2r 2 + 4r + 4 = 0
r 2 + 2r + 2 = 0
в€’ 2В± 4в€’8
r=
= в€’1 В± j1
2
v n (t ) = K 2 e в€’t cos t + K 3 e в€’ t sin t
v (t ) = K 1 + K 2 e в€’ t cos t + K 3 e в€’ t sin t
14FE-2 If all initial conditions are zero in the network in Fig. 14PFE-2, find the transfer
function Vo(s)/Vs(s).
4
Vs ( s ) = I ( s ) + 2 sI ( s ) + 2 I ( s )
s
пЈ®4
пЈ№
Vs ( s ) = пЈЇ + 2 s + 2пЈє I ( s )
пЈ°s
пЈ»
V0 ( s ) = 2 I ( s )
V0 ( s )
2
пЈ®4
пЈ№ пЈ®V ( s ) пЈ№
Vs ( s ) = пЈЇ + 2 s + 2пЈє пЈЇ 0 пЈє
пЈ°s
пЈ»пЈ° 2 пЈ»
V0 ( s )
s
= 2
Vs ( s ) s + s + 2
I (s) =
14FE-3 The initial conditions in the circuit in Fig. 14PFE-3 are zero. Find the transfer function
Io(s)/Is(s).
пЈ«
пЈ¶
пЈ¬ s+4 пЈ·
пЈ· I (s)
I 0 (s) = пЈ¬
3пЈ· s
пЈ¬
пЈ¬s+4+ пЈ·
sпЈё
пЈ­
I 0 (s)
s ( s + 4)
= 2
I s ( s ) s + 4s + 3
14FE-4 In the circuit in Fig. 14PFE-4, use Laplace transforms to find the current I(s). Assume
zero initial conditions and that vs(t)=4cos t u(t).
4s
4
пЈ«
пЈ¶
KVL: 2
= пЈ¬ s + + 2 пЈ· I (s)
s
s +1 пЈ­
пЈё
2
4s
I (s) = 2
( s + 1)( s 2 + 2s + 4)
14FE-5 Assuming that the initial inductor current is zero in the circuit in Fig. 14PFE-5, find the
transfer function Vo(s)/Vs(s).
KVL: Vs ( s ) = 4 I ( s ) + 2 sI ( s )
V s ( s ) = ( 2 s + 4) I ( s )
V0 ( s ) = 2sI ( s )
I (s) =
V0 ( s )
2s
V s ( s ) = ( 2 s + 4)
V0 ( s )
s
=
Vs ( s ) s + 2
V0 ( s )
2s
Solutions to FE Problems
Chapter 15
15FE-1 Given the waveform in Fig. 15PFE-1, determine if the trigonometric Fourier coefficient
an has zero value or nonzero value and why.
By observation, the waveform has odd symmetry. Therefore, an = 0 for all n due to odd
symmetry.
15FE-2 Given the waveform in Fig. 15PFE-2, describe the type of symmetry and its impact on
the trigonometric Fourier coefficient bn.
By observation, the waveform has half-wave symmetry. Therefore, bn =0 for n even due to halfwave symmetry and bn is nonzero for n odd.
15FE-3 Determine the first three nonzero terms of the voltage vo(t) in the circuit in Fig. 15PFE-3
if the input voltage vs(t) is given by the expression
1 в€ћ 30
v s (t ) = + в€‘
cos 2nt V
2 n =1 nПЂ
пЈ« jП‰ пЈ¶
пЈ·пЈ·Vs (nП‰ )
V0 (nП‰ ) = пЈ¬пЈ¬
пЈ­ 1 + jП‰ пЈё
V0 (0) = 0
пЈ« j 2 пЈ¶пЈ« 30
пЈ¶
пЈ·пЈ·пЈ¬ в€ 0В° пЈ·
V0 (П‰ ) = пЈ¬пЈ¬
1
+
j
2
ПЂ
пЈё
пЈ­
пЈёпЈ­
V0 (П‰ ) = 8.54в€ 26.57В°V
пЈ« j 4 пЈ¶пЈ« 30
пЈ¶
пЈ·пЈ·пЈ¬
V0 (2П‰ ) = пЈ¬пЈ¬
в€ 0В° пЈ·
пЈё
пЈ­ 1 + j 4 пЈёпЈ­ 2ПЂ
V0 (2П‰ ) = 4.63в€ 14.04В°V
пЈ« j 6 пЈ¶пЈ« 30
пЈ¶
пЈ·пЈ·пЈ¬ в€ 0В° пЈ·
V0 (3П‰ ) = пЈ¬пЈ¬
пЈё
пЈ­ 1 + j 6 пЈёпЈ­ 3ПЂ
V0 (3П‰ ) = 3.14в€ 9.46В°V
пЈ« j8 пЈ¶пЈ« 30
пЈ¶
пЈ·пЈ·пЈ¬
V0 (4П‰ ) = пЈ¬пЈ¬
в€ 0В° пЈ·
пЈё
пЈ­ 1 + j8 пЈёпЈ­ 4ПЂ
V0 (4П‰ ) = 2.37в€ 7.13В°V
v0 (t ) = 8.54 cos (2t + 26.57В°) + 4.63 cos(4t + 14.04В°) + 3.14 cos(6t + 9.46В°) + ... V
15FE-4 Find the average power absorbed by the network in Fig. 15PFE-4. The source voltage is
as follows:
v s (t ) = 20 + 10 cos(377t + 60В°) + 4 cos(1131t + 45В°) V .
Vs
I (nП‰ ) =
2 + jn3.77
20
I DC =
= 10 A
2
10в€ 60В°
I1 =
= 2.34в€ в€’ 2.05В° A
2 + j 3.77
4в€ 45В°
I3 =
= 0.35в€ в€’ 34.97В° A
2 + j11.31
i(t ) = 10 + 2.34 cos(377t в€’ 2.05В°) + 0.35 cos(1131t в€’ 34.97В°) A
в€ћ
V I
P = V DC I DC + в€‘ n n cos(Оё vn в€’ Оё in )
2
n =1
10(2.34)
4(0.35)
cos(45В° + 34.97В°)
P = 20(10) +
cos(60В° + 2.05В°) +
2
2
P=205.61 W
15FE-5 Find the average value of the waveform shown in Fig. 15PFE-5.
2
a0 =
4
1
1
10 dt + в€« в€’ 2 dt
в€«
40
42
1
[10t ]02 + 1 [ в€’ 2t ]42
4
4
1
1
a 0 = (20) + [в€’ 2(4) в€’ (в€’2)(2)]
4
4
a 0 = 4V
a0 =
Solutions to FE Problems
Chapter 16
16FE-1 A two-port network is known to have the following parameters:
1
1
1
y11 =
S
y12 = y 21 = в€’
S
y 22 = S
14
21
7
If a 2-A current source is connected to the input terminals as shown in Fig. 16PFE-1, find the
voltage across this current source.
I1 = 2 A , I 2 = 0
I 1 = y11V1 + y12V2
I 2 = y 21V1 + y 22V2
1
1
V1 в€’ V2 = 2
14
21
1
1
в€’ V1 + V2 = 0
21
7
21V1 в€’ 14V2 = 588
в€’ V1 + 3V2 = 0
V1 = 36V
V2 = 12V
16FE-2 Find the Thevenin equivalent resistance at the output terminals of the network in Fig.
16PFE-1.
I 1 = y11V1 + y12V2
I 2 = y 21V1 + y 22V2
y11V1 + y12V2 = 0
V1 = в€’
y12
V2
y11
пЈ® y
пЈ№
I 2 = y 21 пЈЇв€’ 12 V2 пЈє + y 22V2
пЈ° y11 пЈ»
пЈ®в€’ y y
пЈ№
I 2 = V2 пЈЇ 21 12 + y 22 пЈє
пЈ° y11
пЈ»
Z TH =
V2
y11
=
I 2 в€’ y 21 y12 + y11 y 22
Z TH
1
14
=
= 9в„¦
пЈ« в€’ 1 пЈ¶пЈ« в€’ 1 пЈ¶ пЈ« 1 пЈ¶пЈ« 1 пЈ¶
в€’ пЈ¬ пЈ·пЈ¬ пЈ· + пЈ¬ пЈ·пЈ¬ пЈ·
пЈ­ 21 пЈёпЈ­ 21 пЈё пЈ­ 14 пЈёпЈ­ 7 пЈё
16FE-3 Find the Y parameters for the two-port network shown in Fig. 16PFE-3.
3
V1
I1
3
56
y11 =
=
=
S
V1 V =0
V1
56
2
2
V1
14 = в€’ 2 V
1
4
56
2
в€’ V1
2
1
= 56 = в€’ S = в€’ S
V1
56
28
V в€’V
I2 = 2
=
4
y 21 =
I2
V1
V2 = 0
0в€’
16
в„¦
3
16
16
4
V = 3 (V2 ) = V2 = V2
16
28
7
+4
3
4
0 в€’ V2
V в€’V
1
7
I1 = 1
=
= в€’ V2
16
16
28
1
в€’ V2
I1
1
y12 =
= 28 = в€’ S
V1 V =0
V2
28
16 8 =
1
4
V2 в€’ V2
V в€’V
3
7
I2 = 2
=
= V2
4
4
28
3
V
I 2 28 2
3
y 22 =
=
=
S
V2
V2
28
16FE-4 Find the Z parameters of the network shown in Fig. 16PFE-4.
The 3в„¦ and 9в„¦ are in series.
9(12) 36
z11 =
=
в„¦
9 + 12 7
12
4
I=
( I1 ) = I1
12 + 9
7
4
пЈ«
пЈ¶ 12
V2 = 3I = 3пЈ¬ I 1 пЈ· = I 1
пЈ­7 пЈё 7
12
I1
V2
12
7
z 21 =
=
= в„¦
I 1 I =0
I1
7
2
3
3
1
I=
(I 2 ) =
I2 = I2
3 + 18
21
7
пЈ« 1 пЈ¶ 12
V1 = 12 I = 12пЈ¬ I 2 пЈ· = I 2
пЈ­7 пЈё 7
12
I2
V1
12
= 7
= в„¦
z12 =
I 2 I =0
I2
7
1
z 22 =
3(18) 18
= в„¦
3 + 18 7
16FE-5 Calculate the hybrid parameters of the network in Fig. 16PFE-5.
h11 = (2 4) + 8 =
28
в„¦
3
4
(I 1 )
4+2
2
I 2 = в€’ I1
3
I2 = в€’
I
h21 = 2
I1
I2 =
в€’
=
V2 = 0
V2
V
= 2
2+4 6
2
I1
3 =в€’2
I1
3
I
h22 = 2
V2
I1 = 0
V2
1
= 6 = S
V2 6
пЈ«V пЈ¶ 2
V1 = 4 I 2 = 4пЈ¬ 2 пЈ· = V2
пЈ­ 6 пЈё 3
2
V2
V1
2
3
=
=
h12 =
V2 I = 0
V2
3
1
```