W ee network_theory_10-06-17_ls2-sol

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8 Electrical Engineering
DetailedExplanations
1.1.1.1.1. (a)(a)(a)(a)(a)
At zero frequency, the capacitors are open circuited.
∴ Z(0) = R1 + R2 = 3 Ω
At ∞ frequency the capacitors are short-circuited.
∴ Z(∞) = R1 = 2 Ω
This gives R2 = 3 – 2 = 1 Ω
Hence, R1 = 2 Ω and R2 = 1 Ω
2.2.2.2.2. (b)(b)(b)(b)(b)
+
–
+ +– –
+
–
VR
I VC
VL
V
VR
VC
V
For a quick analysis, RV in phase with IR leads CV by 90° and hence capacitive. A series resonant circuit
is capacitive below resonant frequency. So (b) is correct.
Mathematical AnalysisMathematical AnalysisMathematical AnalysisMathematical AnalysisMathematical Analysis
RV = R j L
j C
+ ω +
ω
I
I I
RV = RI
CV =
j Cω
I
or R
C
V
V
= jω RC
which means that RV leads CV by 90° as shown in figure (b).
and RV
V
=
R
j
R j L
C
+ ω −
ω
=
( )2
1
RC
RC j LC
ω
ω + ω −
or, RV
V
=
( )
2
1
22 2 2 2
1
tan
1
RC LC
RC
R C LC
−
 ω − ω
∠  ω 
ω + ω −
From the above equation RV will lead V when
1 – ω2 LC > 0 or, ω2 LC < 1 or,
1
LC
ω < = ω0, the resonance frequency.
Hence, choice (b) is correct.
3.3.3.3.3. (c)(c)(c)(c)(c)
From maximum power transfer theorem;
P = I2 × RL
or P =
2
2
( )
Th
L
Th L
V
R
R R
×
+
...(i)

9CTEE17 • NetworksTheory
Hence Pmax at RL = RTh
Pmax =
2
4
Th
Th
V
R
From equation (i), we see first as RL increases the power delivered to the load (P) is increased upto a
maximum value Pmax then P starts declining for further increament in RL.
4.4.4.4.4. (d)(d)(d)(d)(d)
At t = 0 voltage on capacitor = 3 × 1 = 3 V
R = 2 Ω, C = 1F
∴ RC = 2, τ =
1
2
sec; v( ∞) = 0
Now, Vc(t) = Vc(∞) – [Vc(∞) – Vc(0)] × e–t/τ
v(t) = 0 – (0 – 3)e–t/2 = 3 e–t/2
When the switch S is closed after t > 0, the circuit becomes source free and the capacitor will only
discharge.
5.5.5.5.5. (d)(d)(d)(d)(d)
Infinite network having equivalent resistance RAB. If we reduce one step the effective resistance will not
change, so replace remaining network with Req ( = RAB).
R
R⇒
A
B
R R R R
R R R R → ∞
Req
RAB = Req
R
ReqRRAB = Req ⇒
A
B
Req = eq
eq
R R
R
R R
×
+
+
Req – R =
eq
eqR
R R
R
×
+
Req
2 – R2 = R Req
Req
2 – R Req – R2 = 0
we get, Req =
2 2
4
2 1
R R R+ ± +
×
Therefore, the possible value of Req =
( 5 1)
2
R+
= 1.62 × 2 = 3.24 Ω

6.6.6.6.6. (d)(d)(d)(d)(d)
At resonance the current Is will flow through resistor.
Vs = 10 cos ω0 t × 1 = 10 cos ω0 t V
Vs =
d
L
dt
i
LR CIs
+
–
Vs
i
10 cos ω0 t =
d
L
dt
i
d
dt
i
=
10
L
cos ω0 t
d∫ i = 0
10
cos .t dt
L
ω∫
i(t) =
0
10
Lω
sin ω0 t A
7.7.7.7.7. (a)(a)(a)(a)(a)
8.8.8.8.8. (c)(c)(c)(c)(c)
V1 = nV2 ; since 1
2 1
V n
V
 
=  
and I1 =
1
2
2
1 1
;since
n n
 
=  
I
I
I
∴
A B
C D
 
 
 
=
0
0 1/
n
n
 
 
 
9.9.9.9.9. (c)(c)(c)(c)(c)
Here, current through 12 Ω resistor will be zero.
+
–
5 Ω
5 Ω
a
i1
20 V
Loop - 1
+ –
b
5 Ω5 Ω
20 V
Loop - 2
i2
In loop (1) and loop (2),
i1 =
20
10
= 2 A, i2 =
20
10
= 2 A
Va = 5 × 2 = 10 V, Vb = 5 × 2 = 10 V
– + + – + –
10 V 5 V 10 V
a b
Voltage between a and b points
= + 10 + 5 – 10 = 5 V

10.10.10.10.10. (d)(d)(d)(d)(d)
Current is lagging by 30°
θ =
1
tan 30LX
R
−  
= °  
tan 30° = LX
R
⇒
1
3
=
1
R
⇒ R = 3 1.732= Ω
11.11.11.11.11. (b)(b)(b)(b)(b)
For power transferred to load to be maximum. RL should be equal to RTh.
Voc:
+
–
v04 Ω
+
–
+
– 2 0i
5 Ω
6 Ω
+ –
4 V
2 Ω
v0
i0
Voc
+
–
v0 = 0 as current through 6 Ω resistor is zero
Applying KVL
4 Ω
+
–
+
– 2i0
5 Ω2 Ω
i0
Voc
+
–
4 V
2i0 + 4i0 – 2i0 + 5i0 = 4
i0 =
4
9
A
Voc =
4 4
5 2
9 9
− × + × =
12
Volt
9
−
iscscscscsc ;
+
–
v0
4 Ω
+
–
+
– 2 0i
5 Ω
6 Ω
+ –
2 Ω
v0
i0
v0
isc
Short circuited
4 V
Applying nodal analysis
0 0 0 0 00 2 4
6 5 6
v v v v− − + −
+ +
i
= 0 ...(i)
i0 =
2
3
...(ii)

Substituting this value of I0 equation (i)
v0 =
12
Volts
11
−
isc = 0 12 2
A
6 11 6 11
v − −
= =
×
RTh =
12 11
9 2
oc
sc
V − ×
=
× −i = 7.33 Ω
12.12.12.12.12. (c)(c)(c)(c)(c)
We open circuit the load resistance to obtain VTh,
+
–
2 A
4 V
+ –
V0
1 Ω
1 Ω 2 Ω
4V0
X
Y
Open circuit
VTh
+
–
+ –
I
– +
Hence when the load is removed, current from the 2 A source flows in the closed path.
So, voltage V0 = 2 × 2 = 4 V
KVL in the loop:
–VTh + 4 + (2 × 1) + 4 + (4V0) = 0
or, VTh = 26 V
RTh can be obtained by short circuiting all voltage sources (except dependent source) and open circuiting
current source. Take 1V voltage source across load and Imagine ‘I’ current is flowing from source.
Here, 01 5
1
V−
= I
Since V0 = 2I , therefore
1 5(2 )
1
− I
= I
+
–
+ –
V0
1 Ω
1 Ω 2 Ω
4V0
I
Y
1 V
+
–
or, I =
1
11
A
∴ RTh =
( )
1 1
11
1/ 11
= = Ω
I
13.13.13.13.13. (b)(b)(b)(b)(b)
+
–
I3
I1
I2+ +
V2 6 Ω
3 Ω
3 Ω
141 V
I II
N
III
–
V1
–

1
2
V
V
 
 
 
=
1 1 1 2
2 2 1 2
22 1
41 4
V
V
= +  
⇒   = +   
I I I
I I I
Applying KVL
Loop I
3(I1 + I3) + V1 = 141
3I1 + 3I3 + 2I1 + I2 = 141
5I1 + I2 + 3I3 = 141 ...(i)
Loop II
V2 = (I3 – I2) ⋅ 6
V2 = I1 + 4I2
I1 + 4I2 = 6I3 – 6I2
I1 + 10I2 – 6I3 = 0 ...(ii)
Loop III
3(I1 + I3) + 3I3+ 6(I3 – I2) = 141
3I1 – 6I2 + 12I3 = 141 ...(iii)
By equation (i), (ii) and (iii)
I1 = 24 A I2 = 1.5 A I3 = 6.5 A
14.14.14.14.14. (d)(d)(d)(d)(d) +
–
AV
4 Ω 5 Ω
B
j 8.66 Ω– 4j Ω
X
VAB = VAX – VBX
VAX =
4
4 4
j
V
j
−
×
−
VAX = V × 0.707∠–45°
VBX = 8.66
0.866 30
5 8.66
j
V V
j
 
× = × ∠ °  +
VAB = ( )0.707 45 (0.866 30 [ ]AX BXV V V − = ∠ − ° − ∠ ° 
VAB = V × 0.9658∠–105°
Given VAB = 48.3∠30° volts
∴ V =
48.3 30
0.9658 105
∠ °
∠ − °
V = 50∠135° volts
15.15.15.15.15. (c)(c)(c)(c)(c)
Drawing equivalent in s-domain
sL1 sL2
+
–
V2
+
–
V1
R I2I1
+
–
+
–
1/Cs
sM 2I sM 1I
Applying KVL
in loop I
V1 = 1 1 2
1
sL sM
sC
 
+ +  
I I

in loop II
V2 = sMI1 + (R + sL2)I2
comparing with standard equations
Z11 = 1
1
sL
sC
+ Z12 = sM
Z21 = sM Z22 = R + sL2
16.16.16.16.16. (a)(a)(a)(a)(a)
From the circuit diagram, we have:
I1 = 1
2
V
R+
20 V
R
4Ω
2Ω4Ω
+
–
Vx
–+ Vx
V1
+
–
I1
Power absorbs by RΩ resistor is,
PR = I1
2 R
=
2
1 5
2
V
R
R
 
× =  +
(Given) ...(1)
The nodal equations can be written as,
1 120
4 4
V V V− −
+x
= 1
1
2
V
R
=
+
I ...(2)
and Vx = 1
1
2
2
2
V
R
=
+
I ...(3)
From equation (2) and (3), we can write
1 1
5
4 4 4
V V V
− + −x
= 1
2
V
R+
or,
( )
1 12 2
5
4 2 4
V V
R
− +
+ =
( )
1
2
V
R+
or,
( )1
1 1
2 2 2
V
R
 
+ 
+ 
= 5
or, V1 =
( )10 2
3
R
R
+
+
Now, from equation (1) we obtain;
( )2
100
3
R
R
 
 
 + 
= 5
or, R2 – 14R + 9 = 0
Solving for R yields; R = 13.325 Ω or 0.675 Ω
17.17.17.17.17. (c)(c)(c)(c)(c)
at t = 0–, the circuit will be
2 Ω
2 Ω
3 Ω3 Ω
+
–
C
20V
3 Ω
+
–
VC
I 2 Happlying KVL
2I + 3I – 20 = 0
I =
20
4 A
5
=

VC = 3 × 4 = 12 Volts
at t = 0+ the circuit will be 2 Ω
2 Ω
3 Ω3 Ω
+
–
12 V
20 V
3 Ω
2 H
Va
i
iL(0–) = iL(0+) = 0 A
⇒ Va =
3 12
2 3
×
+
= 7.2 V
Va = 3 2
d
dt
+
i
i
at t = 0, i = 0
7.2 = 3 0 2
d
dt
+
i
×
⇒
d
dt
i
= 3.6 A/s
18.18.18.18.18. (d)(d)(d)(d)(d)
h-parameters equations are:
V1 = h11 I1 + h12 V2
I2 = h21I1 + h22 V2
Here, h21 =
2
2
1 0V =
I
I
The circuit for V2 = 0 can be redrawn as shown below.
+
–
V1
I1
I2
R1
R2
αI1
( )αI I1 2+
( )I I1 2+
1
sC
KVL in the above loop gives,
( )
( )1 2
2 1 2R
sC
α +
+ +
I I
I I = 0
1 2
2 1 2 2R R
sC sC
α
+ + +
I I
I I = 0
1 2 2 2
1
R R
sC sC
α   
+ +      
I + I = 0
2 2
1
R
sC
 
+  
I = 1 2R
sC
α 
− +  
I
h21 = 2 2
1 21
R Cs
R Cs
 α +
= −  + 
I
I
19.19.19.19.19. (a)(a)(a)(a)(a)
B =
1 0 0 1 0 0
0 1 0 0 0 1
0 0 1 1 1 1
 
 − 
 − − 
Fundamental loop matrix is always written in the form
B = [I : BT]
Identity
Matrix
Loop matrix with
respect to given tree
Ql = –[BT]T

and Q = [Ql : I]
QL =
1 0 0
0 0 1
1 1 1
T
 
 − − 
 − − 
=
1 0 1
0 0 1
0 1 1
− − 
 
 
  
Q =
1 0 1 1 0 0
0 0 1 0 1 0
0 1 1 0 0 1
− − 
 
 
  
20.20.20.20.20. (c)(c)(c)(c)(c)
Let us consider one L-C unit as shown below.
L
V1 V2
+
–
1
Cs
C
+
–
I1 I2
sL
I1 I2
Applying KVL in input loop,
⇒ –V1 + I1Ls + V2 = 0
V1 = V2 + Ls I1 ...(i)
Applying KVL in the output loop,
⇒ 1 2
2V
sC
− 
− +   
I I
= 0
–V2 Cs + I1 – I2 = 0
I1 = I2 + V2 Cs ...(ii)
Substituting equation (ii) in equation (i)
V1 = V2 + Ls(I2 + V2 Cs)
= V2 + LsI2 + V2 LCs2
or, V1 = V2 (1 + LCs2) + LsI2 ...(iii)
I1 = V2 Cs + I2 ...(iv)
When comparing above two equations with
V1 = AV2 + BI2
I1 = CV2 + DI2
A B
C D
 
 
 
=
( )2
1
1
LCs Ls
Cs
 +
 
  
with n such units connected as shown in given problem, the transmission matrix is the product of each
unit.
∴ [T] =
2
1
1
n
s LC sL
sC
 +
 
 

21.21.21.21.21. (b)(b)(b)(b)(b)
Redrawing the circuit
+
–
+
– 0.2 V s2( )
2 Ω
V1( )s 10I ( )s1
+
–
V2( )sZ
Z = 1 5 1. s =
1 5 1
1 5 1
. s
. s
×
+
=
1 5
1 5 1
. s
. s +
Applying KVL in loop I
V1(s) = 2I1(s) – 0.2 V2(s) ...(i)
V2(s) = 1
1.5
10 ( )×
1.5 +1
s
s
s
− I
V2(s) =
15 ( )
1.5 +1
1s s
s
− I
...(ii)
From equations (i) and (ii)
Input admittance = 1
1
( )
( )
s
V s
I
=
1 5 1
6 2
. s
s
+
+
22.22.22.22.22. (d)(d)(d)(d)(d)
L, B, P
A
1ba 2
1′ 2′
+
–20V
+
– 15V
RL
We can determine the norton’s equivalent across the terminals a and b.
TTTTTo detero detero detero detero determineminemineminemine INNNNN :::::
L, B, P
A
1ba 2
1′ 2′
+
–20V
+
– 15V
IN
We can use superposition theorem to determine IN.
When only 20V source is acting alone.When only 20V source is acting alone.When only 20V source is acting alone.When only 20V source is acting alone.When only 20V source is acting alone.
L, B, P
A
1ba 2
1′ 2′
+
–20V
IN
′
From figure A in the question, we can calculate N′I .
N′I = 8 A
When only 15 V source is acting alone:

L, B, P
A
1ba 2
1′ 2′
+
– 15V
IN
′′
By applying the Reciprocity principle to the figure. A in the given question, we can calculate N′′I .
N′′I =
2
15 3A
10
− × = −
15 10
2N
 
= ′′ I
Thus, IN = 8 3 5AN N+ = − =′ ′′I I (Using superposition theorem)
TTTTTo detero detero detero detero determineminemineminemine RRRRRNNNNN :::::
L, B, P
A
1ba 2
1′ 2′
RN
Equivalently, we can write the above circuit as show below.
L, B, P
A
1 2
1′ 2′
RN
From figure A in the given question, we can calculate RN as
RN =
10V
2.5
4A
= Ω
So, the Norton’s equivalent across the terminals a and b is will be as shown below.
RN = 2.5ΩIN = 5A
IL
RL
a
b
Maximum power will be delivered to RL, when RL will be equal to RN = 2.5Ω
When RL = RN = 2.5 Ω,
5
2.5 A
2L = =I
Hence, Pmax = IL
2RL = IL
2RN = (2.5)2 × 2.5 = 15.625 W

23.23.23.23.23. (b)(b)(b)(b)(b)
3 Ω
2 Ω
4 Ω
j5 Ω
j 5Ω
j3 Ω
+
–
A
20 0° I1
I2
B
Applying Mesh Law
Mesh I
3I1 + j5I1 + j3I2 + 2I1 – 2I2 = 20∠0°
(5 + j5)I1 + (–2 + j3)I2 = 20∠0° ...(i)
Mesh II
2I2 + j5I2 + j3I1 + 4I2 – 2I1 = 0
(6 + j5)I2 + (–2 + 3j)I1 = 0 ...(ii)
From equations (i) and (ii)
I1 = 2.30∠–41.70° A
I2 = 1.064 ∠–137.8° A
VTh = VAB = I2 × (4) = 4.256∠–137.8° V
24.24.24.24.24. (c)(c)(c)(c)(c)
Since the voltage across the 5 Ω resistor is VR, therefore current flowing through it is:
I =
5
RV
...(i)
–
+ 5 VR
+
–
VR
1A
5Ω
500Ω
+ –
100V
1A
+
–
I
Now, writing the mesh equations, we have:
100 = 500 (I – 1) + VR – 5 VR ...(ii)
Substituting equation (i) into equation (ii) yields,
500 + 100 = 500 5
5
R
R R
V
V V+ −
or VR =
600 5
6.25 V
480
×
=

and I =
6.25
1.25 A
5
=
Now, power absorbed by 500Ω resistor is
P500Ω = (Ι – 1)2 × 500 = (1.25 – 1)2 × 500 = 31.25 W
and power absorbed by 5Ω resistor,
5P Ω =
( ) ( )2 2
6.25
7.8125 W
5 5
RV
= =
∴ Total power dissipated = 39.06 W
25.25.25.25.25. (d)(d)(d)(d)(d)
For current to be in phase with applied voltage imaginary part of impedance should be zero.
j LX j LX
R R
–j CX
–j CXRR
Zeq
Zeq =
( ) ( )
2
L CR jX R jX+ + −
=
2
C
L
C
jR X
R jX
R jX
 − ⋅
+ + − 
=
2 2
( )
2
C
L C
C
jR X
R jX R jX
R X
 ⋅
+ − + 
+ 
equating imaginary part to zero
2
2 2
2
C
L
C
R X
X
R X
 
− 
+ 
= 0
R2XC = ( )2 2
L CX R X⋅ +
⇒ XL =
2
2 2
C
C
R X
R X+
26.26.26.26.26. (d)(d)(d)(d)(d)
With ‘S’ open:
Ceq =
( )1 2 6
2 F
1 2 6
+ ×
= µ
+ +
∴ i = i3 = eq
dV
C
dt
1 Fµ
2 Fµ 6 Fµ
C
–+
V = 100 sin 1000 t (v)
i
i2
i1 i3
V1 V2
+ –+ –
i3 = ( ) ( )6
2 10 100 sin 1000−
×
d
t
dt
= (2 × 10–6) × 100 × 1000 × (cos 1000 t)
i3 = 0.2 cos (1000t) A

V2 = 3 6
1 1
0.2cos(1000 )
6 10
dt t dt
C −
=
×
∫ ∫i
V2 = ( )6
0.2 sin1000
33.33 sin 1000 V
10006 10
t
t−
 
=  ×
V1 = V – V2 = 100 sin (1000 t) – 33.33 sin (1000 t)
V1 = 66.67 sin (1000 t) V
27.27.27.27.27. (a)(a)(a)(a)(a)
Converting the circuit into s-domain
Z(s) =
1
1 1
sC
R sL
 
+ +  
+
–
1/sC
+
–
sLR VC1/s
Z(s)Z(s) = 2
1
1 11
s
s ss
s
=
+ ++ +
VC = Voltage across capacitor
= Voltage across Z(s)
= Z(s) ⋅ I(s)
Z(s) VC
+
_
I( )s
1/s
= 2
1
1
s
ss s
⋅
+ +
VC(s) = 2
1
1s s+ +
Taking inverse Laplace
vC(t) = 22 3
sin
23
t /
e t−
 
 
 
28.28.28.28.28. (b)(b)(b)(b)(b)
+
–
+
–
V1 V2
10 Ω
5 Ω
2 Ω4 Ω
25 V
50 V
2 Ω
21
With two principal nodes 1 and 2 and the third chosen as the reference node, by using KCL, the net current
out of node 1 must be equal to zero.

Applying KCL at node-1, we have:
1 1 1 225
2 5 10
V V V V− −
+ + = 0
or, 8V1 – 1V2 = 50 ...(i)
KCL at node 2 gives,
2 2 2 150
2 4 10
V V V V+ −
+ + = 0
–8V1 + 68V2 = –2000 ...(ii)
By solving both the equations:
∴ V1 = 2.61 V
and V2 = –29.1 V
∴ I2Ω resistor = 1 2.61
1.30 A
2 2
V
= =
29.29.29.29.29. (b)(b)(b)(b)(b)
at t = 0–
vC = I × R2
R2
I
+
–
vC( )tC
at t = 0+
vC(0–) = vC(0+) = IR2
Req
IR2R2R1
vC(∞) = 0
(∵ capacitor will discharge fully)
Req = 1 2
1 2
R R
R R+
vC(t) = [ ](0) ( ) ( )
t
C C Cv v e v
−
τ− ∞ + ∞
vC(t) =
1 2
1 2
( )
2 Volts
t R R
R R C
R e
− +
⋅
⋅I

30.30.30.30.30. (b)(b)(b)(b)(b)
Let the additional parallel resistance required be Rx as shown below.
8081Ω Rx 3640ΩIs
3.2 mH
15.7 nF
Given, new bandwidth after adding an additional resistance Rx = 6 kHz.
Quality factor QP =
3
3
22.3 10
3.72
6 10
rf
BW
×
= ≈
×
Since, QP = ωrCReq , therefore
3.72 = 2π (22.3 × 103) (15.7 × 10–9) Req
⇒ Req = 1691 Ω
∵
eq
1
R
=
1 1 1
8081 3640 R
+ +
x
or,
1
1691
=
1 1 1
8081 3640 R
+ +
x
After solving, we get; Rx = 5184.20 ≈ 5184 Ω

W ee network_theory_10-06-17_ls2-sol

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