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S.1Electrical Circuits (November/December-2012, R09) JNTU-Hyderabad
B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
Code No.: A109210204
Jawaharlal Nehru Technological University Hyderabad
II B.Tech. I Semester ExaminationsNovember/December - 2012
ELECTRICAL CIRCUITS
( Common to EEE, ECE, ETM )
Time: 3 Hours Max. Marks: 75
Answer any FIVE Questions
All Questions carry equal marks
- - -
1. (a) Name three passive elements of electrical circuit and deduce the relationship between voltage and current for
each passive elements. (Unit-I, Topic No. 1.1)
(b) Distinguish between ideal and practical voltage sources and draw their V-I characteristics. (Unit-I, Topic No. 1.2)(c) Determine the power being absorbed by each of the circuit element shown in figure. [4+4+7]
(Unit-I, Topic No. 1.1)
4.5 mA+
13.7 V + 25 V
+
+ 62 V12 cos 1200t mA
12.5 mA 5 sin 1200t V,t = 4 ms
2ix,ix = 10 A
4.5 mA+
13.7 V + 25 V
+
+ 62 V12 cos 1200t mA
12.5 mA 5 sin 1200t V,t = 4 ms
2ix,ix = 10 A
Figure
2. (a) Using mesh analysis, find the magnitude of the current dependent source and current through the 2 resistor
as shown in figure. (Unit-II, Topic No. 2.3)2
1
3
1 2 A
5 ii
+
2
1
3
1 2 A
5 ii
2
1
3
1 2 A
5 ii
2
1
3
1 2 A
5 ii
+
Figure
(b) Find the power loss in the resistors of the network for the figure shown using nodal analysis, [8+7]
(Unit-II, Topic No. 2.3)
+
4 A 2 A
10 V 3
1
2 2
+
4 A 2 A
10 V 3
1
2 2
Figure
R09Solutions
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S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2013
B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
3. (a) Define the following for alternating quantity,
(i) R.M.S value
(ii) Average value
(iii) Form factor
(iv) Peak factor. (Unit-III, Topic No. 3.1)
(b) For the circuit shown in figure, determine the total impedance, total current and phase angle. [8+7]
(Unit-III, Topic No. 3.3)
~110 V, 50 Hz
10
70
100 F
210 F~110 V, 50 Hz
10
70
100 F
210 F
Figure
4. (a) Define the bandwidth and derive the expressions for bandwidth of series resonating circuit and its relation with
Q-factor. (Unit-IV, Topic No. 4.2.1)
(b) Write the applications of locus diagrams. For the circuit shown in figure, plot the locus of current. [8+7]
(Unit-IV, Topic No. 4.1)
~ R
jXc
~ R
jXc
Figure
5. (a) State and explain Faradays laws of electromagnetic induction. (Unit-V, Topic No. 5.2)
(b) An iron ring of 8 cm dia. and 14 cm2 in cross section is wound with 250 turns of wire for a flux density 1.8 Wb/m2
and permeability 450. Find the exciting current, the inductance and stored energy. Find corresponding quanti-
ties when there is a 1.8 mm air gap. [5+10](Unit-V, Topic No. 5.1)
6. (a) Define and illustrate the following with an example,
(i) Branch
(ii) Tree
(iii) Cut-set
(iv) Tie-set. (Unit-VI, Topic No. 6.2)
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S.3Electrical Circuits (November/December-2012, R09) JNTU-Hyderabad
B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
(b) Figure shown below represents a resistive circuit. Determine the number of branches, number of nodes and
number of links. Write down the incidence matrix for the given network. Also develop the network equilibrium
equations. [8+7](Unit-VI, Topic No. 6.3)
5 2
2 4
3
1
10 V
+
32
4
1
5 2
2 4
3
1
10 V
+
32
4
5 2
2 4
3
1
10 V
+
32
4
1
Figure
7. (a) Is Norton theorem dual of Thevenins theorem? Justify your answer. (Unit-VII, Topic No. 7.5)
(b) Find the current in the 10 resistor as shown in below figure using superposition theorem. [5+10](Unit-VII, Topic No. 7.2)
+
1
5
1
10
1 A
10 V
2 A
A B+
1
5
1
10
1 A
10 V
2 A
A B
Figure
8. (a) State and explain the compensation theorem. (Unit-VIII, Topic No. 8.8)
(b) Apply Thevenins theorem and obtain the current passing through 210 F capacitor of figure. [7+8]
(Unit-VIII, Topic No. 8.4)
110 V, 50 Hz
10
70
100 F
210 F
110 V, 50 Hz
10
70
100 F
210 F
Figure
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B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
Q1. (a) Name three passive elements of electrical circuit and deduce the relationship between voltage
and current for each passive elements.
Answer : Nov./Dec.-12, (R09), Q1(a) M[4]
Passive Element
For answer refer Unit-I, Q2, Topic: Passive Elements.
Relationship between Voltage and Current of Passive Element
For answer refer Unit-I, Q14, Topic: Voltage-current Relationship of Passive Element.
(b) Distinguish between ideal and practical voltage sources and draw their V-I characteristics.
Answer : Nov./Dec.-12, (R09), Q1(b) M[4]
For answer refer Unit-I, Q4, Topic: Ideal Voltage Source, Practical Voltage Source.
(c) Determine the power being absorbed by each of the circuit element shown in figure.
4.5 mA+
13.7 V + 25 V
++
62 V12 cos 1200t mA
12.5 mA 5 sin 1200t V,t = 4 ms
2ix,ix = 10 A
4.5 mA+
13.7 V + 25 V
++
62 V12 cos 1200t mA
12.5 mA 5 sin 1200t V,t = 4 ms
2ix,ix = 10 A
Figure
Answer : Nov./Dec.-12, (R09), Q1(c) M[7]
The given circuit elements are shown in figure,
4.5 mA+
13.7 V + 25 V
++
62 V12 cos 1200t mA
12.5 mA 5 sin 1200t V,t = 4 ms
2ix:ix = 10 A
4.5 mA+
13.7 V + 25 V
++
62 V12 cos 1200t mA
12.5 mA 5 sin 1200t V,t = 4 ms
2ix:ix = 10 A
To determine,
The power absorbed by each element = ?
We know that,
The power absorbed by any element is given as,
P = VINow,
The power absorbed by element-1 is given as,
P = 13.7 4.5 103
= 0.06165 W
The power absorbed by element-2 is given as,
P = 25 ( 12.5 103)
= 25 1000
5.12
= 0.3125 W
SOLUTIONS TO NOV./DEC.-2012, R09, QP
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S.5Electrical Circuits (November/December-2012, R09) JNTU-Hyderabad
B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
The power absorbed by element-3 is given as,
P= (12 cos 1200t) 103 (5 sin 1200t) 103
But, t= 4 ms
Substituting tvalue, we get,
P= [12 cos(1200 4 103) 103
[5 sin(1200 4 103)] 103
= (12 cos 4.8) 103 (5 sin 4.8) 103
= (12 0.9965) 103 (5 0.0837) 103
= (11.958 103) (0.4185 103)
= 5.004 106 Watts
The power absorbed by element-4 is given as,
P= 62 2 ix
But, ix= 10 A
Substituting ix
value, we get,
P = 62 2 10
= 620 2
= 1240 Watts
P = 1.2 kW
Q2. (a) Using mesh analysis, find the magnitudeof the current dependent source andcurrent through the 2 resistor asshown in figure,
2
1
3
1 2 A
i
+
2
1
3
1 2 A
i
+
Figure
Answer : Nov./Dec.-12, (R09), Q2(a) M[8]
Given circuit is shown in figure (1),
2
1
3
1 2 A
i
+
2
1
3
1 2 A
i
+
Figure (1)
To determine,
Using mesh analysis,
Current dependent source = ?
Current through 2 resistor = ?
By using mesh analysis, marking the mesh currents
as shown in figure (2),
2
1
3
1 2 A
i
+
I3
5 i
I1
I2
2
1
3
1 2 A
i
+
I3
5 i
I1
I2
Figure (2)
From mesh-1, we have,
I1= 2 A
Applying KVL to mesh-2, we get,
5i + 3(I2+I
1) + 1I
2= 0
5(I1+I
2) + 3(I
2+I
1) +I
2= 0 [Qi =I1 +I2]
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B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
5I1+ 5I
2+ 3I
2+ 3I
1+I
2= 0
8I1+ 9I
2= 0
8(2) + 9I2 = 016 + 9I
2= 0
9I2= 16
I2=
9
16= 1.778 A ... (1)
Applying KVL to mesh-3, we get,
5i = 2I3+ 1(I
1+I
3)
5(I1 +I2) = 2I3 +I1 +I3 5I1 I1 + 5I2 = 3I3
4I1 + 5I2= 3I3 4(2) + 5(1.778) = 3I3 8 + 8.89 = 3I3
0.89 = 3I3
I3 = 389.0
I3= 0.297 A
I3 0.3 A
The value of current dependent source,
= 5 i
= 5(I1+I
2)
= 5(2 + 1.778)
= 1.11 Volts
And current through 2 resistor is given as,
I3= 0.3 A
(b) Find the power loss in the resistors ofthe network for the figure shown usingnodal analysis.
+
4 A 2 A
10 V 3
1
2 2
V1 V2 +
4 A 2 A
10 V 3
1
2 2
V1 V2
Figure
Answer : Nov./Dec.-12, (R09), Q2(b) M[7]
Given circuit is shown in figure,
+
4 A 2 A
10 V 3
1
2 2
V1 V2 +
4 A 2 A
10 V 3
1
2 2
V1 V2
Figure
Note: Node voltages V1
and V2
are assumed as shown in
figure,
To determine,
Using noda1 analysis, power loss in each resistor,
i.e., P2 = ?
P1 = ?
P3 = ?
P2 = ?Applying KCL at node-1, we get,
4 =13
10
2
21211 VVVVV ++
+
4 =3
101
3
1
1
1
3
1
2
121 +
+
++ VV
3
2
3
4
6
1121 VV = 0 ... (1)
Applying KVL at node-2, we get,
2 =23
10
1
21212 VVVVV +
+
2 =3
10
2
1
3
11
3
11 21
+++
VV
316
6
11
3
421 +
VV = 0 ... (2)
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S.7Electrical Circuits (November/December-2012, R09) JNTU-Hyderabad
B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
On solving equations (1) and (2), we get,
V1= 5.26 Volts
V2= 6.74 Volts
Power loss across 2 resistor connected at node 1.
P2 = RI
22
But, I2= 2
1V
=2
26.5
= 2.63
P2 = (2.63)
2 2
= 13.83 Watts
Power loss across 1 resistor connected betweennode-1 and node-2.
P1= RI .
21
But, I1 = 1
21 VV
= 5.26 6.74= 1.48 A
Here negative sign indicates that the current is in
opposite direction.
P1= RI
21
= ( 1.48)2 1
= 2.19 Watts
Power loss across 2 resistor connected across node-2.
P2= (I2)2
R
But, I2 = 2
2V
=2
74.6
= 3.37 A
P2= (3.37)
2 2
= 22.71 Watts
Power loss in 3 resistor is given by,
P3 = (I3 )
2.R
But, I3 = 3
10 21 VV +
=3
74.61026.5 +
= 2.84
P3 = (2.84)
2 3
= 24.19 Watts
Q3. (a) Define the following for alternating
quantity,
(i) R.M.S value
(ii) Average value
(iii) Form factor
(iv) Peak factor.
Answer : Nov./Dec.-12, (R09), Q3(a) M[8]
For answer refer Unit-III, Q1.
(b) For the circuit shown in figure, determine
the total impedance, total current andphase angle.
~110 V, 50 Hz
10
70
100 F
210 F
~110 V, 50 Hz
10
70
100 F
210 F
Figure
Answer : Nov./Dec.-12, (R09), Q3(b) M[7]Given that,
Voltage, V= 110 V
Frequency,f= 50 Hz
Resistance,R1= 10
Capacitance, C= 100 F
Resistance,R2= 70
Capacitance, C'= 210 F
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B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
To determine,
(i) Total impedance,Ztotal
= ?
(ii) Total current,Itotal = ?(iii) Phase angle, = ?The given circuit is shown in figure,
~110 V, 50 Hz
10
70
100 F
210 F
Z1
Z2
Z3
~110 V, 50 Hz
10
70
100 F
210 F
Z1
Z2
Z3
Figure
Now,
The reactance,
1C
X =fC21
= 610100502
1
= 410502
1
= 410100
1
= 210
1
=
100
= 31.831 And also,
The reactance,
2C
X =Cf 2
1
=6
10210502
1
=06597.0
1
= 15.1584
The impedance,
Z1
=R1j
1CX
= (10 j31.831)
The impedance,
Z2
=R2= 70
The impedance,
Z3= (j15.1584)
Now,
The total impedance is given as,
Ztotal
=Z1+ (Z
2||Z
3)
= (10 j31.831) + (70 || j15.158)
= (10 j31.831) +
158.1570
158.1570
j
j
= (10 j31.831) +1061.06
70 15.158
j
j
= (10 j31.831) + (3.135 j14.479)
= 13.135 j46.31
= 164.741367.48 ohmsWe know that,
The total current is given as,
Itotal
=totalZ
V
Itotal = 164.741367.480110
Itotal = 2.285 164.74 Amps
And also,
The phase angle is given as,
= 74.164o
Q4. (a) Define the bandwidth and derive theexpressions for bandwidth of seriesresonating circuit and its relation withQ-factor.
Answer : Nov./Dec.-12, (R09), Q4(a) M[8]
For answer refer Unit-IV, Q10(i), Q11.
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S.9Electrical Circuits (November/December-2012, R09) JNTU-Hyderabad
B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
(b) Write the applications of locus diagrams.For the circuit shown in figure, plot thelocus of current.
~ R
jXc
~ R
jXc
Figure
Answer : Nov./Dec.-12, (R09), Q4(b) M[7]
The following are application of Locus diagram,
1. The locus diagrams are used to determine the
behaviour or response of the circuit networks.
2. These are also used to pre-determine the operating
characteristics of A.C circuits under various
conditions.
3. These are employed in determining the magnitude
and phase of a resistance, inductance and capacitance
parameters.
For remaining answer refer Unit- IV, Q3, Topic:
VariableXC.
Q5. (a) State and explain Faradays laws of
electromagnetic induction.
Answer : Nov./Dec.-12, (R09), Q5(a) M[5]
For answer refer Unit-V, Q11.
(b) An iron ring of 8 cm dia. and 14 cm2 incross-section is wound with 250 turns ofwire for a flux density 1.8 Wb/m2 andpermeability 450. Find the excitingcurrent, the inductance and stored
energy. Find corresponding quantitieswhen there is a 1.8 mm air gap.
Answer : Nov./Dec.-12, (R09), Q5(b) M[10]
Given that,
Diameter of ring,D = 8 cm
Cross-section of ring,A = 14 cm2
Number of turns,N= 250 turns
Flux density,B = 1.8 Wb/m2
Permeability = 450
To determine,
The exciting current = ?
Inductance,L = ?
Energy stored = ?
And also,
For an air gap of 2 mm,
Determine,
(i) Exciting current = ?
(ii) Inductance,L = ?
(iii) Energy stored = ?
Now,
The length of flux path in the ring is given as,
Length of flux path = D
= 8
= 3.142 8
= 25.136 cm
= 0.25136 m
The flux is given as,
Flux, = Flux density (B) Cross-section area
= 1.8 14 104
= 0.00252 Wb
= 25.2 104 Wb
And also,
The Ampere turns per meter of flux path length, is
given as,
H=r
B
0
H=450104
8.17
H= 3183.09
We know that,
Total Ampere turns required is given as,
Ampere turns required =H Length of flux path
= 3183.09 0.25136
= 800.10
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(i) Exciting Current
The exciting current is given as,
Exciting current,
I =Total ampere turns required
Number of turns on ring
=250
10.800
= 3.2004 Amps
(ii) Inductance
The inductance is given as,
Inductance,
L =I
N
=2004.3
102.252504
=2004.3
63.0
= 0.1968 H
(iii) Energy Stored
The energy stored is given as,
Energy stored=2
2
1LI
=2
)2004.3(1968.02
1
= 1.0079 J
For an Air Gap of 2 mm
Ampere turns per/meter of flux path length in iron
portion is given as,
Ampere turns/meter,
H= 3183.09
Length of flux path in iron portion = 25.136 0.18
= 24.956 cm
= 0.24956 m
~0.25 m
Ampere turns required by iron portion,
ATi= 3183.09 0.25
= 795.773
Ampere turns required by 2 mm air gap = 0.796Blg 106
= 0.796 1.8 1.8 103 106
= 2579.04
Total ampere turns required =ATi+H
= 795.773 + 3183.09
= 3978.863
(i) Exciting current,
I= 250
863.3978
= 15.915 Amps
(ii) Inductance,
L=I
N
=915.15
102.252504
=915.15
63.0
= 0.03958
~ 0.04
~ 40 103
~ 40 mH
(iii) Energy stored=2
2
1LI
=
23
)915.15(10402
1
= 5.066 J
Q6. (a) Define and illustrate the following withan example,
(i) Branch
(ii) Tree
(iii) Cut-set
(iv) Tie-set.
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B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
Answer : Nov./Dec.-12, (R09), Q6(a) M[8]
(i) Branch
The elements of a tree connecting one node to the other are called as branches. These are denoted by b and are
represented by dark lines. The branches of a tree are also called as twigs. The relation between the number of nodes n and
number of branches b in a tree is given as,
b = n 1
Example
Let us consider the graph as shown figure (1),
12
3
4
12
3
4
Figure (1)
The various branches for figure (1) is represented in figure (2),
2
1 3
4
2
1 3
4
1 2 3
4
1 2 3
4
Figure (2)
(ii) Tree
From an oriented connected graph, a subgraph formed with all the nodes connected by it without forming a closed
path is called as tree. The following are the conditions to be satisfied in order to form a tree.
The tree must have (n 1) branches where n is the number of nodes.
The tree must have same number of nodes as that of a graph i.e., n nodes.
All the nodes must be connected.
The tree should not have any closed path.
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The tree for figure (1) is represented as shown in
figure (3).
12
3
4
12
3
4
Figure (3): Tree
(iii) Cut-set
It is defined as the set of branches which when
removed will divide the connected graph into two connectedsub-graphs.
Example
1 3
2
4
56 1 3
2
4
56
1 3
2
4
56 1 3
2
4
56
Figure
When the branches 2, 4 are removed two subgraphs
of elements 1, 6 and 3, 5 are formed 2, 4 are the cut-sets.
(iv) Tie-set
For answer refer Unit-VI, Q9.
(b) Figure shown below represents aresistive circuit. Determine the numberof branches, number of nodes andnumber of links. Write down theincidence matrix for the given network.Also develop the network equilibrium
equations.
5 2
2 4
3
1
10 V
+
32
4
1
5 2
2 4
3
1
10 V
+
32
4
5 2
2 4
3
1
10 V
+
32
4
1
Figure
Answer : Nov./Dec.-12, (R09), Q6(b) M[7]
The given resistive circuit is shown in figure (1),
5 2
2 4
3
1
10 V
+
5 2
2 4
3
1
10 V
+
Figure (1)
The graph for given resisitive circuit is shown in
figure (2),
1
1 2
3
4
4
2
5
6
31
1 2
3
4
4
2
5
6
3
Figure (2): Graph
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S.13Electrical Circuits (November/December-2012, R09) JNTU-Hyderabad
B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )
From figure (2), we have,
Number of nodes, n = 4
Number of branches, b = 6Number of links, l = b (n 1)
= 6 (4 1)
= 6 3
= 3
Incidence Matrix
The incidence matrix is given as,
Incidence matrix,
=
111000
100101
010110
001011
1 2 3 4 5 61
2
3
4
Nodes(n)
branches(b)
111000
100101
010110
001011
1 2 3 4 5 61
2
3
4
Nodes(n)
branches(b)
Network Equilibrium Equations
The fundamental circuit for the given network is
shown in figure (3),
1 2
3
1
4 5
3
L3
6
L1
L2
2
4
1 2
3
1
4 5
3
L3
6
L1
L2
2
4
Figure (3): Fundamental Circuits
LetI1,I
2,I
3be the loop currents,
Now,
The loop matrix is given as,
Loops\Branches
B =
101001
110100
011010
1 2 3 4 5 6L1L2
L3 101001
110100
011010
1 2 3 4 5 6L1L2
L3
From the above matrix, we have,
V2
V4
V5
= 0 ... (1)
V3 + V5 V6 = 0 ... (2)
V1
+ V4+ V
6= 0 ... (3)
And also,
From figure (1), we have,
R1
= 3 , R2= 1 , R
3= 2
R4= 4 , R
5= 2 , R
6= 5
Now,
V1= i
1R
1 v
1
V1= i
1(3) 10
V1= 3i
1 10
V2= i
2R
2
V2= i
2(1)
V2= i
2
V3= i
3.R
3
V3= i
3(2)
V3= 2i3V
4= i
4.R
4
V4= i
4(4)
V4= 4i
4
V5= i
5.R
5
V5= i
5(2)
V5= 2i
5
V6= i
6.R
6
V6= i6(5)
V6= 5i
6
Substituting V1, V
2, V
3, V
4, V
5, V
6values in equations
(1), (2) and (3), we get,
i2
4i4
2i5
= 0 ... (4)
2i3
+ 2i5
5i6
= 0 ... (5)
(3i1 10) + 4i
4+ 5i
6= 0
3i1+ 4i
4+ 5i
6= 10 ... (6)
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Now,
BT =
110
011
101
010
001
100
We know that,
The branch currents in matrix form is given as,
Ib=BT.I
L
6
5
4
3
2
1
i
i
i
i
i
i
=
3
2
1
110
011
101
010
001
100
I
I
I
From the above matrix, we have,
i1=I
3
i2
=I1
i3=I
2
i4= I
1+I
3=I
3I
1
i5= I
1+I
2=I
2I
1
i6= I
2+I
3=I
3I
2
Substituting i2, i
4, i
5values in equation (4), we get,
I1 4(I
3I
1) 2(I
2I
1) = 0
I1 4I
3+ 4I
1 2I
2+ 2I
1= 0
7I1 2I
2 4I
3= 0 ... (7)
Substituting i3, i
5, i
6values in equation (5), we get,
2(I2) + 2(I
2I
1) 5(I
3I
2) = 0
2I2+ 2I
2 2I
1 5I
3+ 5I
2= 0
2I1+ 9I
2 5I
3= 0 ... (8)
Substituting i1, i
4, i
6values in equation (6), we get,
3(I3) + 4(I
3I
1) + 5(I
2I
1) = 10
3I3+ 4I
3 4I
1+ 5I
2 5I
1= 10
9I1+ 5I
2+ 7I
3= 10 ... (9)
The equations (7), (8) and (9) represent the network
equilibrium equations.
Q7. (a) Is Norton theorem dual of Theveninstheorem? Justify your answer.
Answer :
Nov./Dec.-12, (R09), Q7(a) M[5]Considering a Thevenin circuit, a voltage source is
connected in series with a resistance. However, when a
Norton circuit is considered, the source will be a current
source, which is a dual of voltage source and also the series
resistance will be replaced with a parallel resistance, which
is also a dual representation from series connection to parallel
connection. Hence, from the above, it is clear that the Norton
circuit represent a dual of Thevenin circuit. Further, we can
conclude that the Norton theorem is a dual of Thevenin
theorem.
(b) Find the current in the 10 resistor asshown in below figure using superposi-tion theorem.
+
5
1 1
10
1A
2A
10 v A B
+
5
1 1
10
1A
2A
10 v A B
FigureAnswer : Nov./Dec.-12, (R09), Q7(b) M[10]
Given circuit is shown in figure (1),
+
5
1 1
10
1A
2A
10 v A B
+
5
1 1
10
1A
2A
10 v A B
Figure (1)
To determine the current in 10 resistor usingsuperposition theorem,
To find the current through 10 resistor, consideringeach source at a time.
Firstly, considering the 10 V source i.e., open
circuiting the current sources. The circuit diagram when
considering 10 V source is shown in figure (2),
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+
5
1 1
10 10 V+
5
1 1
10 10 V
Figure (2)
The step by step series parallel reduction of the
circuit of figure (2) is shown as,
+
5
1 1
10 10 V+
5
1 1
10 10 V
Figure (3)
+
5
1 + 1 = 2
10 10 V+
5
1 + 1 = 2
10 10 V
Figure (4)
+
10 10 V
=+
43.125
25
(I1)10
+
10 10 V
=+
43.125
25
(I1)10
Figure (5)
Current through 10 resistor is given by,
(I1)10 = 43.110
10
+
= 0.87 A
Now considering the 2 A current source and
neglecting the other sources. Therefore, the circuit of figure (1),
can now be redrawn as shown in figure (6),
5
1 1
10
2 A5
1 1
10
2 A
Figure (6)
The step by step series parallel reduction of the
circuit of figure (6) is shown below,
5
1 1
10
2 A
5
1 1
10
2 A
Figure (7)
1
2A1 10 5
(I2)10
1
2A1 10 5
(I2)10
Figure (8)
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1
2A1 =+
33.3
105
105
1
2A1 =+
33.3
105
105
Figure (9)
Current through (1 + 3.33) branch is given as,
33.41
12
+
= 33.5
2= 0.375 A
Now from figure (8), we have current flowing through
10 resistor is given as,
(I2)
10 = 105
5375.0
+
= 0.125
Now, considering the 1A current source and
neglecting other source. The circuit from figure can now be
drawn as,
5
1 1
10
1 A
5
1 1
10
1 A
Figure (10)
The step by step series parallel reduction of the
circuit of figure (10) is shown in figure (11),
5
1 1
10
1 A
5
1 1
10
1 A
Figure (11)
1
1A1 10 5
(I3)10
1
1A1 10 5
(I3)10
Figure (12)
1
1A1 =
+
33.3105
105
1
1A1 =
+
33.3105
105
Figure (13)
1A 4.33 1
1A 4.33 1
Figure (14)
Current through branch 4.33 is given as,
I4.33 = 33.41
11
+
=33.5
1= 0.188 A
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From figure (12), current through 10 resistorbranch is given as,
(I3)
10 = 1055188.0
+
= 0.063 A
According to the super position theorem, currentflowing through 10 resistor branch is given as,
I10 = (I1)10 + (I2)10 + (I3)10
= 0.87 0.125 0.063
= 0.682.
Q8. (a) State and explain the compensationtheorem.
Answer : Nov./Dec.-12, (R09), Q8(a) M[7]
For answer refer Unit-VIII, Q20.
(b) Apply Thevenins theorem and obtainthe current passing through 210 F
capacitor of figure.
~110 V, 50 Hz
10
70
100 F
210 F
~110 V, 50 Hz
10
70
100 F
210 F
Figure
Answer : Nov./Dec.-12, (R09), Q8(b) M[8]
The given circuit is shown in figure (1),
~
10 100 F
70 210 F~
10 100 F
70 210 F
Figure (1)
From figure (1), we have,
Voltage, V= 110 0 V
Frequency,f= 50 Hz
Resistance,R1
= 10
Capacitance, C1
= 100 F
Resistance,R2= 70
Capacitance, C3
= 210 F
And also,
By analyzing the circuit we have,
X2
= 0
R3
= 0
Now,
The figure (1) is modified as shown in figure (2),
~
A
B
10 100 F
110 V, 50 Hz
Z1= R
1+
Z2= R
2= 70
1cjX
~
A
B
10 100 F
110 V, 50 Hz
Z1= R
1+
Z2= R
2= 70~
A
B
10 100 F
110 V, 50 Hz
Z1= R
1+
Z2= R
2= 70
1cjX
Figure (2)
From figure (2),
The Thevenin voltage is given as,
VTh
=21
2.ZZ
ZV
+... (1)
Now,
The capacitive reactance of branch-1 is given as,
1C
X =fC21
= 610100502
1
=03142.0
1
= 31.827
Now,
Impedance,
Z1
=R1 1C
jX
= 10 j31.827
Impedance,
Z2
=R2
= 70
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Substituting V,Z1,Z
2values in equation (1), we get,
VTh = 110 0 70)827.3110(
70
+j
= 110 0 )827.3180(
70
j
= 110 0 (0.7554 +j0.3005)
= 110 0 0.813 21.692
= 89.43 21.692
Now,
The equivalent Thevenin impedance is given as,
ZTh
=Z1
+Z2
= 10 j31.827 + 70
= (80 j31.827)
The equivalent Thevenin circuit is represented in figure (3),
ZTh
A
B
C = 210 FVTh
ZTh
A
B
C = 210 FVTh
Figure (3)
Now,
The current passing through the 210 F capacitor is given as,
iC
=3Th
Th
ZZ
V
+... (2)
But,
Z3
=3C
jX =32
fC
j
= 610210502
j
=066.0
j
= j 15.1515
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Substituting VTh
,ZTh
,Z3
values in equation (2), we get,
iC
=)1515.15()827.3180(
692.2143.89
jj +
=1515.15827.3180
692.2143.89
jj
=89.43 21.692
80 46.9785j
=422.30773.92
692.2143.89
iC = 0.963952.114
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