1
CHAPTER – 1: NETWORK THEORY Basic Circuit Analysis Analysis and KVL/KCL Resistance: It is the property of a resistor resistor to oppose current. R + V –
i
Case-I:
R
I
Case-II:
Conventional current flow from high to low potential. pot ential.
–
Electron current flow from low to high potential
+
In both case I and case II: V iR Where resistance of Resistor: R
l A
where l = length of o f conductor, conducto r, A = Area of conductor, co nductor, = resistiv r esistivity ity of conductor. conducto r.. So
RA l
1 and conductivity = Resistivity
1
Current:
The phenomenon of transferring charge from one point in a circuit to another is described by the term electric current. An electric current may be defined as the time rate of net motion of electric charge across a crosssectional boundary i
dq dt
= sec, i = Amp q = Coulombs, t =
Voltage:
To move the electrons from one point to other point in particular direction external force is required. In analytical circuit external force is provided by emf and it is given by V
dW dq
where a differential amount of charge dq is given a differential increase in energy dW . Voltage = “Energy per unit charge” = “Work per unit charge” A voltage can exist between a pair of electrical terminal whether a current is flowing or not. An automobile battery, for example, has a voltage of 12V across its terminals even if nothing whatsoever is connected to the terminal.
Ohm’s Law:
At constant temperature, the potential difference V across the terminals of a resistor R as show below, is directly proportional to the current i flowing through it. That is V
i
V
iR
i +
R V
–
Ohm’s law can also be expressed in terms of conductance G as i GV
G
1 R
G = Conductance [mho or Siemens]
NETWORK THEORY
2 Field Interpretation of Ohm’s Law:
“At constant temperature current densi d ensity ty is directly proportional proport ional to electric field field intensity.” intensity.” I
a
J
l
J
E
E J ........... Current density (A/m2),
........... Conductivity [( m) 1 ]
E ........... .... ....... Electric field field intensity (V/m) (V/m) Linearly test of Resistor:
When resistive element obeys ohm’s law then the element is called as linear resistor otherwise it is a non-li no n-linear near resistor. resisto r.
Power Dissipation in a Resistor: P Vi (iR )i i R 2
V2 R
V
2
G
I 2 G
Practical Voltage Sources: It delivers energy at specified (V) which depends on current delivers by
sources. i
I V
R s
R s V(t) Vs
V (t ) Vs IRs
i
Vs(t) AC
DC
Current Sources: Ideal current source delivers energy at a specified (I), which is independent on voltage
across the source. Internal resistance of ideal current source = . i(t)
I
I Is
V
Is(t (t))
V ideal AC
DC
Practical Current Sources: Practical current sources delivers energy at specified current I, which is dependent
on voltage across the source. In real time system current source does not exist. I
i(t)
+
I Is
R 1
V
Is(t (t)) V
DC
–
AC
NETWORK THEORY
3 Capacitance: It is the property of capacitor to hold charge q to confined electric field. +
V
–
d
C
EA
I
V
d
1
idt
C
V
d = distance between plates, A = Area of cross section of each conducting plates Note: When the capacitance of capacitor is independent of current is called as linear capacitor. Q
C
I
Q
ic
V |Constant
V
dV dt
So under steady state i.e. at t
dV dt
i 0
Capacitor acts as open circuit: Inductance:
If the current ‘i’ flowing in an element of figure below changes with time, the magnetic flux ‘ ’ produced by the current also changed which causes a voltage to be induced in the circuit, equal to the rate of flux linkage that is V
d
V
dt
dt dt
V
i.e.
L
di dt
where, L is constant of proportionality and is called self inductance. +
i(t)
i(t) S
(t)
v(t)
– (a) Inductive circuit
AB
G
(b) Mutually coupled circuit
Therefore, there is an induced emf in coil B, which is equal to
d dt
e N
or
di dt
e L
As (di/dt ) is directly proportional to ( d / dt ) . Comparison of these two equation gives N Li
NETWORK THEORY
4
From above it may noted that an inductor is a device, while inductance is the quantity L. L
N i
As V L
di dt
, so in steady state i.e. t
V 0
And circuit behaves like shots circuit. Active Element: When an element is capable of delivering energy for infinite period of time is called as active element. Example: Voltage, source, current source. Passive Element: When an element is not able to deliver energy for infinite time period is called as passive element. Example: Resistor, inductor and capacitor Active Elements: Voltage Source: Ideal voltage source delivers energy at a specified ‘V’. Internal resistance of ideal voltage
source is zero.
Voltage Divider Rule:
V1 R 1
R1 V1 V s , R1 R2
R2 V2 V s R1 R2
Vs ~ V2 R 2
Current Division Rule: I
~
I1
I2
R 1
R 2
1 1 1 R eq R1 R 2
R2 I1 I , I2 R R 1 2
R 1 I R1 R2
Same for Inductor: I
~
I1
I2
L1
L2
1 Leq
1 L1
1 L2
L2 I1 I; I2 L L 1 2
NETWORK THEORY
L 1 I L1 L2
5
For capacitor: I
~
I1
I2
C1
C2
I1
C1 C 2
Ceq
C1 C1 C2
I , I2
C 2
I C1 C 2
Kirchoff’s Voltage Law (KVL):
It states that algebraic sum of all voltages in a closed loop is equal to zero. It is based on conservation of energy. R 1 + V1 – I
+ Vs1 –
+ – Vs2
V2 + R 2 –
Applying KVL:
Vs1 V2
Vs1 V2 V1 Vs 2 0 Q.
Calculate values of V1 and V2 across 4 and 5 resistor by KVL. 4 + V1 –
5V
1V
3V
2
I
6V
Soln.
V1 Vs 2
2V
5 + V2 –
Applying KVL 6 + 3 + 1 – 5 – 4I – 2I – 2 – 5I = 0 3 = 11I, I
3 Amp 11
V1 = 4I = 4 ×
3 11
V
12 volt 11
3 15 = volt 11 11 Calculate VAB in the given circuit V2 = –5I = –5 ×
Q.
2 5V
I1
Y
A 2V
4 X
2
10V
I2
5
B
NETWORK THEORY
6 Soln.
KVL at loop 1: (Left loop) –4I1 – 2I1 + 5 = 0, I1 = 5/6 Amp KVL at loop 2: (Right loop) 10 – 2I2 – 5I2 = 0 I2 =
Q.
10 Amp, VAB = VAX + VXY + VYB 7
5 10 20 20 32 = 4 2 2 = 2 = 6 7 6 7 21 The value of I1 in the circuit is
V AB
1.53Volt
3A 5
7
9
5A
+ 8V –
+ 2V – I1
Soln.
The circuit can be redrawn in the form of voltage source as [Current voltage source] 7
+
15V
– 5 – 45V + 9 + 2V –
+ 8V – I1
Applying KVL 8 – 7I – 15 – 5I + 45 – 9I – 2V = 0 53 – 17 – 21 I = 0, I =
36 21
I 1.71
Kirchoff Current Law (KCL) (i)
KCL states that algebraic sum of all current meeting at a point is equal to zero.
i
c
(ii)
0
KCL worked on principle of law of conservation of charge. I1 I2 I3
I5 I4
I 1 + I 2 + I 3 – I 4 – I5 = 0
I1 +I 2 +I3 =I4 +I5
NETWORK THEORY
7 Q.
Calculate the values of V1 and V2 by KCL: 5
1
V2
1
6A
Soln.
2
V1
3
Applying KCL at node 1: 6 =
3V
V1
1
V1 V 2
... (1)
2
3 V2 V2 V2 V1 1 3 2 By solving equation (1) and (2); We get V1 = 5V and V2 = 3V KCL at node (2) :
... (2)
Limitations of KVL and KCL:
KCL and KVL will fail for the high frequency circuit. KVL and KCL will fail in distributed elements since in distributed element it is not possible to separate effect of R, L, C. Source Transformation: V
R
I
V/R
R
Series Combination of Batteries: R
2R
R V2
V1 – V2
V1
Parallel combination of current sources:
R 1
I1
Q.
R 2
I2
I1+I2
R1 || R 2
Obtain single current source for network shown A
Soln.
+ 18V –
+ – 10V
6
5
3A
6
2A
5
5A
2.73
NETWORK THEORY
8 Q.
Convert given circuit into a single voltage source X 8A
4
12A
6 Y X
Soln.
4 + – 32V
10 + – 104V
6 + – 72V Y
Linear and non-linear Elements:
A linear network shows linear characteristics of voltage versus current. For a non-linear element the current passing through it does not change linearly with the linear change in applied voltage at a particular frequency. Semiconductor devices are usually examples of non-linear element. In V.I. relation if output is zero for zero input and relation is linear then it is called linear network or element obeys ohm’s law. V = ki linear V = ki2 Non-linear V = 2i + 3 Non-linear V = in Non-linear V
V
Linear
I
Non-Linear
i
Simple resistors inductors and print at end capacitors are linear elements and their resistance inductance and capacitances do not change with a change in applied voltage on the circuit current . Active and Passive Elements:
If a circuit element has the capability of enhancing the energy level of a signal passing through it then it is called an active element. Vacuum tubes and semiconductor devices are active elements on the other hand resistors, inductors, capacitors, thermistors etc. are passive elements as they do not have any intrinsic mean of signal boosting. In V-I relation if any portion has V-I value as negative then it is active network. Bilateral and Unilateral Element:
If the magnitude of the current passing through an element is affected due to change in polarity of polarity of the applied voltage then element is called unilateral element. On the other hand if current magnitude remains the same even if the applied voltage’s polarity is changed then it is called a bilateral elements. If by changing variable relation between both dependent variable and independent variable remains same then it is called bilateral network otherwise unilateral network.
NETWORK THEORY
9 Mesh Analysis:
Mesh is a property of a planner circuit and is undefined for a non-planner circuit. We define a mesh as “A loop that does not contain any other loops within it.”
+ – + – Fig. (a) [Planner]
Fig. (b) [Non-Planner]
No branch passes over or under any other branch Network Theorem: Superposition Theorem:
The statement of superposition theorem follows as below: “In any linear bilateral network having more than one source, response in any one of the branches is equal to algebraic sum of the response caused by individual source while rest of the sources are replaced by their internal impedances.
Note:
The principle of superposition is useful for linearly test of the system. This is not valid for power relationship. Sources can be made inoperative by (a) Shot circuiting the voltage sources (b) Open circuiting the current sources A linear network comprises independent sources, linear dependent source and linear passive elements like resistor, inductor, capacitor and transformer. Moreover, the components may either be time varying or time invariant. Other Definition: “The superposition principle states that the voltage across (or current through) on element in a linear circuit is the algebraic sum of the voltage across (or current through) that element due to each independent source acting alone. Q. Calculate value of V for the given circuit by using superposition theorem. 8
+ 6V –
Soln.
+ 4 V –
3
At one time only effect of one source is considered. Voltage source is short circuited and current source is open circuited. Since there are two sources, So, V = V 1 + V2
NETWORK THEORY
10
Let V1 is the voltage across 4 ohm due to 6V voltage source alone (in this case current source is open circuited). 8 + 4 V1 –
+ 6V –
V 1
6 84
4,
V 1
6 4 2V 12
Let V2 is the voltage due to 3A current source alone (in this case voltage source is short circuited). 8 + 4 V2 –
Q.
3A
V2
4
V 2
8
,
3=
V 2
8
{2 1}
V 2
8 volts
So total voltage V is V = V1 + V2 = 2 + 8 = 10 volt Calculate the value of V0 by use of superposition theorem. 3 + V0 2 –
Soln.
3=
5 + – 20V
8A
Since there are two sources so total voltage due to both sources will be V0 V1 V 2 Let V1 is the voltage only due to 8A current source, (In this case 20V source is short circuited) 3
5
+ V1 2 –
3 + V0 2 –
8A
5 + 20V –
8 5 4 A , V1 = 2 × 4 = 8 volt 3 2 5 Let V2 is voltage due to 20V source alone i1
V 2
20 2 , V 2 = 4 volt, V 0 = V 1 + V 2 = 8 + 4 V 0 12volt 10
Thevenin’s Theorem:
Any two terminal linear bilateral network can be replaced by a voltage source in series with impedance. Zth
N/W
A B
A
+ V – th B
NETWORK THEORY
11 Calculation of Thevenin’s equivalent for independent source: For Calculation of Rth: All independent current sources are open circuited and all independent voltage
sources are short circuited. For Calculation of Vth: Voltage across open circuit terminal.
Calculation of Thevenin’s equivalent for dependent source: For calculation of Vth: Voltage is calculated across open circuited terminal by assuming current zero in that
terminal.
For calculation of Rth: First Ise is calculated by assuming short-circuited terminal and then R th is calculated
by RTh VTh / I Sc Q.
For the given circuit calculate the power loss in the 1 ohm resistor by use of thevenin’s theorem. X 1
5
10 2A + – 10V
Soln.
Y
For calculation of Vth branch X–Y terminal is assumed as open circuited and let voltage is VthI1 10
2A
I2 5
+ – 10V
2 = I1 + I 2 , 2 = V x
10 Volts ,
V x
X Vx + –
– + Y V y
V x 10
10
V x
5
1 1 1 , 10 5
, 2 = V x
V y 5 2 10
VTh
3 3 10
V x
10 Volt
For Calculation of RTh: (If dependent source is present) 2A current source is open circuited and 10V
source is short circuited and let resistance is R Th :
10
I Th
Q.
R Th
5
V Th RTh
RL
RTh
10 5 10 3.3 , , RTh 10 5 3
10 2.31A 3.33 I
Find the current through the 5 resistor in the circuit by use of Thevenin’s theorem. 2
2A
1
5
+ – 5V
NETWORK THEORY
12 Soln.
For calculation of VTh = 5 ohm resistor is open circuited and so current in 5 ohm will be zero. VTh
Vab 5V 2 a
1
2A
+ – 5V
b
For Calculating RTh: 2A current source is open circuited and 5V source is short circuited. 2 a
1
5 V Th 03 I 1A 0 R Th = , L R R 5 10 3 0 Th L
b
Norton’s Theorem: Any two terminal linear bilateral network containing active and passive element can be
replaced by an equivalent current source in parallel to an equivalent impedance current source. A A
N/W
Z N
I N
B
Z N
Z Th
B
Note: R Th I N
R N
+ – VTh (a) Thevenin's equivalent circuit
Both of the above circuits are convertible to each other with the relations given as below: RTh
Q.
(b) Norton's equivalent circuit
RN , VTh I N RN I N RTh
Find the Norton’s equivalent circuit across a-b for the network shown in figure: a 5
2
5A 2V
b
5
Soln.
5A
Isc
2V
NETWORK THEORY
13
–2 + 5(5 + Isc) = 0, I sc =
2 5 10 RTh 1.43 25 7 Find Norton’s equivalent to the right of a-b terminal (across 3V source)
RTh
Q.
2 5 = 0.4 – 5 I sc 4.6 A 5
a
10
+ – 3V
5
i0 = 1A
b
Soln.
The equivalent circuit after shorting ab terminal. 10
10
V + – i0 = 1A
5
Isc
i1
(b) [To calculate RTh]
(a)
i0 = i1 + Isc, I = i1 + I=
V
5
V
10
5
,I=
V
10
3V 10 1 ,V= and Isc = = 0.33A RTh 10 5 15 10 3 3
Reciprocity Theorem:
A linear network is said to be reciprocal or bilateral if it remains invariant due to the interchange of position of cause and effect in the network. For verification of the reciprocity theorem following conditions must be satisfied. Circuit should consist of linear, time-invariant bilateral element. Circuit should consist of only a single independent sources. When circuit consist dependent source, reciprocity theorem can not be verified. Maximum Power Transfer Theorem: Maximum power transferred from source to load is only possible when (i) Source impedance = Load impedance (ii) Thevenin impedance = Load impedance For DC Circuits: X DC Source Network
I
R Th
X I + – VTh
R L
R L
Y Y
(Load connected to the do source network)
(Equivalent source network and load)
NETWORK THEORY
14 I
V Th RTh
RL
while the power delivered to the resistance load is 2
V 0 P L I RL RL R R Th L 2
.... (i)
Differentiating equation (i) with respect to R L and equating to zero. We get R L RTh Hence, it has been provided that power transfer from a dc source network to a resistance network is maximum when the internal resistance of the dc source network is equal to the load resistance. Again value of that maximum power is, 2
Pmax
2
VTh RTh
( RTh RTh )2
V Th
4 RTh
Pmax
or
The total power supplied is thus: P 2
V Th2
[ RTh R2 ]
4 R L
V Th2
4 RTh
P
2
V Th
2 RTh
During maximum power transfer the efficiency becomes, 2
V Th
Pmax P
100%
4 RTh V Th2
100%
50%
2 RTh
So that efficiency in this case is 50% i.e. half of the total power is transferred to the load R L. [Maximum Power Transfer Theorem for ac circuits] Consider the Thevenin’s equivalent circuit for an ac network as shown below: ZTh VTh ~
ZL
Z L RL JX Here, ZTh RTh JX Th , L
Now let us consider different cases for maximum power transfer. Case-1: Both R L and XL are variable When both R L and XL are variable then maximum power from source to load will be transferred if load impedances is complex conjugate of internal impedance of the network. * i.e. Z L Z Th
In this case maximum power (active) will be calculated as Pmax
NETWORK THEORY
V Th2
4 R L
15
Also during maximum power transfer efficiency will be 50%. Case-2: R L is variable but XL is constant.
In this case maximum power will be transferred when, R L | ZTh JX L | 2 or R L RTh ( X Th X L )2
R L 100 % Efficiency can be calculated as R R Th L
Efficiency comes out to be greater than 50% 50%
Case-3: Load impedance is purely resistive.
In this case maximum power will be transferred when, R L | Z Th |
Efficiency comes out to be greater than 50%. Case-4: R L and XL are variable but the impedance angle is constant i.e. Z L
and RL JX L
X L Constant R
Q tan 1
In this case maximum power will be transferred when, Z L Z Th Q.
Calculate value of R in circuit such that maximum power transfer takes place and also calculate amount of this power. 1 + 4V –
5 2
R
1 6V + –
R Th
Soln.
R
Above circuit can be solved by Thevenin’s theorem: VTh 1
To find R Th
5 2
+ To find VTh open load R 4V –
1
R Th RTh
1
5 2
(2 ||1) 5 ||1
1
VTh
NETWORK THEORY
RTh
VTh
0.85
6.4V
16 V Th2
(6.4) 2 12W Maximum Power = 4 RTh 4 0.85 Q.
Assuming maximum power transfer from source to load R calculate the value of R and maximum value of power transferred. 1
50V 1 + –
5 10
R Y
3
Soln.
X
100 25 VTh across XY = – V, R Th across XY = 3 3 2
Pmax
(100 / 3)2 100 = Watt 25 4 RTh 3 4 3 V Th
NETWORK THEORY
17 Assignment - Network Theory 1.
A segment of a circuit is shown in figure VR = 5V, VC = 4 sin 2t . The voltage VL is given by Q + 1A 5 VR – 2A
P
+ 2H VL – S
2.
1F +V –
R
c
(a) 3 – 8 cos 2t (b) 32 sin 2t (c) 16 sin 2t (d) 16 cos 2t In the circuit of figure, the magnitudes of VL and VC are twice that of VR . Given that f = 50 Hz, the inductance of coil is VR 5
C VC
5<0 ~
3.
(a) 2.14 mH In the figure value of R is
L
(b) 5.30 H
VL
(c) 31.8 mH
(d) 1.32 H
R 14
1
10A
5A 2
+ 100V –
4.
+ – 40V
(a) 10 (b) 18 (c) 24 In figure, the potential difference between points P and Q is
(d) 12
2A 2
P
R 10V
6
(a) 12V (b) 10V In figure, the value of resistance R in is 10 + 100V –
(a) 10
(b) 20
Q
~
8
5.
4
(c) –6V
(d) 8V
2A R
10
(c) 30
NETWORK THEORY
(d) 40
18 6.
The RMS value of the voltage u(t ) = 3 + 4 cos (3 t ) (a)
7.
17 V
(b) 5V
(c) 7V
(d) (3 2 2)V
Assuming ideal element in the circuit shown below, the voltage Vab will be 2
a
– 5V +
Vab
1A
i b
8.
(a) –3V (b) 0V (c) 3V The current through the 2k resistance in the circuit shown is 1k
C 1k
A
B
2k 1k
(d) 5V
D 1k 6V
9.
(a) 0mA (b) 1mA The time constant for the given circuit will be 1F
1F
10.
(c) 2ma
(d) 6mA
3 3A
3
1F
(a) 1/gs (b) 1/4s (c) 4s (d) gs In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3 is R
10V
+ –
3 Load
6
(b) 3 (c) 6 The r.m.s. value of the current i(t ) in the circuit shown below is (a) 0
11.
1F
(d)
1H 1
i(t)
7
~
(1 sin t) v
(a)
1 A 2
(b)
1 A 2
(c) 1A
NETWORK THEORY
(d)
2A
19 12.
In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at the terminals P– Q, is given by
20
(a) (2V,5) 13.
10 10
4V
(b) (3V,5)
Unknown Network
(c) (4V,5)
(d) (2V, 7.5)
In the circuit shown, the power supplied by the voltage source is 1
1
1
+ 10V –
1A
1
2A
1
14.
(a) 0W (b) 5W The voltage e0 in the figure is:
(c) 10W
(d) 100W
2 10
8A
16V 6
12
+ e0
–
15.
(a) 48V (b) 24V (c) 36V In the circuit of figure, the value of the voltage source E is
(d) 28V
0V + – + 2V – 1V E + + – – V1 4V 10V
V2
16.
(a) –16V (b) 4V The voltage V in figure is equal to:
(c) –6V
(d) 16V
4V + – 2
+ 5V –
+ – 4V
3 + V –
17.
(a) 3V The voltage in figure is
(b) –3V
(c) 5V
(d) None of these
3
+
+ – 10V
V
2A
–
(a) 10V
(b) 15V
(c) 5V
NETWORK THEORY
(d) None of these
20 18.
What is the value of R required for maximum power transfer in network shown above? 4
5 20
25V
19.
R
3A
(c) 8
(a) 2 (b) 4 The voltage across the terminal a and b in figure is: 2
1
a
2
1V
(d) 16
3A b
20.
21.
(a) 0.5V (b) 3.0V The nodal method of circuit analysis is based on (a) KVL and ohm’s law (c) KCL and KVL The current i4 in the current of figure is equal to:
(c) 3.5V
(d) 4.0V
(b) KCL and Ohm’s law (d) KCL, KVL and Ohm’s law
5A
3A
7A
i4
22.
i3 = 4A
(a) 12A (b) –12A (c) 4A (d) None of these Thevenin equivalent voltage VAB and resistance R Th across the terminals AB in the above circuit are 3
2 10V 3
(a) 6V,5 23.
A
(b) 4V,5
B
2
(c) 2V,2.4
(d) 2V,2.5
The differential equation for the current i(t ) in the circuit of the figure is + sin t –
(a) 2 (c) 2
d 2i dt
2
d 2i dt 2
2 2
di
i (t ) sin t
dt di
i (t ) cos t
dt
i(+)
2
2H 1F
(b)
(d)
d 2i dt
2
2
di
2i (t ) cos t
dt
d 2i
di
dt
dt
2 2
NETWORK THEORY
2i(t ) sin t
21 24.
The RC circuit shown in the figure is C
R
+
+
Vi
R
V0
C
–
25.
–
(a) a low-pass filter (b) a high-pass filter (c) a band-pass filter (d) a band-regeat filter In the circuit shown below, the value of R L such that the power transferred to R L is maximum. 10 10 10 + – 5V
26.
1A
(a) 5 (b) 10 (c) 15 Find the value of C to deliver the maximum power to load.
+ – 2 2 sin 2t
2
J2
29.
–J2
ZL
(c) ( J 2 )
2) (b) (2 J
(d) 2
The voltage across a capacitor is triangular in waveform. The waveform of current is (a) triangular (b) trapezoidal (c) Sinusoidal (d) Rectangular The value of R in of the network 1
1 R in
1
1 1
1 1
1 1 (d) 3 2 The maximum power that can be distributed in the load in the circuit shown is (a) 1.62
30.
(d) 4
J2
+ Vs ~ –
28.
4 Load
C
(a) 0.125F (b) 0.5F (c) 2F Find ZL such that maximum power is transferred to it.
2) (a) (2 J
(d) 20
1H
2
27.
R L
+ – 2V
(b) 2
(c)
3
1 6
9V
R L
10
(a) 0.396 W
(b) 6W
(c) 6.75W
NETWORK THEORY
(d) 13.5W
22 31.
Find Vx from the given circuit. 5 + Vx 10 –
10A
32.
50
(a) 42.2 V (b) 83.3 V If Vc(f) = 4 cos (105 t)V in the circuit, find Vs.
(c) 97.3V
(d) 103V
2mH + 80nF – Vc
+ Vs –
33.
(a) –6.4 cos 105 t V (b) 2.4 cos 105 t V (c) 6.4 cos 105 t V (d) –2.4 cos 105 t V Determine the current through the branch AB of network shown below. A
J5
5
5
J5
+ – 10V
J10 B
34.
(a) 3.5 + J 1.5 (b) 3.5 – J 1.5 Thevenin equivalent circuit is given by 2 + 4V –
(c) 1.5 + J 3.5 3
(d) 1.5 – J 3.5
+
Vx/4000
Vx –
35.
(a) 4V and 10K (b) 4V and 6K (c) 8V and 10K (d) 8V and 6K What is the ratio of currents in the circuit due to 3A and 2A source? Current is taken through R. 100
100 = R
3A
(a)
3 2
(b)
100
2 3
2A
(c)
9 4
NETWORK THEORY
(d)
4 9
23 ANSWER KEY
1.
(b)
2.
(c)
3.
(d)
4.
(c)
5. (b)
6.
(a)
7.
(a)
8.
(a)
9.
(c)
10. (a)
11.
(b)
12.
(a)
13.
(a)
14.
(d)
15. (a)
16.
(a)
17.
(a)
18.
(c)
19.
(c)
20. (b)
21.
(b)
22.
(b)
23.
(c)
24.
(c)
25. (c)
26.
(a)
27.
(d)
28.
(d)
29.
(a)
30. (a)
31.
(b)
32.
(d)
33.
(a)
34.
(c)
35. (a)
NETWORK THEORY
