BEEE UNit1 (DC & AC Circuits)
BEEE UNit1 (DC & AC Circuits)
BEEE UNit1 (DC & AC Circuits)
1 PART-A
SOLUTIONS
SHORT QUESTIONS WITH
What is a circuit?
01.
or
Ans a circuit comprises
eneTey
from source to load. In general, elements ana
1s d e f n e d as a path
closed that transfers energy supplyingcurrent
A circuit s o u r c e s are batteries,
generators (or) any
and loads usually. The energy is shown in figure.
elements. The.circuit
Wires
sources, connecting motors (or) any current absorbing
(or) lamps (or)
the loads may lights
be
>Connecting
wire
Energy Load
source
Connecting.
wire
Figure
w
Figure (1)
G Inductor
which stores the energy shown
in
the tormin of magnetic field whenever a current is passed through
Inductor is a passive element representation is figure
(2).
It is denoted by L'and its symbolic L Unls(A)
Figure (2)
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Ans
In general an ideal practical sóurces are classified as shown in figure (1).
Types of sources
Figure (1)
Circuits)
When indepenaent source and 1.3
dependent source are considered,
then the sources are classified as
Types of sources snoW
Independent Dependent
Source
source
Figure 12)
What is source transformation?
ns:
ource Transformation
Source transformation is a network reduction technique, in which oneform of source is replaced with its another equivalent
into a simple form and hence, calculation can be made easier.
m. Using this technique, a complicated networkcan be converted
source. A practical voltage source always
Basically, there are two types of sources i.e., the voltage source and thecurrent Using
with it. transformation
the source
current source will have aresistor in parallel
avea resistor in series with it and a practicalresistor can be converted to a current source in parallel with the same resistor.
chnique, a voltage source in series with a
source transformation technique.
Two operations can be performed using
source.
source to a current
Converting a given voltage
source.
current source to a voltage
Converting a given
State Kirchhoff's laws.
Model Paper-i, Q1(a)
ns:
rchboff's Current Law (KCL) that, the sum ot the
currents entering any node is equal to the sum ofcurrents leavino
Kirchhofs Current Law (KCL) states
hat node. OR
and leaving the node is zero.
Figure (1)
-Y
-B -B
2. In star connection the relation between load
voltage 2 In delta connection the relation between load vote
and phase voltage is given as, and phase voltage is given as,
3. In star connection, the relation between load current 3. In delta connection the relation between load
and phase current is given as, cu
and phase current is given as,
3
A23. Write the relation between phase values and line values in star and delta connected systems.
Ans
Star Connection
The relation between line voltage and phase voltage is,
Whenever a D.C. supply is given across a capacitor, it acts as open circuit.
a26. Distinguish between active and passive elements with suitable examples.
Ans:
Model Pape
Active Elements Passive Elements
1. Active element is an element which can energize a 1. A passive element is an element which consumetheen
circuit or network, i.e., it delivers a net amount of. being delivered by an active element. Some passive
energy. elements stores energy.
2. Active elements supply energy. 2 Passive elements utilize electrical energyeither by
converting or storing it.
3. Active elements can supply an average power for an 3. Passive elements cannot supply average power greater
infinitetimetoexternaBcircuit. than zero for an infinite time to external circuit.
4. Examples of active elements are, 4. Examples of passive elements are,
v, R(2)
Figure (1)
The current/
entering the
I,11 and I, which flows out ofnode P isP.divided into currents Figure (3)
Kirchhoff's current law, we node Therefore, with the
have, The voltage
current passes
drop occurs acrOss each
I=1+1,+I,+1, i.e., current in node Pis equal total through the circuit. The sum resistor
of the voltsa
current around the loop will be
leaving node P. to The voltages at equal to the total votage inltags
points 1, 3, 5, 7 will be more th
If nodeQ is considered, Kirchhoff's points 2, 4, 6,8.
same as current law will be
as
comE
applied to node P.
Therefore, from the Kirchhoff's
+1+ +I,=I voltage law, we
Consider the following
V V,+V,+ V, +V
figure (2). The current
supplied by the voltage
voltage law. sourve
obtained through Kirchhoff's
Using Ohm's law, the voltage across
given by, each resa
voltage source and enters the negative terminal as the current The series combination of the three equaid
as shown in figure (1). Let a voltage Vis apple
always flows from higher to lower potential. 1s
flowing through the circuit.
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3 S2
Current I'is the sum of the currents of1, wW-
and 7,
OW applying
the current and
current division rute for figure (3) to find
7, I,
109
Current. 1,=1x R4R 10v
SA
Current, I,=I'x
19.89 V-V-R
Soiving equation
(2) a
F1.989 Amps
120 V - R
Current in 5 2 branch =
V
5
100 V, -2R
10.84 20 R
5 Internal resistance, R
10V 2.168 Amps Now, substituting R
Current in 3 2 branch= 3
120 V,-20
V,= 120+2
19.89-10.84
V-10
Current in 12 branch= 1
10,84-10 Q33. Determine v, andi
figure.
known.Hen
=0.84Amps
voltage
20
terminal
load
current is
is 100 V.
the voltage
of120 Vwhen terminal
of the
voltage
2A, the
Current is
internal
resistance
short
circuit 12v
and
u l a t e the voltage
circuit
source
Cur
open
lgidlune-17, (R16), Q3(b)
Current.
. E p i a i n in detail about different representations of sinusoidal quantities.
ARS: (May/June-17, (R16), a3(a) | Model Papera
v n s oms that are employad for representing sinusoidal quantity are.
Compiex form (or) rectangular formn
Tngonometric form
Eponential faom
Polar form
Complex Form or Rectangular Form
i s foem he AC quantiy is represented as sum of real and imaginary part.
Ai)a-jb
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Nd A.C Circui
of
This formn of cuits) 1.29
and
representation is basically used
Ifa, +jb a,+jb, are two whenever have
we have to add or subtract two A.C quantities.
We have,
=1+ , , f.
= 1 +j0-
At)=4|eA
form of representing sinusoidal
quantity
called the exponential
The above expression is
Where,
and
A = Magnitude
Product-4|,+;
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What is form 1.31
Q42. of an tor
Explain its significance.
alternating quantity? In figure (2) shown above, the
sinusoidal wave A4
wave B with
Ans: sinusoidal
represents the reference wave and the 90°.
to the right by
Form Factor dotted line is the whose phase is shifted
wave
waveform B is saia to
the phase difference is 90° and the
nce other words
wavetorm
For answer refer Unit-I, Q10, ing behind waveform A by 90°. In
A Is leading waveform B by 90.
significance of Form Factor its
also be represents
in terms of
sine wave can
ne phasor diagram.
gives the peakiness of the
It and angular position by using a
whose
waveform. agnitude
of 60° and 120
the greate
will be the value
of form factor andPeakier the wave Consider
two cases i.e., for angles
vice-versa. phasor diagrams are shown below.
For example, if we compare the
sine and square waves. 900
o sine wave is peakier than square wave.
So the sîne wave
alue
has greater valu of form factor than
square wave. The form
factor is useful in rectifier circuits.
270
31t
2t Figure (3)
x (rad) 90
2
UCasurèment
that
referencewhereas,
theterrm phase two_waves.
to a 270°
vave)relative phase-between
difference in considered
a s the
sdefined figure (1) is Figure (4)
shown in shifted
left
sinusoidal
wave wave
is
Tf the s i n u s o i d a l
shifts
of the a r r o w
a phasephase above figures, the length
as the reference wave
and any
there
occurs
In both the sinusoidal wave with
then this the magnitude ofthe
wave
and
due to
ating 1 and 2 is
respectively. In this way the
eference
or right to this wave
shifts
formingand
angle diference of 60° and 120°
a phase
shifted
ie., the phase of t
f the in the
phase
at instant can be expressed as a positive
d i f f e r e n c e
d+ierence. position ofa phasor any m e a n s that when the phase
occut a
shift, thereOcCurs as phase
or negative angle. Negative angle i.e.,
example-90°. For this condition,
known
from 0.
is superposed
Q44. Prove that if a D.C current ofmos maximum
e(degrees) currént of
in a conductor by an A.C
value of the resultant is
value Iamps, the r.m.s.
90
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2 Analysis of 1.39
e-Phase
Three- Signal Phanase A.
and Current Balanced Circuits,ircuits.
Relations
(iv) Admittance
in Star andVoltage of impedance. It is
denoted
Rationalizing the
denominator, we get, vice-versa.
an A.C through R-L
Q48. Explain
the behavior of
(RTjX) series circuit.
(Rt jX)(RTJA)
Ans:
(RTiX) R-L Series Circuit
Y= (G F jB)U
-V -V
Where, susceptance,
X
B =
(R+X*)
( R + 2 (nduetive
susceptance)
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