JRE SCHOOL OF Engineering
PRE UNIVERSITY EXAMINATIONS MAY15
Subject Name
Electrical Engineering
Roll No. of Student
Date
30th April 2015
For EE, EC and IT only
Subject Code
Max Marks
Max Duration
Time
NEE-201
Set-A
100 Marks
3 Hrs.
10.00AM-1.00PM
SECTION – A
1. ATTEMPT ANY FOUR QUESTIONS.
(4*5=20)
(a) Define ideal dc voltage source & ideal current source? Draw the characteristics of ideal and
practical dc voltage and current sources.
Ideal voltage source is defined as the energy source which gives constant voltage across its
terminals irrespective of the current drawn through its terminals. The symbol for ideal voltage
source is shown in the fig. This is connect to the load as shown in Fig. At any time the value of
voltage at load terminals remains same. This is indicated by V-I characteristics shown in the Fig.
Practical Voltage Source:
But practically, every voltage source has small internal resistance shown in series with voltage
source and is represented by Rse as shown in the Fig.
Key Point: For ideal voltage source, Rse=0
Current Source
Ideal current source is the source which gives constant current at its terminals irrespective of the
voltage appearing across its terminals. The symbol for ideal current source is shown in the Fig. This is
connected to the load as shown in the Fig. At any time, the value of the current flowing through load
IL is same i.e. is irrespective of voltage appearing across its terminals. This is explained by V-I
characteristics shown in the Fig.
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But practically, every current source has high internal resistance, shown in parallel with current
source and it is represented by Rsh. This is shown in the Fig.
Because of Rsh, current through its terminals decreases slightly with increase in voltage at its
terminals.
Key Point: For ideal current source, Rsh= ∞
(b) State Thevenin’s Theorem. Explain it with the help of a simple example. Mention its
limitations.
Ans: Thevenin’s theorem:- Any active, linear, bilateral network can be replaced by a voltage source V th
(Thevenin’s voltage) and a resistance R th(Thevenin’s resistance) both are always connected in series across
the load terminals.
The concept of Thevenin's equivalent across the terminals of interest can be explained by considering the
circuit shown in the Fig. The terminals A-B are the terminals of interest across which RL is connected.
Then Thevenin's equivalent across the load terminals A-B can be obtained as shown in the Fig.
The voltage VTH is obtained across the terminals A-B with RL removed. Hence VTH is also called open
circuit Thevenin's voltage. The circuit to be used to calculate VTH is shown in the Fig., for the network
considered above. While Req is the equivalent resistance obtained as viewed through the terminals A-B
with RL removed, voltage sources replaced by short circuit and current sources by open circuit. This is
shown in the Fig.
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(a) Calculation of VTH
(b) Calculation of Req or Rth
While obtaining VTH, any of the network simplification techniques can be used.
When the circuit is replaced by Thevenin’s equivalent across the load resistance, then the load current
can be obtain as.
IL 
VTH
RL  Rth
By using this theorem, current through, any branch of the circuit can be obtained, treating that branch
resistance as the load resistance and obtaining Thevenin's equivalent across the two terminals of that
branch.
(c) Derive the quality factor Q of the series RLC circuit at resonance. A series circuit has R=10
Ω, L=0.01mH, and C=10 µF; Calculate its bandwidth and Q-Factor.
Ans: Reactive components such as capacitors and inductors are often described with a figure
of merit called Q. While it can be defined in many ways, it’s most fundamental
description is
It can also expressed in other way
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(d) Prove with the help phasor diagram that the power and power factor of a balanced threephase load can be measured by two wattmeters. Also draw the connection diagram and
phasor diagram.
Ans:
Expression of power and power factor using two wattmeter method in a balanced 3 phase star connected
load
Consider a star connected load and two wattmeter connection as shown in figure
Then readings of wattmeter are
W1 = VRBIRCos(VRB^IR)
W2= VYBIYCos(VYB^IY)
To find the angle between (VRB & IR) and (VYB & IY) let us draw the phasor diagram assuming load p.f be
CosΦ.
Thus angle (VRB^IR) = 30 – Φ
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And angle (VYB^IY) = 30 + Φ
Therefore,
W1 = VRBIRCos(30 - Φ)
Or W1 = VLILCos(30 - Φ) ………. (i)
W2 = VYBIYCos(30 + Φ)
or W2 = VLILCos(30 + Φ) ……….. (ii)
Adding equations (i) & (ii),
W1+ W2 = VLIL( Cos(30 - Φ) + Cos(30 + Φ ))
= VLIL( cos30cos Φ + sin30sin Φ + cos30cos Φ –sin30sin Φ)
= 2VLILcos30cos Φ
= √3 VLILcos Φ
W1+ W2 = √3 VLILcos Φ = Total 3-phase power ……….. (iii)
Subtracting equiation (ii) from (i)
W1 - W2 = VLIL( Cos(30 - Φ) - Cos(30 + Φ ))
= VLIL( cos30cos Φ + sin30sin Φ - cos30cos Φ + sin30sin Φ)
= 2VLILsin30sin Φ
= VLILsin Φ
W1 - W2 = VLILsin Φ …………. (iv)
Hence, tan Φ = √3 (W1 - W2)/ (W1 + W2)
(a) List various losses occurring in a transformer and mention the condition for the maximum
efficiency.
Ans: Losses in a Transformer In a transformer, there exists two types of losses.
i) The core gets subjected to an alternating flux, causing core losses.
ii) The windings carry currents when transformer is loaded, causing copper losses.
Core or Iron Losses: There are two types of losses presents in core of the transformer.
Hysteresis losses: Due to alternating flux set up in the magnetic core of the transformer, it undergoes a
cycle of magnetization and demagnetization. Due to hysteresis effect there is loss of energy in this process
which is called hysteresis loss. It is given by, hysteresis loss = Kh Bm1.67 f v watts where, Kh = Hysteresis
constant depends on material. Bm = Maximum flux density. f = Frequency. v = Volume of the core.
Eddy current losses:
The induced e.m.f. in the core tries to set up eddy currents in the core and
hence responsible for the eddy current losses. The eddy current loss is given by,
Eddy current loss = Ke Bm2 f2 t2 watts/ unit volume
Where, Ke = Eddy current constant t = Thickness of the core As seen earlier, the flux in the core is almost
constant as supply voltage V1 at rated frequency f is always constant. Hence the flux density Bm in the
core and hence both hysteresis and eddy current losses are constants at all the loads. Hence the core or
iron losses are also called constant losses. The iron losses are denoted as Pi. The iron losses are
minimized by using high grade core material like silicon steel having very low hysteresis loop by
manufacturing the core in the form of laminations.
Copper Losses The copper losses are due to the power wasted in the form of I 2 R loss due to the
resistances of the primary and secondary windings. The copper loss depends on the magnitude of the
currents flowing through the windings.
Total Cu loss = I12 R1 + I22 R2
= I12 ( R1 + R2' )= I22 ( R2 +R1' )
= I12 R1e = I22 R2e
The copper looses are denoted as Pcu. If the current through the windings is full load current, we get
copper losses at full load. If the load on transformer is half then we get copper losses at half load which
are less than full load copper losses. Thus copper losses are called variable losses. For transformer VA
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rating is proportional to current and voltage. As voltage is constant, we can say that copper losses are
proportional to the square of the KVA rating. So, Pcu is proportional to I2 or (KVA)2
Thus for a transformer,
Total losses = Iron losses + Copper losses = Pi + Pcu
Condition for Maximum Efficiency When a transformer works on a constant input voltage and
frequency then efficiency varies with the load. As load increases, the efficiency increases. At a certain load
current, it achieves a maximum value. If the transformer is loaded further the efficiency starts
decreasing.
The load current at which the efficiency attains maximum value is denoted as I2m and maximum
efficiency is denoted as ηmax.
The efficiency is a function of load i.e. load current I 2 assuming cos Φ constant. The secondary terminal
voltage V2 is also assumed constant. So for maximum efficiency, dη /d I 2 = 0 Now η = (V2 I2 cos Φ2 )/(V2 I2
cos Φ2 + Pi + I22 R2e)
(V2 I2 cos Φ2 + Pi + I22 R2e)(V2 cos Φ2) - (V2 I2 cos Φ2)(V2 cos Φ2 + 2I2 R2e) = 0 Cancelling (V2 cos Φ2) from
both the terms we get, V2 I2 cos Φ2 + Pi +I22 R2e - V2 I2 cosΦ2 - 2I22 R2e = 0 ... Pi - I22 R2e= 0 ... Pi = I 22 R2e =
Pcu So condition to achieve maximum efficiency is that, Copper losses = Iron losses Load Current I2m at
Maximum Efficiency For ηmax, I22 R2e = Pi but I2 = I2m2 R2e = Pi.
(b) Derive the emf equation in DC machine.
E.M.F. Equation of a DC machine:
Let

= Flux per pole in Weber
Z
= Total Number of armature conductors
= Number of slots x Number of conductors per slot
P
= Number of generator pole
A
= Number of Parallel path in Armature
N
= Armature rotation in revolution per minute (i.e., in rpm)
E
= e.m.f. induced in any parallel path in armature
According to the Law of electromagnetic Induction, the generated or induced e.m.f is proportional to rate
of change of flux with respect to time and if only one conductor is taken in to consideration then the
induced e.m.f is given by
E
d
dt
(1)
Initially let us consider only one conductor, which is rotating inside a magnetic field having ‘P’ number of
poles. If the conductor makes one complete rotation then to total flux cut by conductor
Total Flux  P
(2)
If ‘N’ is the speed in rpm at which the conductor is rotating in that magnetic field then time taken by the
conductor for one complete rotation
Time for one complete rotation =
1
60
min utes 
sec onds
N
N
(3)
There fore
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Average e.m. f induce in one conductor 
Or
E
P
60 / N
total flux cut in one revolution
time taken for one complete revolution
PN

(4)
(5)
60
If the machine has total ‘Z’ number of conductors and has ‘A’ number of parallel path, number of
conductors in each parallel path is
Number of conductor / Parallel Path 
Z
A
(6)
The conductors in each parallel path are connected in series hence the total e.m.f is added. Total e.m.f
induced in all conductors will be
E
Or
E
PN
60

Z
A
(7)
ZNP
(8)
60 A
SECTION – B
2. ATTEMPT ANY THREE QUESTIONS.
(3*10=30)
(a) Using nodal analysis, calculate the currents I 1 and I2 in fig.1.
fig.1
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(b) State and explain superposition theorem. Mention its limitations. Find current ‘I’ in fig.2 using
superposition theorem.
Statement of Superposition Theorem: In any active, linear, bilateral containing two or more than two
energy sources the total response (voltage or current) will be the algebraic sum of responses produced by
each energy source individually. When one energy source is considered then all remaining energy sources
are replaced by their internal resistances. Ideal voltage source is replaced by short circuit and ideal
current source by an open circuit.
Suppose there are two voltage source s V1 and V2 acting simultaneously on the circuit. Because of
these two voltage sources, say current I flows through the resistance R.
Now replace V2 by short circuit, keeping V1 at its position and measure current through the resistance, R.
Say it is I1.
Then replace, V1 by short circuit, reconnect V2 to its original position and measure current through the
same resistance R and say it is I2.
Now if we add these two currents, I 1 and I2 we will get the current which is equal to the current - was
actually flowing through R, when both voltage source s V1 and V2 were acting on the circuit
simultaneously. That is I1 + I2 = I
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Limitations:a) Minimum two energy sources must be present in the network.
b) Network should not contain any non linear and unilateral elements.
fig.2
(c) Define average value and RMS value. Calculate the average value, RMS value, form factor and
peak factor of the output of (i) Half-wave rectifier (ii) Full-wave rectifier.
Average Value: The average value of an alternating current is that value of steady current which
transfers the same charge as the alternating current flowing for the same time.
The arithmetic average of all the values of an alternating quantity over one cycle is called its
average value.
Average value = Area under one cycle/Base
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For Symmetrical waveforms, the average value calculated over one cycle becomes equal to zero
because the positive area cancels the negative area. Hence for symmetrical waveforms, the average
value is calculated for half cycle.
Average value = Area under one half cycle/ Base
RMS Value: The effective or RMS value of an alternating quantity is that steady current (dc) which
when flowing through a given resistance for a given time produces the same amount of heat
produced by the alternating current flowing through the same resistance for the same time . It is
also defined as the root of the mean of the squares of the instantaneous values of the alternating
quantity.
When the input voltage is going through its positive half cycle, output voltage is almost the same
as the input voltage and during the negative half cycle no voltage is available across the load. This
explains the unidirectional pulsating dc waveform obtained as output. The process of removing one half
the input signal to establish a dc level is aptly called half wave rectification.
Vav the average or the dc content of the voltage across the load is given by
RMS voltage at the load resistance can be calculated as
Form Factor
The ratio of RMS value to the average value of an alternating quantity is known as Form Factor
Peak Factor or Crest Factor
The ratio of maximum value to the RMS value of an alternating quantity is known as the peak
factor
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A Full Wave Rectifier is a circuit, which converts an ac voltage into a pulsating dc voltage using both half
cycles of the applied ac voltage. It uses two diodes of which one conducts during one half cycle while the
other conducts during the other half cycle of the applied ac voltage.
During the positive half cycle of the input voltage, diode D1 becomes forward biased and D2 becomes
reverse biased. Hence D1 conducts and D2 remains OFF. The load current flows through D1 and the
voltage drop across RL will be equal to the input voltage.
The average voltage or the dc voltage available across the load resistance is
RMS value of the voltage at the load resistance is
Form Factor
Form factor is defined as the ratio of the rms value of the output voltage to the average value of the output
voltage.
Peak Factor
Peak factor is defined as the ratio of the peak value of the output voltage to the rms value of the output voltage.
(d) For delta -connected system in a three phase circuit prove that IL =
3 Iph and VL = Vph.
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A three-phase 400V supply is connected to a 3-phase delta connected load of phase impedance
(15+j20) Ω. Find (i) the phasor current in each line. (ii) what is the power consumed per phase?
(iii) What is the phasor sum of three line currents?
ANS:
A balanced delta or mesh connected 3 phase system is shown in figure.
From the circuit diagram it is obvious that
VRY= VR; VYB=VY; VBR=VB
Hence it can be concluded that line voltages are equal to phase voltages
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i.e. VL=VP
By writing KVL for the mesh RYB, we have
VR+VY+VB=0
Also line current IR could be written as
IR= IYR – IRB
IR= IYR + (-IRB)
From the phasor diagram it could be written as
Hence it is clear that line current is √3 times the phase current
i.e. IL=√3IP
Also if the power factor of the load is given by cosΦ
Then Power output per phase = VPIPCosΦ
Total three phase power output; P3Φ = 3 VPIPCosΦ = √3 VLILCosΦ
Total three phase reactive power output; Q 3Φ = 3 VPIPSinΦ = √3 VLILSinΦ
Total three phase apparent power output; S 3Φ = 3 VPIP = √3 VLIL
(e) Draw exact equivalent circuit and the corresponding phasor diagram for capacitive load of a
single phase transformer on load and explain them.
Ans: The term equivalent circuit of a machine means the combination of fixed and variable resistances
and reactance, which exactly simulates performance and working of the machine. For a transformer, no
load primary current has two components, I m = Io sinΦo = Magnetizing component Ic = Io cosΦo = Active
component Im produces the flux and is assumed to flow through reactance X o called no load reactance
while Ic is active component representing core losses hence is assumed to flow through the reactance R o.
Hence equivalent circuit on no load can be shown as in the Fig. 1. This circuit consisting of R o and Xo in
parallel is called exciting circuit. From the equivalent circuit we can write, R o = V1/Ic and Xo= V1/Im
When the load is connected to the transformer then secondary current I2 flows. This causes voltage drop
across R2 and X2. Due to I2, primary draws an additional current I 2' = I2/ K. Now I1 is the phasor addition
of Io and I2'. This I1 causes the voltage drop across primary resistance R 1 and reactance X1. Hence the
equivalent circuit can be shown as in the Fig.
But in the equivalent circuit, windings are not shown and it is further simplified by transferring all the
values to the primary or secondary. This makes the transformer calculation much easy. So transferring
secondary parameters to primary we get, R 2'= R2/K2 , X2' = X2/K2 , Z2' = Z2/K2 While E2' = E2/K, I2' = K/I2
Where K = N2 /N1 While transferring the values remember the rule that Low voltage winding High current
Low impedance High voltage winding Low current High impedance Thus the exact equivalent circuit
referred to primary can be shown as in the Fig.
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Similarly all the primary value can be referred to secondary and we can obtain the equivalent circuit
referred to secondary. R1' = K2 R1 , X1' = K2 X1, Z1' = K2 Z1, E1'= K E1, Io' = I1 /K' Io' = Io /K Similarly the
exciting circuit parameters also gets transferred to secondary as R o' and Xo '. The circuit is shown in the
Fig.
1. Consider flux Φ as reference
2. E1 lags Φ by 90o. Reverse E1 to get -E1.
3. E1 and E2 are in phase.
4. Assume V2 in a particular direction
5. I2 is in phase with V2.
6. Add I2 R2 and I2 X2 to get E2.
7. Reverse I2 to get I2'.
8. Add Io and I2' to get I1.
9. Add I1 R1 and to -E1 to get V1. Angle between V1 and I1 is Φ1 and cosΦ1 is primary power factor.
Remember that I1X1 leads I1 direction by 90o and I2 X2 leads I2 by 90o as current through inductance lags
voltage across inductance by 90o. The phasor diagram is shown in the Fig. below.
Leading Power Factor Load, cos Φ2 -- As load power factor is leading, the current I 2 leads V2 by angle
Φ2. So change is to draw I2 leading V2 by angle Φ2. All other steps remain same as before. The complete
phasor diagram is shown in the Fig. below.
(f) Explain the starting methods of single phase induction motor.
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Ans: Single phase induction motor: A single phase induction motor is very similar to a 3-phase squirrel
cage induction motor. It has (i) a squirrel cage rotor identical to a 3- phase motor and (ii) a single phase
winding on the stator.
Unlike a three phase induction motor, a single phase induction motor is not a self starting but requires
some starting means. The single phase stator winding produces a pulsating magnetic field (sinusoidal in
nature). The field polarity reverses after each half cycle but the field does not rotate. Consequently, the
alternating flux cannot produce rotation in a stationary rotor. However, if rotor of 1-phase induction
motor is rotated in one direction by some mechanical means, it will continue to run in the direction of
rotation.
Making self starting: To makes a single phase induction motor self starting, we should somehow
produce a revolving stator magnetic field. This may be achieved by converting a single phase supply into
two phase supply through the use of an additional winding. When the motor attains sufficient speed, the
starting means (i.e. additional winding) may be removed depending upon the type of the motor. As a
matter of fact, single phase induction motor are classified and named according to the method employed
to make them self starting.
Split phase motor: The stator of a split phase induction motor is provided with an auxiliary or starting
winding in addition to the main or running winding. The starting winding is located 90 degree electrical
from the main winding. The two windings are so designed that the starting winding has a high resistance
and relatively small reactance while the main winding has relatively low resistance and large reactance,
the current flowing in the two windings have reasonable phase difference.
Operation: When the two stator windings are energized from a single phase supply, the main winding
carries current Im while starting winding carries Ia. Since main winding is made highly inductive while the
starting winding highly resistive, the current I m and Ia have a reasonable phase difference between
them.
Consequently, a weak revolving field approximation to that of a 2-phase machine is produced which
starts the motor.
When the motor reaches about 75% of synchronous speed, the centrifugal switch opens the circuit of the
starting winding. The motor operates as a single phase induction motor and continues to accelerate till it
reaches the normal speed.
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SECTION – C
NOTE: ATTEMPT ALL QUESTIONS.
(5*10=50)
3. Attempt any one part:
(a) Using Maximum Power Transfer Theorem, find the value of load resistance R L for Maximum Power
flow through it in fig.3. Also find the maximum power.
fig.3
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(b) Determine i1 and i2 in fig.4 using mesh current analysis. Justify your result using source
transformation.
fig.4
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4. Attempt any two parts:
(a) The alternating voltage is given by v = 141.4 sin 314t. Find (i) Frequency (ii) RMS value of the
voltage (iii) The instantaneous value when ‘t’ is 3 msec. (iv) The time taken for the voltage to reach
100 V for the first time after passing through zero value.
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(b) Define bandwidth and derive the formula for series RLC circuit at resonance.
Ans:
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(c) Two impedances Z1= (10+j5) Ω and Z2= (8+j6) Ω are connected in parallel across a voltage of 200V.
Calculate the circuit current, power factor and reactive power.
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5. Attempt any two parts:
(a) A balanced Y-connected load of impedance (16+j12) Ω per phase is connected to a 3 phase 400V
supply. Find phase current, line current, power factor, power, Reactive VA and total VA.
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(b) Two wattmeters connected to measure power in a balanced 3-phase circuit, reads 10kW at unity pf.
Find the readings of the each wattmeter, when the power factor is 0.866 (lagging), the total power
remaining unchanged.
Ans;
W1+ W2 = 10kW
tan Φ = √3 (W1 - W2)/ (W1 + W2)
cosΦ = 0.866, Φ= 30˚, tan 30˚ = 1/√3
W1 - W2 = (W1 + W2)* tan Φ/√3
= 10*(1/√3)/ √3
=3.34
W1=6.67kW, W2 =3.33kW
(c) Explain in brief the working principle of permanent magnet moving coil instruments. Why is the
scale linear?
Ans: Permanent magnet moving coil:
The operating principle of PMMC instruments is based on the principle that when a current carrying
conductor is placed in a magnetic field it experiences a mechanical force, which provides the necessary
deflection torque. This deflecting torque is responsible to the deflection of the pointer. Pointer is attached
to the rectangular shape of coil which moves over is used in PMMC instrument. Mechanical force,
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F = NiBl
Deflecting torque, Td = NBil x b =NBIA (where A=l x b)
Tc  
In equilibrium,
Td  Tc
 I
Advantage:
(i) PMMC have uniform Scale.
(ii) Sensitivity is very high
(iii) No Hysteresis loss.
Disadvantages:
(i)
(ii)
Cost of PMMC instruments are high.
It can not be used in AC measurement.
6. Attempt any two parts:
(a) Mention any 3 similarities and 2 dissimilarities between Electrical circuit & Magnetic Circuit.
Ans: Similarities between magnetic circuit and electric circuit

Magnetic circuit follows equation Ni = (Ф) ( l / μA) or

m.m.f (magneto motive force) = (Flux) (reluctance).

Electric circuit follows ohm’s law that is E = I.R

e.m.f(electro motive force) = (current) (Resistance)

From above point m.m.f in magnetic circuit is like an e.m.f in electrical circuit.

Flux in magnetic circuit is similar as current in electrical circuit.

Reluctance in magnetic circuit, S = ( l / μA) is similar to resistance R = (ρl/A) in electric circuit.

Permeance (= 1/reluctance) in magnetic circuit is equivalent to conductance (=1/resistance) in
electric circuit.
Differences between magnetic circuit and electric circuit
In magnetic circuit flux establishes but not flow like as current in magnetic circuit.

In magnetic circuit energy needed only to establish the flux but no consistent energy need to
maintain it whereas in electric circuit continuous energy needed to flow of current.

Resistance of an electric circuit is constant (for same temperature) and is independent of current
but reluctance of magnetic circuit is not constant because it depends on μ (=B/H) which is not
constant and depends on B/H.
(b) State the various safety precautions to be observed while using electric supply.
Ans:
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(c) A 200kVA transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at
three quarter of full load, calculate the efficiency at half full load. Assume negligible magnetising
current and pf of 0.8 lagging at all loads.
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7. Attempt any two parts:
(a) Derive principle of working of a 3-phase synchronous motor. Why is starting torque not produced in
this motor?
Ans: Synchronous motor is a doubly excited machine i.e two electrical inputs are provided to it. It is
inherently not self starting motor. The stator winding which consists of a three phase winding is provided
with three phase supply and rotor is provided with DC supply. The three phase stator winding carrying
three phase currents, produces three phase rotating magnetic flux. The three phase rotating magnetic
flux is rotating with a high speed. So at a particular instant rotor and stator poles might be of same
polarity (N-N or S-S) causing repulsive force on rotor and the very next second it will be N-S causing
attractive force. But due to inertia of the rotor, it is unable to rotate in any direction due to attractive or
repulsive force and remain in standstill condition. Hence it is not self starting.
To overcome this inertia, the rotor is first rotated with some auxiliary means like a DC motor. When
the speed of the rotor reaches near synchronous speed magnetic locking happens and rotor starts
rotating with synchronous speed.
(b) 4-pole DC generator with wave connected armature has 41 slots, 12conductors/slot. Armature
resistance Ra = 0.5 Ω, Shunt resistance Rsh = 200 Ω, Flux per pole is 125 mWb, Speed N = 1000 rpm.
Calculate voltage across 5 Ω load resistance connected across the armature terminals.
(c) Draw following characteristics for DC shunt motor.
(i) Torque Vs Armature current (ii) Speed Vs Armature current (iii) Speed Vs Torque
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