Question 1 |

Figure Shows the waveform of the current passing through an inductor of
resistance 1\Omega and inductance 2 H. The energy absorbed by the inductor in the first four seconds is

144J | |

98J | |

132J | |

168J |

Question 1 Explanation:

For 0 \lt t \lt 2s current varies linearly with time and given as i(t)=3t and for 2s \lt t \lt 4s current is constant, i(t)=6A.

The energy absorbed by the inductor (Resistance neglected) in the first 2 sec,

E_L=\int_{0}^{T}Li\frac{di}{dt}=E_{L_1}+E_{L_2}

E_{L_1}=\int_{0}^{2}Li \left (\frac{di}{dt} \right )dt

\;\;=\int_{0}^{2} 2 \times 3t \times 3\; dt

\;\;=18\int_{0}^{2}t\; dt

\;\;=18 \times \left.\begin{matrix} \frac{t^2}{2} \end{matrix}\right|^2_0

\;\;= 18 \times \left [ \frac{4}{2}-0 \right ]=36J

The energy absorbed by the inductor in (2\rightarrow 4) second

E_{L_2}=\int_{2}^{4}Li\left ( \frac{di}{dt}dt \right )

\;\;=\frac{2}{4}2 \times 6 \times 0 \;dt=0 J

A pure indictor does not dissipate enegy but only stores it. Due to resistance, some energy is dissipated in the resistor. Therefore, total energy stored in the inductor and the energy dissipated in the resistor.

The energy dissipated by the resistance in 4 sec.

E_R=\int_{0}^{T}i^2 R\;dt

\;\;=\int_{0}^{2}(3t)^2 \times 1 \;dt+\int_{2}^{4}6^2 \times 1 \; dt

\;\;=9\int_{0}^{2}t^2\;dt +36\int_{2}^{4}1 \; dt

\;\;=9 \times \left.\begin{matrix} \frac{t^3}{3} \end{matrix}\right|_0^2+36t|_2^4

\;\;=9 \times \left ( \frac{8}{3} \right )+36 \times 2

\;\;=24+72=96J

The total energy absorbed by the inductor in 4 sec

=96 J +36 J =132 J

The energy absorbed by the inductor (Resistance neglected) in the first 2 sec,

E_L=\int_{0}^{T}Li\frac{di}{dt}=E_{L_1}+E_{L_2}

E_{L_1}=\int_{0}^{2}Li \left (\frac{di}{dt} \right )dt

\;\;=\int_{0}^{2} 2 \times 3t \times 3\; dt

\;\;=18\int_{0}^{2}t\; dt

\;\;=18 \times \left.\begin{matrix} \frac{t^2}{2} \end{matrix}\right|^2_0

\;\;= 18 \times \left [ \frac{4}{2}-0 \right ]=36J

The energy absorbed by the inductor in (2\rightarrow 4) second

E_{L_2}=\int_{2}^{4}Li\left ( \frac{di}{dt}dt \right )

\;\;=\frac{2}{4}2 \times 6 \times 0 \;dt=0 J

A pure indictor does not dissipate enegy but only stores it. Due to resistance, some energy is dissipated in the resistor. Therefore, total energy stored in the inductor and the energy dissipated in the resistor.

The energy dissipated by the resistance in 4 sec.

E_R=\int_{0}^{T}i^2 R\;dt

\;\;=\int_{0}^{2}(3t)^2 \times 1 \;dt+\int_{2}^{4}6^2 \times 1 \; dt

\;\;=9\int_{0}^{2}t^2\;dt +36\int_{2}^{4}1 \; dt

\;\;=9 \times \left.\begin{matrix} \frac{t^3}{3} \end{matrix}\right|_0^2+36t|_2^4

\;\;=9 \times \left ( \frac{8}{3} \right )+36 \times 2

\;\;=24+72=96J

The total energy absorbed by the inductor in 4 sec

=96 J +36 J =132 J

Question 2 |

A segment of a circuit is shown in figure. V_{R} = 5V, V_{C} = 4 \sin 2t. The voltage V_{L} is
given by

3-8 \cos 2t | |

32 \sin 2t | |

16 \sin 2t | |

16 \cos 2t |

Question 2 Explanation:

By KCL,

I_P+I_Q+I_C+I_L=0

2+2+I_C+I_L=0

But, I_C=C \times \frac{dv}{dt}

\;\; =1 \times \frac{d}{dt}(4 \sin 2t)

\;\;=(8 \cos 2t)

\therefore \;\; I_L=-(2+1+8 \cos 2t)

\;\; \;\;=-3-8 \cos 2t

\therefore \;\; V_L=L\left ( \frac{di}{dt} \right )

\;\;\;=2 \times 2 \times 8 \sin 2t

\;\;\;=32 \sin 2t

NOTE: KCL is based on the law of conservation of charges.

I_P+I_Q+I_C+I_L=0

2+2+I_C+I_L=0

But, I_C=C \times \frac{dv}{dt}

\;\; =1 \times \frac{d}{dt}(4 \sin 2t)

\;\;=(8 \cos 2t)

\therefore \;\; I_L=-(2+1+8 \cos 2t)

\;\; \;\;=-3-8 \cos 2t

\therefore \;\; V_L=L\left ( \frac{di}{dt} \right )

\;\;\;=2 \times 2 \times 8 \sin 2t

\;\;\;=32 \sin 2t

NOTE: KCL is based on the law of conservation of charges.

Question 3 |

In the figure,

Z_{1}=10\angle -60 ^{\circ} , Z_{2}=10\angle 60^{\circ} , Z_{3}=50\angle 53.13^{\circ}.

The venin impedance seen form X-Y is

Z_{1}=10\angle -60 ^{\circ} , Z_{2}=10\angle 60^{\circ} , Z_{3}=50\angle 53.13^{\circ}.

The venin impedance seen form X-Y is

56.66\angle 45^{\circ} | |

60\angle 30^{\circ} | |

70\angle 30^{\circ} | |

34.4\angle 65^{\circ} |

Question 3 Explanation:

By Thevenin's theorem

Z_{Th}=Z_{X-Y}=Z_1||Z_2+Z_3

\;\;=\frac{Z_1 \times Z_2}{Z_1+Z_2}+Z_3

\;\;=\frac{10\angle -60 \times 10\angle 60}{10\angle -60 + 10\angle 60}+50\angle 53.13

\;\;=56.66\angle 45^{\circ}\Omega

Z_{Th}=Z_{X-Y}=Z_1||Z_2+Z_3

\;\;=\frac{Z_1 \times Z_2}{Z_1+Z_2}+Z_3

\;\;=\frac{10\angle -60 \times 10\angle 60}{10\angle -60 + 10\angle 60}+50\angle 53.13

\;\;=56.66\angle 45^{\circ}\Omega

Question 4 |

Two conductors are carrying forward and return current of +I and -I as
shown in figure. The magnetic field intensity H at point P is

\frac{I}{\pi d}\vec{Y} | |

\frac{I}{\pi d}\vec{X} | |

\frac{I}{2 \pi d}\vec{Y} | |

\frac{I}{2 \pi d}\vec{X} |

Question 4 Explanation:

\bar{H}=\bar{H_1}+\bar{H_2}=\frac{I}{2 \pi d}\vec{y}+\frac{I}{2 \pi d}\vec{y}=\frac{I}{ \pi d}\vec{y}

Question 5 |

Two infinite strips of width w m in x -direction as shown in figure, are carrying
forward and return currents of +I and -I in the z - direction. The strips
are separated by distance of x m. The inductance per unit length of the
configuration is measured to be L H/m. If the distance of separation between
the strips in snow reduced to x/2 m, the inductance per unit length of the
configuration is

2L H/m | |

L/4 H/m | |

L/2 H/m | |

4L H/m |

Question 5 Explanation:

Inductance is proportional to separation between the strips so when separation is reduced to x/2 then inductance will become L/2 H/m.

Question 6 |

A sinple phase transformer has a maximum efficiency of 90% at full load and
unity power factor. Efficiency at half load at the same power factor is

86.70% | |

88.26% | |

88.90% | |

87.80% |

Question 6 Explanation:

\begin{aligned}
\text{Efficiency} &=\frac{\text{Output}}{\text{Input}} \\
&= \frac{\text{Output}}{\text{Output}+\text{Losses}}\\
\text{Output}&= x S \cos \phi \\
\text{Where,}\;\;S &=\text{Rating of the machine} \\
x&= \% \text{of the full load}\\
\cos \phi &= \text{Power factor} \\
P_i &= \text{Iron or core loss}\\
P_{cu} &=\text{Full load copper loss} \\
&\text{For maximum frequency}\\
P_i &=x^2P_{cu} \\
\text{Efficiency} =\eta &= \frac{xS \cos \phi }{xS \cos \phi +P_i+x^2 P_{cu}} \\
\text{Max efficiency}=\eta _{max}&=\frac{xS \cos \phi }{xS \cos \phi +2P_i}\\
\Rightarrow \;\;0.9&=\frac{1 \times S \times 1}{1 \times S \times 1+2P_i}\\
P_i&=0.055S
\end{aligned}

Full load copper loss = 1^2 P_{cu}=P_i

P_{cu}=0.055S

Efficiency of halh load, upf

x=\frac{1}{2}

\eta =\frac{\frac{1}{2} \times S \times 1 }{\frac{1}{2} \times S \times 1 +0.055S+\left ( \frac{1}{2} \right )^2 \times 0.055S} \times 100

\;\;\;\approx 87.8\%

Full load copper loss = 1^2 P_{cu}=P_i

P_{cu}=0.055S

Efficiency of halh load, upf

x=\frac{1}{2}

\eta =\frac{\frac{1}{2} \times S \times 1 }{\frac{1}{2} \times S \times 1 +0.055S+\left ( \frac{1}{2} \right )^2 \times 0.055S} \times 100

\;\;\;\approx 87.8\%

Question 7 |

Group-I lists different applications and Group-II lists the motors for these
applications. Match the application with the most suitable motor and choose
the right combination among the choices given thereafter

P-3 Q-6 R-4 S-5 | |

P-1 Q-3 R-2 S-4 | |

P-3 Q-1 R-2 S-4 | |

P-3 Q-2 R-1 S-4 |

Question 7 Explanation:

Universal motors are used where light weight is important., as in vacuum cleaners and portable tools e.g. food mixer which usually operate at high speed (1500-15000 rpm). Domestic water pump are usually of low rating. So single phase induction motor can be used for such application. Three phase induction motor is suitable for escalator as it has high starting torque.

Question 8 |

A stand alone engine driven synchronous generator is feeding a partly inductive
load. A capacitor is now connected across the load to completely nullify the
inductive current. For this operating condition.

the field current and fuel input have to be reduced | |

the field current and fuel input have to be increased | |

the field current has to be increased and fuel input left unaltered | |

the field current has to be reduced and fuel input left unaltered |

Question 8 Explanation:

Assuming resistance of the armature to be zero. In first case, the generator is feeding a partly inductive load. It means that generator is supplying lagging power. The generator supplies a lagging power factor current when it is overexcited which is represented by E_{f_1}.

In second case, a capacitor is connected across the load to completely nullifythe inductive current. It means the generator supplies no reactive power and unity power factor current is drawn from the generator. The excitation (E_{f_2}) corresponding to unity power factor is known as normal excitation.

From the phasor diagram E_{f_1} \gt E_{f_2} and as the field current approximately directly proportional to excitation, the field current has to be reduced. From phasor diagram E_{f_1} \sin \delta _1=E_{f_2} \sin \delta _2=\text{constant}

P_e(\text{ power delivered})=\frac{E_fV_t}{X_s}\sin \delta

Fuel input remains unaltered.

Question 9 |

Curves X and Y in figure denote open circuit and full-load zero power
factor(zpf) characteristics of a synchronous generator. Q is a point on the zpf
characteristics at 1.0 p.u. voltage. The vertical distance PQ in figure gives the
voltage drop across

Synchronous reactance | |

Magnetizing reactance | |

Potier reactance | |

Leakage reactance |

Question 9 Explanation:

The vertical distance PQ between O.C. (curve X) and ZPF characteristic (curve Y) in above characteristic is leakage reactance.

Question 10 |

No-load test on a 3-phase induction motor was conducted at different supply
voltage and a plot of input power versus voltage was drawn. This curve was
extrapolated to intersect the y-axis. The intersection point yields

Core loss | |

Stator copper loss | |

Stray load loss | |

Friction and windage loss |

Question 10 Explanation:

Power input at no-load (P_0) provided losses only as the shaft output is zero. These losses comprise P_i(Iron/core loss) and P_{wf}(windage and friction loss).

As the voltage is reduced below the rated value, the core-loss decreases as the square of voltage. Since the slip does not increase significantly, the windage and friction loss remains almost constant. The plot of P_0 versus V is extrapolated at V=0 which gives, P_{wf} as P_{i}=0 art zero voltage.

As the voltage is reduced below the rated value, the core-loss decreases as the square of voltage. Since the slip does not increase significantly, the windage and friction loss remains almost constant. The plot of P_0 versus V is extrapolated at V=0 which gives, P_{wf} as P_{i}=0 art zero voltage.

There are 10 questions to complete.