85
At present there are 153 (Elect85, Mech68) vacancies; however, in future more vacancies are likely to occur till one year from announcement of result.
SC  ST  SEBC  EWS  UR  PWD  
M  F  M  F  M  F  M  F  M  F  
3  0  18  7  14  4  12  2  65  28  15 
Year  Remuneration  Remarks 

1st Year  17,500/ p.m. 
No other allowance or benefits would be admissible. Reimbursement of TA/DA as per GSO333 dated 03.02.2003. Pay scale of Rs. 2600056600 subject to satisfactory completion of Three years as Vidyut Sahayak. 
2nd Year  19,000/ p.m.  
3rd Year  20,500/ p.m. 
Post  Min. Qualification 

VSJE (Electrical) 
Full time/Regular Diploma in Electrical Engineering from recognized University with minimum 55% in Last Year / 5th & 6th semester. 
Category of candidates  Upper age limit 

Unreserved  36 years 
Reserved & EWS  41 years 
Category of candidates  Relaxation 

Female Candidate  05 Years 
Person with Disability candidate  10 Years 
Ex. Armed Force Personnel  10 Years 
Dependent of Retired Employee of GUVNL & Subsidiaries  Up to age of 40 years (Considered only on submission of undertaking) 
Post  Disability Level 

VS PA GRI (Electrical)  Person with Disability (PWD) candidate  The PWD (Person with Disability) candidates with disability of Acid Attack Victim (AAV) and Specific Learning Disability (SLD) can apply and shall have to submit Certificate of Civil Surgeon /Government Designated Authority, indicating existing Percentage of disability. Their applications will be considered as per rules of the Company. 
Sr. No.  Events  Dates 

1  Online submission of application commences  03012023 
2  Last date for Online submission of application  23012023 
3  Important Dates  The last date of Online application is 23/01/2023, 06.00 p.m. 
Categories  Fees Payable 

UR/SEBC/EWS  Rs.500.00 (inclusive of GST) 
ST/SC/PWD  Rs.250.00 (inclusive of GST) 
The tentative syllabus for the exam will be including but not limited to following topics and emphasis could differ.
Section  Topics  Weightage  Content 

I 
Gujarati Language &Grammar 
10%  
II 
General Knowledge 
10%  
III 
English Language 
10%  
IV 
Computer Knowledge 
10%  
V  Electrical Engineering  60% 

“The question paper will be in English & Gujarati Language only”
Kindly visit https://www.gsecl.in/ for an official documents published by GSECL and read instructions carefully before apply. For any query you may contact on our Help Desk No. +917353259222 which will be available between 10 am to 6 pm on working days. You may also send an Email for your query on recruit.gsecl@gebmail.com
Subject  Number of questions 

POWER SYSTEM  6 
ELECTRICAL CIRCUIT ANALYSIS  13 
GUJARATI  10 
GENERAL KNOWLEDGE  10 
ENGLISH  10 
ENGINEERING MATHEMATICS  8 
ELECTRICAL MACHINES  9 
BASICS OF COMPUTER  10 
POWER ELECTRONICS & DRIVES  4 
BASIC ELECTRONICS  2 
DIGITAL ELECTRONICS  3 
ELECTROMAGNETICS  3 
MEASUREMENT & INSTRUMENTATION  3 
CONTROL SYSTEM ENGINEERING  7 
SIGNALS & SYSTEMS  2 
Total  100 
Correct Ans: D
Correct Ans: C
Correct Ans: D
Correct Ans: B
Correct Ans: C
Correct Ans: D
Correct Ans: B
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: D
Correct Ans: C
Correct Ans: B
Correct Ans: B
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: C
Correct Ans: C
Correct Ans: D
Correct Ans: C
Correct Ans: D
Correct Ans: C
Correct Ans: C
Correct Ans: C
Correct Ans: C
Correct Ans: C
Correct Ans: D
Correct Ans: D
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: D
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: B
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: C
Correct Ans: C
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: B
Correct Ans: C
Correct Ans: A
Correct Ans: D
Correct Ans: B
Correct Ans: C
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: D
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: D
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: D
Correct Ans: D
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: D
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: D
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: D
Correct Ans: B
Correct Ans: A
Correct Ans: C
Subject  Number of questions 

GENERAL KNOWLEDGE  10 
ENGLISH  10 
ENGINEERING MATHEMATICS  7 
ELECTRICAL CIRCUIT ANALYSIS  12 
SIGNALS & SYSTEMS  4 
CONTROL SYSTEM ENGINEERING  7 
ELECTROMAGNETICS  4 
ELECTRICAL MACHINES  9 
POWER ELECTRONICS & DRIVES  5 
BASIC ELECTRONICS  1 
DIGITAL ELECTRONICS  3 
MEASUREMENT & INSTRUMENTATION  3 
POWER SYSTEM  5 
BASICS OF COMPUTER  10 
GUJARATI  10 
Total  100 
Correct Ans: D
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: D
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: C
Correct Ans: D
Correct Ans: D
Correct Ans: C
Correct Ans: B
Correct Ans: A
Correct Ans: B
Correct Ans: B
Correct Ans: C
Correct Ans: A
Correct Ans: C
Correct Ans: C
Correct Ans: A
Correct Ans: D
Correct Ans: B
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: D
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: D
Correct Ans: D
Correct Ans: B
Correct Ans: A
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: B
Correct Ans: C
Correct Ans: A
Correct Ans: C
Correct Ans: D
Correct Ans: D
Correct Ans: A
Correct Ans: A
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: B
Correct Ans: C
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: C
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: A
Subject  Number of questions 

POWER ELECTRONICS & DRIVES  8 
BASIC ELECTRONICS  3 
DIGITAL ELECTRONICS  2 
POWER SYSTEM  4 
ENGINEERING MATHEMATICS  6 
CONTROL SYSTEM ENGINEERING  4 
ELECTRICAL MACHINES  10 
ELECTROMAGNETICS  6 
MEASUREMENT & INSTRUMENTATION  4 
ELECTRICAL CIRCUIT ANALYSIS  9 
ENGLISH  10 
GENERAL KNOWLEDGE  10 
GUJARATI  10 
BASICS OF COMPUTER  10 
SIGNALS & SYSTEMS  4 
Total  100 
Correct Ans: C
Solution:
\( \int_0^1{\int_0^{x^2}{(x^2+y^2)}} \: dy \: dx \)
= \( \int_0^1 [x^2 y + \frac{y^3}{3}]_0^{{x}^{2}} dx \)
= \( \int_0^1 [x^4 + \frac{x^6}{3}] dx\)
= \([\frac{x^5}{5} + \frac{x^7}{21}]_0^1\)
= \([\frac{1}{5} + \frac{1}{21}]\)
=\(\frac{26}{105}\)
Correct Ans: C
Solution:
\(\Rightarrow\) In spherical coordinate system 'r' is the distance of point P from origin and so \( 0 \leq r < \infty \)
Correct Ans: A
Solution:
\(\Rightarrow\) The pushpull amplifier increases the output voltage and current delivering capability of the operational amplifier.
Correct Ans: A
Solution:
\(\mathbf{Given \:data:}\)
\(W_1=W_2=600\:W\)
\(\mathbf{ Measurement\:of\:Power\:and\:Power\:factor\:by\:Two\:Wattmeter\:Method\:in\:a\:balanced\:three\:phase\:load}\)
\(Wattmeter\: reading\: W_1=V_LI_L\cos(30^\circ\phi)....(1)\)
\(Wattmeter\: reading \:W_2=V_LI_L\cos(30^\circ+\phi)....(2)\)
\(Using\: equation\:(1)\:and\:(2)\)
\(Total\:power\:absorbed\:in\:a\:three\:phase\:W=W_1+W_2=\sqrt3V_LI_L\cos\phi....(3)\)
\(W_1W_2=V_LI_L\sin\phi....(4)\)
\(Using\: equation\:(3)\:and\:(4)\)
\(\frac{W_1W_2}{W_1+W_2}=\frac{V_LI_L\sin\phi}{\sqrt3V_LI_L\cos\phi}\)
\(\tan\phi=\sqrt3[\frac{W_1W_2}{W_1+W_2}]....(5)\)
\(\phi=\tan^{1}[\frac{\sqrt3(W_1W_2)}{W_1+W_2}]....(6)\)
Values \(W_1=W_2=600\:W\) put in equation (6)
\(\phi=\tan^{1}(0)\implies \phi=0\)
\(Power\:factor=\cos\phi=\cos0=1\)
Therefore,In measurement of 3phase power by two wattmeter method, both wattmeter readings are equal. The power factor of the load is unity.
Correct Ans: C
Solution:
\(\Rightarrow \mathbf{Energy\: Signal}\) is a signal whose energy is finite and power is zero.
\(\Rightarrow \mathbf{Power\: Signal}\) is a signal whose power is finite and energy is infinite.
Correct Ans: D
Solution:
\(\Rightarrow\) Starting current of DC motor is given by
\( I_{ast} = \frac{V}{R_a} \)
\( I_{ast} = \frac{230}{0.5} = 460 \: A \)
Correct Ans: B
Solution:
\(\Rightarrow\) The half controlled converter can provide positive output voltage and current only.
\(\Rightarrow\) Hence it can provide single quadrant operation only.
Correct Ans: D
Solution:
\(\Rightarrow\) Snubber circuit is used to limit high dv/dt in SCR
\(\Rightarrow\) High dv/dt can cause malfunctioning of SCR.
Correct Ans: D
Solution:
\(\Rightarrow\) Emitter follower has unity gain.
\(\Rightarrow\) So emitter follower only provides impedance transformation without affecting the magnitude of signal.
Correct Ans: A
Solution:
\(\mathbf{Advantages \:of \:Three\: Phase\: System:}\)
\(\mathbf{1.}\) Three phase system requires only 75% weight of conducting materials of that required by single phase system to transmit the same amount of power at a given voltage and over a given distance.
\(\mathbf{2.}\) Three phase system is more capable and reliable than single phase system.
\(\mathbf{3.}\) A threephase machine gives more output compared to a singlephase machine of the same size.
\(\mathbf{4.}\) A threephase system has better voltage regulation.
Correct Ans: A
Solution:
\(\Rightarrow\) Maximum Efficiency occurs when Copper Loss = Iron Loss
Correct Ans: D
Solution:
Given,
\(A = \begin{bmatrix}0 & 5 \\5 & 0 \end{bmatrix}\)
\(\Rightarrow\) Condition for Skewsymmetric matrix is \(A^T = A\)
For matrix \(A =\begin{bmatrix}0 & 5 \\5 & 0 \end{bmatrix}\)
\(\implies A^T = \begin{bmatrix}0 & 5 \\5 & 0 \end{bmatrix} = (1) \begin{bmatrix}0 & 5 \\5 & 0 \end{bmatrix} = A \)
\(\therefore Skewsymmetric \: matrix
=\begin{bmatrix}0 & 5 \\5 & 0 \end{bmatrix}\)
Correct Ans: C
Solution:
Reluctance \(S=\frac{l}{\mu A}\)
Where,
\(S\)  Reluctance in \(AT/Wb\)
\(l\)  Length of magnetic circuit in \(m\)
\(A\)  Area of crosssection of the magnetic path in \(m^2\)
\(\mu\)  Absolute permeability of the materials in \(H/m\)
\(S=\frac{l}{\mu A} \implies \mu=\frac{l}{S\times A}=\frac{m}{\frac{AT}{Wb}\times m^2}=\frac{Wb}{AT \times m}\)
Correct Ans: C
Solution:
\(\mathbf{Given \:data:}\)
An electric heater rating \(P=1\:kW=1000\:W,\:V=200\:V\)
\(P=\frac{V^2}{R}\implies R=\frac{V^2}{P}=\frac{(200)^2}{1000}=40\:\Omega\)
Current drawn by the heating element \(I=\frac{V}{R}=\frac{200}{40}=5\:A\)
Correct Ans: A
Solution:
\(\Rightarrow\) In given figure, the field winding is excited saperately by a DC source and so it is saperately excited DC machine.
\(\Rightarrow\) Here the current is flowing away from the armature and so it is generator.
Correct Ans: D
Solution:
Correct Ans: A
Solution:
\(\Rightarrow\) A ripple counter with \(n\) flipflops will able to count \(2^n1\) counts.
\(\Rightarrow\) Here, the counter needs to count 3000. \(\quad \therefore 2^n1 \geq 3000\).
\(\quad \therefore 2^n \geq 3001 \implies n = \frac{\log 3001}{\log 2} \approx 12.\)
Correct Ans: C
Solution:
Types of Shunt type of faults:
\(\rightarrow\) Line to Ground fault (LG)
\(\rightarrow\) Line to Line fault (LL)
\(\rightarrow\) Double Line to Ground fault (LLG)
\(\rightarrow\) Three phase fault
The shunt type faults characterized by:
\(\rightarrow\) Fall in voltage level
\(\rightarrow\) Fall in frequency
\(\rightarrow\) Increase in current value
Correct Ans: A
Solution:
Self induced e.m.f. in coil 1 \(E_{S1}=L_1\frac{dI_1}{dt}\)
Self induced e.m.f. in coil 2 \(E_{S2}=L_2\frac{dI_2}{dt}\)
Mutually induced e.m.f. in coil 1 \(E_{m1}=M\frac{dI_2}{dt}\)
Mutually induced e.m.f. in coil 1 \(E_{m2}=M\frac{dI_1}{dt}\)
Total induced e.m.f. in coil 1 \(E_1=L_1\frac{dI_1}{dt}M\frac{dI_2}{dt}\)
\(\mathbf{Note:}\)
Total induced e.m.f. in coil 2 \(E_2=L_2\frac{dI_2}{dt}M\frac{dI_1}{dt}\)
Total induced e.m.f. in the combination \(E=E_{S1}+E_{S2}+E_{m1}+E_{m2}\)
Correct Ans: B
Solution:
\( y=a\cos{nx}+b\sin{nx}\)
\(\frac{dy}{dx} = [an ( \sin nx) + b n \cos nx]\)
\(\frac{d^2y}{dx^2} = an^2 \cos nx  bn^2 \sin nx\)
\(\frac{d^2y}{dx^2} = n^2 [a \cos nx + b \sin nx]\)
\( \frac{d^2y}{dx^2} = n^2 y \)
Correct Ans: B
Solution:
For the DC Shunt motor back emf is given by
\(E_b = V  I_aR_a \)
\(E_b = 500  60 \times 0.2 = 488 \: V\)
Back emf can also be written as
\(E_b = \frac{\phi Z N P}{60A} \)
For the wave winding number fo parallel paths \( A = 2 \)
So above equation can be written as
\(E_b = \frac{\phi Z N P}{60\times 2} \)
\(\implies\) \( N = \frac{E_b \times 60\times 2}{\phi Z P} \)
\(\implies\) \( N = \frac{488 \times 60\times 2}{0.03 \times 720 \times 4} \)
\(\implies\) \( N = 677.77 \: RPM \)
Correct Ans: C
Solution:
The passive transducer converts change in non electrical quantity to a change in some passive electrical quantity, such as capacitance, resistance, or inductance.
Example: RTD, Strain Gauge
\(\Rightarrow\) Both RTD and strain gauge are resistive transducers but RTD is a temperature transducer and strain gauge is a pressure transducer.
Correct Ans: A
Solution:
\(\mathbf{Series\:Resonance:}\)
\(\mathbf{Resonance\:Curve}\)
\(\Rightarrow\) Two points A and B on the resonance curve are known as halfpower points.
\(\Rightarrow\) \(f_1\)  Lower cutoff frequency (Lower halfpower frequency)
\(\Rightarrow\) \(f_2\)  Upper cutoff frequency (Upper halfpower frequency)
\(\Rightarrow\) \(f_r\)  Resonance frequency is the Geometric mean of the two half power frequencies.
\(\Rightarrow\) Bandwidth (B.W) \(=f_2f_1\)
\(\Rightarrow\) At below resonance frequency, Circuit behaves like a capacitive circuit \((X_C > X_L)\)
\(\Rightarrow\) At above resonance frequency, Circuit behaves like an inductive circuit \((X_L > X_C)\)
\(\Rightarrow\) At resonance frequency, Circuit behaves like a purely resistive circuit \((X_L = X_C)\)
Correct Ans: D
Solution:
\(\Rightarrow\) Given single line diagram is equivalent circuit of generator because here current flows outwards, it is also a synchronous machine because it contains synchronizing reactance \( jX_s \).
Correct Ans: A
Solution:
\(\Rightarrow \)For Simple RC High Pass Filter, the circuit will be:
\(\Rightarrow\) As per voltagedivider relationship,
\(V_{out} = \frac{R}{RjX_c}\:V_{in}\)
\(\implies \) Transfer Function \(=H(j\omega) =\frac{V_{out}}{V_{in}}= \frac{R}{RjX_c}\)
\(\quad \quad = \frac{R}{R+\frac{1}{j\omega C}}\)
\(\quad \quad = \frac{j\omega RC}{1+j\omega RC}\)
Correct Ans: C
Solution:
Correct Ans: D
Solution:
\(\Rightarrow \mathbf{Mason's\: Gain\: Formula:}\) The relation between an input variable and an output variable of a Signal Flow Graph is given by Mason's Gain Formula.
\(\implies\) The overall system gain is given by:
\(G(s)=\frac{C(s)}{R(s)} = \frac{\sum_{k=1}^n P_k\Delta_k}{\Delta} \)
where,
\(\rightarrow P_k =\) Forward path gain of the \(K^{th}\) forward path.
\(\rightarrow \Delta = 1  \) [Sum of the loop gain of all individual loops] + [Sum of gain products of all possible two nontouching loops] + [Sum of gain products of all possible three nontouching loops] + .......
\(\rightarrow \Delta_k \) is obtained from \(\Delta\) by removing the loops which are touching the kth forward path.
Correct Ans: B
Solution:
\(\Rightarrow\) The symbol shown in figure is of TRIAC.
\(\Rightarrow\) TRIAC is a controlled bidirectional switch.
Correct Ans: D
Solution:
\(\Rightarrow\) Differential form of Gauss law states that the divergence of electric field E at any point in space is equal to \(\frac{1}{\varepsilon_0}\) times the volume charge density, \( \rho \), at that point.
\( \Big ( \frac{dE^x}{dx} + \frac{dE^y}{dy}+\frac{dE^z}{dz}\Big)=\frac{\rho}{\varepsilon_0}\)
Correct Ans: B
Solution:
Given,
\(\frac{dy}{dx}=\sqrt{\frac{1y^2}{1x^2}} \)
\(\Rightarrow\) \(\sqrt{1x^2} \: dy = \sqrt{1y^2} \: dx\)
\(\Rightarrow\) \(\frac{dy}{\sqrt{1y^2}} = \frac{dx}{\sqrt{1x^2}} \)
Apply Integration both side,
\(\Rightarrow\) \(\int \frac{dy}{\sqrt{1y^2}} = \int \frac{dx}{\sqrt{1x^2}} \)
\(\implies \sin^{1} y = \sin^{1} x + C\)
Correct Ans: A
Solution:
\(\Rightarrow\) The OAI (OR AND Invert) gate is used to realize the product of sum form of Boolean function.
\(\Rightarrow\) It is used in CMOS technology.
\(\Rightarrow\) It reduces the number of transistor needed to implement a Boolean function.
Correct Ans: B
Solution:
\(\Rightarrow\) Wave number \((k)\), when the wave is traveling through any material other than the vacuum is given by:
\(K=2\pi f \sqrt{ \varepsilon_r \varepsilon_0 \mu_r \mu_0} \)
Correct Ans: A
Solution:
\(\Rightarrow\) Type A chopper \(\rightarrow\) First quadrant operation (+ve voltage and +ve current)
\(\Rightarrow\) Type B chopper \(\rightarrow\) Second quadrant operation (+ve voltage and ve current)
\(\Rightarrow\) Type C chopper \(\rightarrow\) First and second quadrant operation (+ve voltage and current in both direction)
\(\Rightarrow\) Type D chopper \(\rightarrow\) First and fourth quadrant operation (ve and +ve voltage and +ve current)
\(\Rightarrow\) Type E chopper \(\rightarrow\) Four quadrant operation
Correct Ans: C
Solution:
\(\Rightarrow \mathbf{Forward\: paths\: for\: given\: SFG:}\) From R(s) to C(s)
\(\implies 1.\quad G_1 \rightarrow G_2 \)
\(\implies 2.\quad G_7 \rightarrow G_8 \)
\(\implies 3.\quad G_4\)
\(\implies 4.\quad G_7 \rightarrow G_6 \rightarrow G_2\)
\(\implies 5.\quad G_1 \rightarrow G_5 \rightarrow G_8\)
Correct Ans: D
Solution:
\(\Rightarrow\) Ampere's circuital law states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.
\(\Rightarrow\) So this law is applicable for the closed path only.
Correct Ans: B
Solution:
Correct Ans: A
Solution:
\(\Rightarrow\) The passive transducer converts change in non electrical quantity to a change in some passive electrical quantity, such as capacitance, resistance, or inductance.
Example: RTD, Strain Gauge, Thermistor
\(\Rightarrow\) The active transducer converts change in non electrical quantity to a change in some active electrical quantity, such as current or voltage without any auxilary power supply.
Example: Thermocouple, Photo voltaic transducer, Piezoelectric transducer
Correct Ans: D
Solution:
\(\Rightarrow\) If V is the electric potential in an electromagnetic field, then find the notation of \(dV\) is given by
\(dV = \frac{\partial V}{\partial x} dx+ \frac{\partial V}{\partial y} dy+ \frac{\partial V}{\partial z} dz\)
Correct Ans: C
Solution:
\(\Rightarrow\) To measure \(0100\:V\) by \(100\: \Omega\) instrument having full scale at \(10\: mA\) suppose resistance R is connected in series with the instrument.
\(\Rightarrow\) If \(0 \: V \) is applied across the combination \( 0 \: A\) flows through it and so displacement will be zero.
\(\Rightarrow\) If \( 100 \: V \) is applied across the combination \(10\: mA\) should flow through it.
So \( 10\: mA = \frac{100 \: V}{R + R_i} \)
\(\Rightarrow\) \( 10\: mA = \frac{100 \: V}{R + 100} \)
\(\Rightarrow\) \( R + 100 = \frac{100 \: V}{10\: mA} = 10000 \)
\(\Rightarrow\) \( R = 9.9 \: k\Omega \)
Correct Ans: D
Solution:
Equivalent electrical circuit of short shunt compound wound generator is given by
From above circuit it is clear that the series current of short shunt compound generator is equal to load current.
Correct Ans: B
Solution:
\(\Rightarrow\) A system is said to be \(\mathbf{Controllable}\), if it is possible to transfer the system state from any initial state to any desired state in finite interval of time.
\(\Rightarrow\) A system is said to be \(\mathbf{Observable}\), if every state can be completely identified by measurements of the outputs at the finite time interval.
\(\Rightarrow\) For given options, only option B indicated finite time interval.
Correct Ans: C
Solution:
As per figure, One line diagram is drawn for basic power system network.
\(\Rightarrow\) To calculate the base voltage for Transformer HT side, one example is taken.
Assume ratings:
Generator : \(11 KV, 25 KVA\)
Transformer 1: \(11 / 66 KV, 25 KVA\)
Transformer 2: \( 66/3.3 KV, 25 KVA\)
\(\Rightarrow\) Take base voltage at Generator side \(11 KV\) as base voltage.
\(\rightarrow\) \(Base \: voltage \: on \: Transformer \: HT \: side \: (Transmission \: line) = \frac{66}{11} \times Base \: voltage \: LT \: side = 66 kV\)
\(\therefore\) \(kV_b \: (HT) \: connection = \frac{HT \: rating}{LT \: rating} \times kV_b \: (LT) \: connection\)
Correct Ans: B
Solution:
Correct Ans: B
Solution:
Correct Ans: A
Solution:
\(\mathbf{Bilateral \:circuit:}\)
\(\Rightarrow\) A circuit, whose properties or characteristics are the same in either direction is known as Bilateral circuit.
\(\mathbf{Unilateral \:circuit:}\)
\(\Rightarrow\) A circuit, whose properties or characteristics are not the same in either direction is known as Unilateral circuit.
\(\mathbf{Bilateral \:element:}\)
\(\Rightarrow\) The voltagecurrent relation is the same irrespective of direction of flow of current. Examples: R, L,C.
\(\mathbf{Linear \:element:}\)
\(\Rightarrow\) Linear elements are those through which the flow of current changes linearly with the changing of the applied voltage across them.
Correct Ans: B
Solution:
\( \Rightarrow\) \(Qfactor = \frac{Resonance\:frequency(F_r)}{Band\:Width(BW)}\)
\( \Rightarrow\) \(Qfactor = 2\pi (\frac{maximum\:energy\:stored}{energy\:dissipated\:per\:cycle})\)
\( \Rightarrow\) \(Qfactor =\frac{1}{R}\sqrt{\frac{L}{C}}\)
\( \Rightarrow\) \(QFactor =\frac{Potential\:drop\:across\:capacitor\:or\:inductor}{Potential\:drop\:across\:R}\)
\( \Rightarrow\) \(QFactor = \frac{\omega L}{R}\)
\(\mathbf{Note:}\)
\(Qfactor\) ia also known as Quality Factor.
Correct Ans: B
Solution:
Equation of synchronous motor when operating in leading power factor is given by
\(E_f^2 = (V_t \cos \phi  I_aR_a)^2 +(V_t \sin \phi + I_aX_s)^2\)
Equation of synchronous motor when operating in lagging power factor is given by
\(E_f^2 = (V_t \cos \phi  I_aR_a)^2 +(V_t \sin \phi  I_aX_s)^2\)
Correct Ans: NA
Solution:
\(\mathbf{Given \:data:}\)
\(\Rightarrow\) \(V_s=250\:V,R=25\:\Omega,\:\cos\phi=0.8\)
\(\cos\phi=\frac{R}{Z}\implies Z=\frac{R}{\cos\phi}=\frac{25}{0.8}=31.25\:\Omega\)
\(I_s=\frac{V_s}{Z}=\frac{250}{31.25}=8\:A\)
Total power \(P=V_sI_s\cos\phi=250\times 8\times 0.8=1600\:W\)
Correct Ans: D
Solution:
Given,
\(2y \: dx  (3y  2x)dy = 0\)
\(\mathbf{Homogeneous \: equation}\) : The degree of all the terms in the equation is same.
\(\Rightarrow\) \(2y \: dx  (3y  2x)dy = 0\), it is Homogeneous equation.
\(\mathbf{Linear \: equation}\): A linear equation or polynomial, with one or more terms, consisting of the derivatives of the dependent variable with respect to one or more independent variables is known as a linear differential equation.
\(\Rightarrow\) \(2y \: dx  (3y  2x)dy = 0\) , this equation can be write as,
\(\implies \frac{dx}{dy} + \frac{x}{y} = \frac{3}{2}\), it is Linear equation.
\(\mathbf{Exact \: equation}\): \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), for the differential equation \(M dx + N dy = 0\)
\(\Rightarrow\) \(2y \: dx  (3y  2x)dy = 0\) , by comparison
\( M = 2y, \quad N = 2x3y\)
\(\implies \frac{\partial M}{\partial y} = \frac{\partial 2y}{\partial y} = 2\)
\(\implies \frac{\partial N}{\partial x} = \frac{\partial (2x3y)}{\partial x} =2 \)
\(\therefore\) \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), given equation is Exact.
\(\therefore\) \(2y \: dx  (3y  2x)dy = 0\) equation is \(\mathbf{Homogenous, \: Linear \: and \: Exact \: equation}\).
Correct Ans: B
Solution:
\(\Rightarrow\) Form Factor = Ratio of RMS value to average value.
\(\Rightarrow\) Pitch Factor = Ratio of emf generated in short pitch coil to the emf generated in full pitch coil.
\(\Rightarrow\) The Distribution Factor or the Breadth Factor = Ratio of the actual voltage obtained to the possible voltage if all the coils of a polar group were concentrated in a single slot.
Correct Ans: A
Solution:
As the square wave is odd function,
\(a_0 = a_n = 0\) and \(L=\pi\)
and
\(b_n = \frac{1}{L} \int_0^{2L} f(x). \sin (\frac{n \pi x}{L}).dx \)
here, \( L=\pi\)
\(\Rightarrow\) \(b_n = \frac{2}{L} \int_0^{L} f(x). \sin (\frac{n \pi x}{L}).dx \)
\(b_n = \frac{4}{n \pi} . \sin^2 (\frac{1}{2} n\pi) \)
\(b_n = \frac{2}{n \pi} [1  (1)^n]\)
\(\Rightarrow\) \(b_n = \begin{cases}0 & n = even\\ \frac{4}{n \pi} & n = odd\end{cases}\)
\(\Rightarrow\) Now, the Fourier series is,
\(SW(x) = \frac{4}{\pi} \displaystyle\sum\limits_{n=1,3,5,...}^\infty \frac{1}{n}. \sin (\frac{n \pi x}{L}) \)
\(\Rightarrow SW(x) = \frac{4}{\pi} [\frac {\sin x}{1} + \frac {\sin 3x}{3} + \frac {\sin 5x}{5} + ....]\) for \(L =\pi\)
Correct Ans: C
Solution:
\(\Rightarrow\) Pole changing method is mainly applicable to cage motors, because rotor of cage motor develops a number of poles, which is equal to the poles of the stator winding.
\(\Rightarrow\) In the stator frequency control scheme, the variable frequency control can be achieved by cycloconverter.
Correct Ans: B
Solution:
\(\Rightarrow\) Snubber circuit is used to prevent SCR from high dv/dt.
\(\Rightarrow\) A capacitor is connected across thyristor, which allows voltage to rise exponentially across SCR.
\(\Rightarrow\) In order to discharge this capacitor safely, a resistor is connected in series with capacitor.
Correct Ans: A
Solution:
Applying Gauss' theorem to cylindrical Gaussian surface of radius \(r\), the radial component of electric field intensity \(D_r\) is given by :
\(D_r = \frac{q}{2r}\)
Correct Ans: C
Solution:
\(\Rightarrow \mathbf{Step\:response\:of\:High\:Pass\:Filter:}\)
\(\implies\) For step input \(v_i(t) = u(t) \Rightarrow V_I(s) = \frac{1}{s}\)
\(\implies\) Output \(V_r(s) = \frac{1}{s+\frac{1}{RC}}\)
\(\implies v_r(t) = e^{\big ( \frac{t}{RC} \big )}\) for t\(\geq\)0
\(\implies v_r(t) = e^{\big ( \frac{t}{RC} \big )}\:u(t)\)
Correct Ans: B
Solution:
\(\Rightarrow\) The passive transducer converts change in non electrical quantity to a change in some passive electrical quantity, such as capacitance, resistance, or inductance.
Example: RTD, Strain Gauge, Thermistor
\(\Rightarrow\) The active transducer converts change in non electrical quantity to a change in some active electrical quantity, such as current or voltage without any auxilary power supply.
Example: Thermocouple, Photo voltaic transducer, Piezoelectric transducer
Correct Ans: C
Solution:
\(\Rightarrow\) \(\mathbf{Phase \: Crossover \: Frequency \: ( \omega_{pc})}\) is the frequency at which phase angle of \(G(s)H(s)\) is \(180^o\).
\(\Rightarrow\) \(\mathbf{Gain \: Crossover \: Frequency \: ( \omega_{gc})}\) is the frequency at which the magnitude of the \(G(s)H(s)\) becomes 1.
Correct Ans: C
Solution:
\(\Rightarrow\) The average output voltage of singlephase full converter is \(V_{dc} = \frac{2V_m}{\pi}\cos \alpha\)
Correct Ans: B
Solution:
\(\Rightarrow\) Permanent Split Capacitor (PSC) Induction Motor does not have any starting switch.
Correct Ans: D
Solution:
Correct Ans: D
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Correct Ans: D
Solution:
Correct Ans: A
Solution:
Correct Ans: B
Solution:
Correct Ans: B
Solution:
Correct Ans: B
Solution:
Correct Ans: B
Solution:
Correct Ans: D
Solution:
Correct Ans: C
Solution:
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Correct Ans: C
Solution:
Correct Ans: B
Solution:
Correct Ans: A
Solution:
Correct Ans: A
Solution:
Correct Ans: D
Solution:
Correct Ans: B
Solution:
Correct Ans: C
Solution:
Correct Ans: B
Solution:
Correct Ans: D
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Correct Ans: C
Solution:
Correct Ans: B
Solution:
Correct Ans: C
Solution:
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Correct Ans: D
Solution:
Correct Ans: A
Solution:
Correct Ans: A
Solution:
Correct Ans: B
Solution:
Correct Ans: A
Solution:
Correct Ans: D
Solution:
Correct Ans: B
Solution:
Correct Ans: B
Solution:
Subject  Number of questions 

POWER ELECTRONICS & DRIVES  8 
DIGITAL ELECTRONICS  3 
BASIC ELECTRONICS  3 
POWER SYSTEM  4 
ENGINEERING MATHEMATICS  6 
CONTROL SYSTEM ENGINEERING  3 
ELECTRICAL MACHINES  11 
ELECTROMAGNETICS  6 
MEASUREMENT & INSTRUMENTATION  4 
ELECTRICAL CIRCUIT ANALYSIS  10 
ENGLISH  10 
GENERAL KNOWLEDGE  10 
GUJARATI  10 
SIGNALS & SYSTEMS  2 
BASICS OF COMPUTER  10 
Total  100 
Correct Ans: D
Solution:
\(\Rightarrow\) Compound DC generator is having both series and shunt windings.
\(\Rightarrow\) If the series field flux opposes the shunt field flux in a DC generator, the generator is said to be a differentially compounded generator.
\(\Rightarrow\) If the series field flux supports the shunt field flux in a DC generator, the generator is said to be a cumulatively compounded generator.
Correct Ans: D
Solution:
Correct Ans: B
Solution:
The efficiency (%\( \eta\)) of transformer in case of fractional load, where n (Actual load/full load) is less than 1 is given by:
%\( \eta = \frac{n (VA \: rating) \: \cos{\phi_2}}{n^2P_{cu} F.L. + P_i + n (VA \: rating) \: \cos{\phi_2}} \times 100\)
Correct Ans: C
Solution:
\(\Rightarrow\) For an induction motor
Rotor input or airgap power \( P_{in} = \frac{3 I_2^2 R_2}{s} \)
Rotor copper losses \( P_{cu} = s \times P_{in} = 3 I_2^2 R_2 \)
Mechanical output power \( P_o = P_{in}  P_{cu} = \frac{3 I_2^2 R_2}{s}  3 I_2^2 R_2 \)
\(\implies\) \( P_o = 3 I_2^2 R_2 (\frac{1s}{s}) \)
\(\implies\) \( P_o = P_{cu} (\frac{1s}{s}) \)
\(\implies\) \( P_{cu} = P_o (\frac{s}{1s}) \)
\(\implies\) \( P_{cu} = 1920 \times (\frac{0.04}{0.096}) \)
\(\implies\) \( P_{cu} = 80 \: W \)
Correct Ans: B
Solution:
\(\mathbf{Given \:data:}\)
\(\Rightarrow\) \(I=3\:A,\:R=8\:\Omega\)
As per Figure mention in question, Voltage to Current Source Transformation
\(I=\frac{V_s}{R}\implies V_s=I \times R=3 \times 8=24\:V\)
Internal resistance of the equivalent current source has the same value as the internal resistance of the voltage source. Therefore, \(R=8\:\Omega\)
\(\mathbf{Note:}\)
\(\mathbf{According\: to \:Source \:transformation:}\)
\(\Rightarrow\) A practical voltage source of constant voltage V and internal resistance \(R_{int}\) is equivalent to a current source of current \(I=\frac{V}{R_{int}}\) and internal resistance \(R_{int}\) in parallel with current source.
\(\Rightarrow\) Internal resistance of the equivalent current source has the same value as the internal resistance of the voltage source.
Correct Ans: A
Solution:
Correct Ans: A
Solution:
Given,
\( \frac{dy}{dx} = \frac{x}{1+x^2} \)
\(\implies dy = \frac{x}{1+x^2} dx\)
Apply integrate both side,
\(\implies\int dy = \int \frac{x}{1+x^2} dx\)
Take \(t = x^2, dt = 2x \: dx \implies x dx = \frac{1}{2} dt\)
\(\implies y = \int \frac{dt}{2(1+t)}\)
\(\implies 2y = log(1+t) + C\)
\(\implies 2y = log(1+x^2) + C\)
Correct Ans: C
Solution:
\(\Rightarrow\) Potential at every point on the spherical surface having radius R containing charge Q inside is:
\( V_R = \frac{Q}{4R\pi \varepsilon_0 } \)
Correct Ans: A
Solution:
\(\Rightarrow\) The hall effect sensor is a type of transducer used for measuring the magnetic field by converting it into an emf.
\(\Rightarrow\) The transducer converts the magnetic field into an electric quantity which is easily measured by the analogue and digital meters.
Correct Ans: D
Solution:
\(\Rightarrow\) Equation of motion of the threephase synchronous alternator rotor, that is driven by a prime mover is given by:
\( J\frac{d^2\theta}{dt^2} = T_m  T_e \)
Correct Ans: C
Solution:
\(\Rightarrow\) As per the sampling theorem, the sampling frequency \(\omega_s\) must be atleast twice of the maximum frequency or cutoff frequency \(\omega_M\).
\(\Rightarrow\) \(\omega_s > 2\:\omega_M\)
Correct Ans: D
Solution:
\(\Rightarrow\) Armature reaction is the effect of the flux set up by the currents in the armature winding, on the main field flux.
\(\Rightarrow\) Interpoles windings are connected in series with the armature so that any changes in the armature current due to armature reactance, loading, or selfinductance pass through the interpole windings.
\(\Rightarrow\) This creates a changing magnetic field equal to and opposite to that of the armature, thereby cancelling the effect.
Correct Ans: A
Solution:
Evaluate \( y(t) \) from \( y(s) = \frac{2s}{(s+1)(s+2)} \)
Here the denominator has nonrepeated linear factors.
Let \(\frac{2s}{(s+1)(s+2)}=\frac{A}{s+1}+\frac{B}{s+2} \)
Multiplying both side by (s+1)(s+2)
We get 2s=A(s+2)+B(s+1)
Putting S=2, We get \(B=4\)
Putting S=1, We get \(A=2\)
\(\therefore\) \(\frac{2s}{(s+1)(s+2)}=\frac{2}{s+1}+\frac{4}{s+2} \)
Taking both side InverseLaplace transform
\(\therefore\) \(L^{1}\lbrace\frac{2s}{(s+1)(s+2)}\rbrace=2L^{1}\lbrace\frac{1}{s+1}\rbrace +4L^{1}\lbrace\frac{1}{s+2}\rbrace=2e^{t}+4e^{2t}\)
Correct Ans: C
Solution:
\(\Rightarrow\) Kelvin’s bridge is a DC bridge which is used for measurement of low resistance.
Correct Ans: D
Solution:
Equivalent circuit diagram of long shunt DC compound generator is as shown below,
\(P_L = 110 \:kW\) and \( V_L = 220 \: V\),
So, \(I_L = \frac{P_L}{V_L}= \frac{110000}{220} = 500 \: A\)
Shunt field current \( I_{sh} = \frac{220}{110} = 2\: A \)
Armature current \( I_a = I_L + I_{sh} = 502 \: A \)
Induced EMF \( E_g = V_L + I_a (R_{se} + R_A) = 220 + 502 \times (0.002 + 0.01)\)
\( E_g = 220 + 502 \times (0.012 ) =226.024 \: V\)
Correct Ans: C
Solution:
\(\Rightarrow\) Figure shows the inverting amplifier using opamp.
\(\Rightarrow\) The output of inverting amplifier is \(V_{O} = \frac{R_f}{R_1} V_{in}\)
\(\implies V _{O} = (\frac{25\:k}{10 \:k}) 0.5\)
\(\implies V _{O} =1.25\: V\)
Correct Ans: A
Solution:
\(\Rightarrow\) In a slip ring induction motor
\(\rightarrow\) The rotor winding is similar to the stator winding.
\(\rightarrow\) Rotor copper losses are high and hence less efficiency.
\(\rightarrow\) Construction is complicated due to presence of slip ring and brushes.
Correct Ans: D
Solution:
Given,
\(A = B^{1}\)
\(\rightarrow\) Post Multiply with \(B\) matrix,
\(\implies AB = B^{1}B = I\)
\(\rightarrow\) Pre Multiply with \(B\) matrix,
\(\implies BA = BB^{1} = I\)
\(\therefore AB=BA=I\)
Correct Ans: D
Solution:
\(\Rightarrow\) Maxwell’s inductance bridge cannot be used for measurement of high Q values.
Correct Ans: B
Solution:
\(\mathbf{Given \:data:}\)
\(\Rightarrow\) Resonance Frequency \(f_r=850\:Hz\)
\(\Rightarrow\) Bandwidth \(BW=150\:Hz\)
\(f_2\)  Upper cutoff frequency (Upper halfpower frequency)
\(f_2=f_r+\frac{1}{2}B.W=850+\frac{150}{2}=925\:Hz\)
\(f_1\)  Lower cutoff frequency (Lower halfpower frequency)
\(f_1=f_r\frac{1}{2}B.W=850\frac{150}{2}=775\:Hz\)
\(\mathbf{Note:}\)
\(\mathbf{Series\:Resonance:}\)
\(\mathbf{Resonance\:Curve}\)
\(\Rightarrow\) Two points A and B on the resonance curve are known as halfpower points.
\(\Rightarrow\) \(f_1\)  Lower cutoff frequency (Lower halfpower frequency)
\(\Rightarrow\) \(f_2\)  Upper cutoff frequency (Upper halfpower frequency)
\(\Rightarrow\) \(f_r\)  Resonance frequency is the Geometric mean of the two half power frequencies.
\(\Rightarrow\) Bandwidth (B.W) \(=f_2f_1\)
\(\Rightarrow\) At below resonance frequency, Circuit behaves like a capacitive circuit \((X_C > X_L)\)
\(\Rightarrow\) At above resonance frequency, Circuit behaves like an inductive circuit \((X_L > X_C)\)
\(\Rightarrow\) At resonance frequency, Circuit behaves like a purely resistive circuit \((X_L = X_C)\)
Correct Ans: D
Solution:
\(\Rightarrow\) As shown in the characteristics, the circuit offers zero gain to the signals with frequency less than the cut off frequency \(f_c\) and offers high gain to the signals with frequency more than cut off frequency \(f_c.\)
\(\Rightarrow\) That means the characteristics shown in figure is of high pass filter, which allows high frequency signals to pass through.
Correct Ans: A
Solution:
Slip at maximum torque:
\(s_{max} =\frac{R_m}{X_m} \)
\(\rightarrow\) By varying the rotor resistance, a maximum torque can be obtained at any required slip.
Maximum torque is given by
$$ T_{max} = \frac{180}{2 \pi N_s} \frac{V^2}{2X_2} $$
\(\rightarrow\) The maximum torque is directly proportional to square of rotor induced EMF at the standstill.
\(\rightarrow\) The maximum torque is inversely proportional to the rotor reactance.
Correct Ans: C
Solution:
Correct Ans: B
Solution:
\(\Rightarrow\) Power rating of device = Current rating \(\times\) Voltage rating
\(\Rightarrow\) Power rating of device = 15 \(\times\) 250 = 3750 W
Correct Ans: A
Solution:
\(\int \sec^2 (3 + 4x) \:dx\)
\(\rightarrow\) Take \(u = 3+4x\)
\(\implies du = 4 \: dx\)
\(\implies dx = \frac{1}{4} du\)
\(\rightarrow\) Put value in main equation,
\(\int \sec^2 (3 + 4x) \:dx\)
\(=\int \frac{1}{4} \sec^2 u du\)
\(=\frac{1}{4} tan \:u = \frac{1}{4} tan(3+4x)\)
Correct Ans: B
Solution:
\(\Rightarrow\) Boolean expression of sum of half adder = \(A \oplus B\)
\(\Rightarrow\) This can be directly implement using ExOR gate.
Correct Ans: C
Solution:
\(\mathbf{For \:delta\: connected \:load:}\)
\(\Rightarrow\) A threephase threewire supply feeds a delta connected load consisting of three equal resistors. \(33.33 \%\) of reduction in load if one of the resistor is removed.
\(\mathbf{For \:star\: connected \:load:}\)
\(\Rightarrow\) A threephase threewire supply feeds a star connected load consisting of three equal resistors. \(50 \%\) of reduction in load if one of the resistor is removed.
Correct Ans: C
Solution:
\(\Rightarrow\) The differential amplifier amplifies the difference of input signals.
\(\Rightarrow\) It also provides high input impedance.
Correct Ans: A
Solution:
\(\Rightarrow\) SFG is applicable for Linear Timeinvarient Systems only.
Correct Ans: A
Solution:
\(\Rightarrow\) Maximum phase lead frequency: \(\omega_m = \frac{1}{\tau \sqrt{\beta}}\)
\(\Rightarrow\) Maximum phase lead: \(\phi_m=\sin^{1} \Big (\frac{a1}{a+1} \Big ) \)
Correct Ans: B
Solution:
Correct Ans: A
Solution:
\(\mathbf{Series\:Resonance:}\)
\(\mathbf{Resonance\:Curve}\)
\(\Rightarrow\) Two points A and B on the resonance curve are known as halfpower points.
\(\Rightarrow\) \(f_1\)  Lower cutoff frequency (Lower halfpower frequency)
\(\Rightarrow\) \(f_2\)  Upper cutoff frequency (Upper halfpower frequency)
\(\Rightarrow\) \(f_r\)  Resonance frequency is the Geometric mean of the two half power frequencies.
\(\Rightarrow\) Bandwidth (B.W) \(=f_2f_1\)
\(\Rightarrow\) At below resonance frequency, Circuit behaves like a capacitive circuit \((X_C > X_L)\)
\(\Rightarrow\) At above resonance frequency, Circuit behaves like an inductive circuit \((X_L > X_C)\)
\(\Rightarrow\) At resonance frequency, Circuit behaves like a purely resistive circuit \((X_L = X_C)\)
Correct Ans: C & D
Solution:
\(\Rightarrow\) \(\mathbf{Correct \: answer : \: C \: \& \: D}\)
At node P, applying KCL
Consider current \(I_2\) leaving the junction
\(I_2+I_5=I_1+I_3+I_4\implies I_2=I_1+I_3+I_4I_5\)
Consider current \(I_2\) entering the junction
\(I_5=I_1+I_2+I_3+I_4\implies I_2=I_5I_1I_3I_4\)
\(\mathbf{Note:}\)
\(\mathbf{Kirchhoff's \:Current\: Law\: (KCL):}\)
KCL law states that the total current entering a circuits junction is exactly equal to the total current leaving the same junction.This law represents conversation of charge.
Correct Ans: B
Solution:
\(\Rightarrow\) In asynchronous circuit, the clock pulse is given to the first flip flop only.
\(\Rightarrow\) Then output of first flip flop will drive the second flip flop and so on...
Correct Ans: C
Solution:
\(\mathbf{Given \:data:}\)
\(K=0.7,\:M=70\:mH,\:L_1=80\:mH\)
Coefficient of coupling \(K=\frac{M}{\sqrt{L_1L_2}}\)
Where
\(L_1 =\)selfinductance of the first coil
\(L_2 =\)selfinductance of the second coil
Mmutual inductance between first coil and second coil
\(K=\frac{M}{\sqrt{L_1L_2}}\implies L_2=\frac{M^2}{K^2\times L_1}=\frac{(70)^2}{(0.7)^2\times 80}=125\:mH\)
Correct Ans: A
Solution:
\(\Rightarrow\) Equation of synchronous motor when operating in leading power factor is given by
\(E_f^2 = (V_t \cos \phi  I_aR_a)^2 +(V_t \sin \phi + I_aX_s)^2\)
\(\Rightarrow\) Equation of synchronous motor when operating in lagging power factor is given by
\(E_f^2 = (V_t \cos \phi  I_aR_a)^2 +(V_t \sin \phi  I_aX_s)^2\)
Correct Ans: A
Solution:
Applying KVL to the given circuit
\(V=V_R+V_L\)
\(V=iR+L\frac{di}{dt}\)
\(\frac{di}{ViR}=\frac{1}{L}dt\)
Multiplying both sites by (R)
\(\frac{(R) di}{ViR}=\frac{R}{L}dt\)
Integration on both sides
\(\int \frac{(R) di}{ViR} =\int \frac{R}{L}dt +C\)
\(\log_e(ViR)=\frac{Rt}{L}+C\)........(1)
\(i=0\) when \(t=0\) put in equ(1)
\(\log_e V=C\) value put in equ(1)
\(\log_e(ViR)=\frac{Rt}{L}+\log_eV\)
\(\log_e(ViR)\log_eV=\frac{Rt}{L}\)
\(\log_e(\frac{ViR}{V})=\frac{Rt}{L}\)
\((\frac{ViR}{V})=e^{\frac{Rt}{L}}\)
\(ViR=Ve^{\frac{Rt}{L}}\)
\(iR=V[1e^{\frac{Rt}{L}}]\)
\(i=\frac{V}{R}[1e^{\frac{Rt}{L}}]\)
Correct Ans: C
Solution:
\(\Rightarrow\) As per coulomb's law, electric field intensity E at a distance of R for a simple point charge is:
\(\frac{1}{4\pi \varepsilon_0} \Big (\frac{Q}{R^2} \Big ) \:V/m \)
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Given,
\(y=me^{5x}\)
\(\implies \frac{dy}{dx} = \frac{d}{dx} me^{5x}= m \: e^{5x} \: 5 = 5y\)
Correct Ans: A
Solution:
\(\Rightarrow\) The voltage equation for the Piezoelectric transducer is given by \(E_0 = gtP\).
g = Voltage sensitivity of the crystal
t = Thickness of crystal
P = Pressure on the crystal
Correct Ans: A
Solution:
\(\Rightarrow\) In resistance triggering method, SCR is triggered using a diode and resistor network.
\(\Rightarrow\) When diode is forward biased the SCR will be triggered.
\(\Rightarrow\) The forward bias voltage is controlled by a potential divider using variable resistor.
\(\Rightarrow\) Triggering angle can be controlled up to 90° because the diode forward voltage can be controlled up to maximum value of input only.
Correct Ans: D
Solution:
\(\Rightarrow\) Hysteresis losses in a transformer are given by:
\(W_h = K_h \: B_m^{1.6}\:f\:V\)
Correct Ans: D
Solution:
\(\Rightarrow\) The rectangular coordinate system is an orthogonal coordinate system with coordinate axis defined x, y, and z, since its coordinate surfaces \(x = constant\), \(y = constant\), and \(z = constant\) are planes that meet at right angles to one another.
\(\Rightarrow\) Same way cylindrical and spherical coordinate systems are also orthogonal.
Correct Ans: C
Solution:
\(P = \frac{EV}{X}sin \delta\)
\(\rightarrow \frac{dp}{d\delta} \propto cos \delta\)
\(\implies \delta\) = \(90^o \: to \: 90^o \) \(\implies\) \(\cos \delta\) is positive
\(\therefore\) \(\frac{dp}{d\delta}\) is positive for \(90^o \: to \: 90^o \) angle range.
Correct Ans: C
Solution:
\(\Rightarrow\)Consider a propagation of an electromagnetic wave through a uniform dielectric medium of dielectric constant \(\varepsilon\).
\(\Rightarrow\)The dipole moment per unit volume \(P\) induced in the medium by the wave electric field \(E\) is:
\(P = \varepsilon_0 (\varepsilon  1)E\)
Correct Ans: A
Solution:
\(\Rightarrow\) The controller has only Proportionality constant. So it is P Controller.
Correct Ans: B
Solution:
\(\Rightarrow\) The output load voltage in phase controlled converter can be varied by changing the thyristor firing angle.
\(\Rightarrow\) This method is known as delay firing angle method.
\(\Rightarrow\) As firing angle increases, the output voltage decreases and visa versa.
Correct Ans: B
Solution:
Given,
\(f(x,y) = x^2 + xyz +z\)
\(\implies f_x = 2x + yz \)
\(\implies f_{{x}_{(2,2,2)}} = (2 \times 2) + (2 \times 2) = 8\)
Correct Ans: A
Solution:
Given,
\(y = \cos(23x)\)
\(Take, \:f(x) = \cos \: x, g(x) = 23x\)
\(\Rightarrow\) Use Chain rule, \(\frac{d}{dx} f(g(x)) = f^{'}(g(x))g^{'}(x) \)
\(\rightarrow\) Take \(u = 23x\)
\(\implies \frac{d}{du}[\cos \: u] \frac{d}{dx}[23x]\) \((\because Chain \: rule ) \)
\(\frac{dy}{dx}=[ \sin \: u ] \times [3] = 3 \sin (23x)\)
Correct Ans: A
Solution:
\(\Rightarrow\) Ampere's law for the electromagnetic field can be expressed as:
\(\oint B.dl =\mu_0I\)
Correct Ans: C
Solution:
\(\Rightarrow\) Single line diagram: It is the representation of a power system network using the symbol for each component.
\(\implies\) The transmission line parameters are represented by series impedance in Short Transmission Line. \((\because\) Capacitance effect is neglected in short transmission line)
Correct Ans: B
Solution:
\(\Rightarrow\) The electric potential at point \(P(V_p)\) due to a uniform volume charge is given by:
\(V_p = \frac{1}{4\pi\varepsilon_0}\int_{v} \frac{\rho_v dv}{R}V\)
Correct Ans: B
Solution:
\(\Rightarrow\) As we know that inductor current opposes the rate of change of current, it is used to limit the rate of change of anode current in Thyristor.
\(\Rightarrow\) High rate of change of anode current may create local hot spots, which may damage the thyristor.
Correct Ans: C
Solution:
\(L\lbrace f(t)\rbrace = L\lbrace\cos 3t\rbrace=\frac{s}{s^2+3^2}=\frac{s}{s^2+9}\)
\(L\lbrace f(t)\rbrace = L\lbrace\sin 3t\rbrace=\frac{3}{s^2+3^2}=\frac{3}{s^2+9}\)
Correct Ans: B
Solution:
\(\Rightarrow\) In half wave AC voltage controller the output voltage is controlled by controlling the voltages in any one half cycle (generally positive half cycle)
\(\Rightarrow\) Therefore it uses one SCR to control the voltage of half cycle and 1 diode to provide path to the current in other half cycle.
\(\Rightarrow\) So aniparallel connection of 1 diode and 1 SCR is is used in half wave AC voltage controller.
Correct Ans: A
Solution:
\(\mathbf{Magnetic\:flux\:density:}\)
\(\Rightarrow\) It is defined as the magnetic flux per unit area of crosssection at the right angle to the flux.
\(\Rightarrow\) It is denoted by the symbol \(B\).
\(\Rightarrow\) Mathematically, \(B=\frac{\phi}{A}\)
\(\Rightarrow\) Its unit is \(Wb/m^2\) or Tesla (T)
Correct Ans: C
Solution:
For given function \)f(x)\) with period \(T=2\pi \implies \omega_0 = \frac{2\pi}{T}=1\)
Also, due to symmetric condition, \(a_n=0\)
\(\Rightarrow\) For \(a_0:\)
\(a_0 = \frac{1}{T} \int_0^T f(x)\:dx \)
\(\quad \: =\frac{1}{2\pi} \int_{\pi}^0 1\:dx \)
\(\quad \: =\frac{1}{2\pi} [0(\pi)]\)
\(\quad \: =\frac{1}{2}\)
\(\Rightarrow\) For \(b_n:\)
\(b_n = \frac{2}{T} \int_0^T f(x)\sin(n\omega_0 x)\:dx \)
\(\quad \: =\frac{2}{2\pi} \int_{\pi}^0 \sin(n\omega_0 x)\:dx \)
\(\quad \: =\frac{1}{\pi} [\frac{\cos(nx)}{n}]_{\pi}^0\)
\(\quad \: =\frac{1}{\pi n}[1\cos (n\pi)]\)
\(\quad \: =\frac{2}{\pi n} \:for\: n=1,3,5,...\)
\(\Rightarrow\) Fourier series representtion of f(x):
\(f(x) = a_0+\sum_{n=1}^\infty [a_n\cos(n\omega_0 x)+b_n \sin(n \omega_0 x)] \)
\(\implies f(x) = \frac{1}{2} \frac{2}{\pi}[\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+.....]\)
Correct Ans: B
Solution:
\(\Rightarrow\) For stepdown chopper output voltage \(V_{av} = DV\)
\(\Rightarrow\) Where, \(D = \frac{T_{on}}{T_{on} + T_{off}}\)
\(\quad \therefore V_{av} = \frac{V\:T_{on}}{T_{on} +T_{off}}\)
Correct Ans: A
Solution:
Equivalent circuit model of induction motor is as follows:
Correct Ans: B
Solution:
Correct Ans: D
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Correct Ans: D
Solution:
Correct Ans: D
Solution:
Correct Ans: B
Solution:
Correct Ans: D
Solution:
Correct Ans: C
Solution:
Correct Ans: C
Solution:
Correct Ans: B
Solution:
Correct Ans: B
Solution:
Correct Ans: D
Solution:
Correct Ans: D
Solution:
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Correct Ans: B
Solution:
Correct Ans: C
Solution:
Correct Ans: C
Solution:
Correct Ans: A
Solution:
Correct Ans: B
Solution:
Correct Ans: B
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Correct Ans: B
Solution:
Correct Ans: A
Solution:
Correct Ans: D
Solution:
Correct Ans: B
Solution:
Correct Ans: A
Solution:
Correct Ans: D
Solution:
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Correct Ans: A
Solution:
Correct Ans: A
Solution:
Correct Ans: C
Solution:
Correct Ans: D
Solution:
Subject  Number of questions 

POWER ELECTRONICS & DRIVES  5 
DIGITAL ELECTRONICS  5 
BASIC ELECTRONICS  1 
POWER SYSTEM  4 
ENGINEERING MATHEMATICS  6 
CONTROL SYSTEM ENGINEERING  4 
ELECTRICAL CIRCUIT ANALYSIS  10 
ENGLISH  10 
GENERAL KNOWLEDGE  10 
GUJARATI  10 
SIGNALS & SYSTEMS  5 
ELECTRICAL MACHINES  11 
ELECTROMAGNETICS  5 
MEASUREMENT & INSTRUMENTATION  4 
BASICS OF COMPUTER  10 
Total  100 
Correct Ans: B
Solution:
\(N_s\: = \:\frac{120f}{P}\: = \:\frac{120 \times 50}{4} = \:1500 \: RPM\)
\(s\: = \:\frac{N_s \:\: N_r}{N_s} = \:\frac{1500 \:\: 600}{1500} =0.6\)
Rotor frequency \(f_r\: = \:s.f_s\: = 0.6 \times 50 = 30 \: Hz\)
Correct Ans: B
Solution:
None of above
\(\Rightarrow\) CMRR = \(\frac {A_{ol}}{A_{cm}} = \frac{100000}{0.2} = 500000\)
Correct Ans: D
Solution:
\(\Rightarrow\) For the given System,
Forward Path Gain \(G(s) = \Big ( K_p+\frac{K_i}{s}\Big )\Big ( \frac{1}{s+1} \Big)=\frac{sK_p+K_i}{s(s+1)}\)
Now, \(\frac{E(s)}{R(s)}=\frac{1}{1+G(s)}=\frac{s(s+1)}{s^2+s(1+K_p)+K_i} \)
Also, \(\frac{U(s)}{R(s)}=\frac{(s+1)(sK_p+K_i)}{s^2+s(1+K_p)+K_i}=1\)
\(\Rightarrow \frac{C(s)}{R(s)}=\frac{(sK_p+K_i)}{s^2+s(1+K_p)+K_i} =\frac{1}{s+1}\)
\(\Rightarrow\) For Step input, \(r(t) = u(t) \implies R(s) = \frac{1}{s}\)
\(\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \implies C(s) = \frac{1}{s}  \frac{1}{s+1}\)
\(\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \implies c(t) =1e^{t}\)
Correct Ans: D
Solution:
\(\mathbf{Given \:data:}\)
\(V_L=210\:V\)
Impedance per phase \(Z_{ph}=7\sqrt{3}\:\Omega\)
Power factor \(\cos\phi=\frac{1}{\sqrt{3}}\)
\(V_L=\sqrt{3}V_{ph}\) (For Star connected System)
\(V_{ph}=\frac{V_L}{\sqrt{3}}=\frac{210}{\sqrt{3}}=121.24\:V\)
Current per phase \(I_{ph}=\frac{V_{ph}}{Z_{ph}}=\frac{121.24}{7\times \sqrt{3}}=10\:A\)
Power \(P=3\times V_{ph}\times I_{ph} \times \cos\phi=3\times 121.24\times 10 \times \frac{1}{\sqrt{3}}=2100\:W\)
Correct Ans: A
Solution:
\( u = y^2 \sec{x} \)
\(\implies \frac {du}{dy} = \frac {d}{dy} [y^2 \sec{x}] = 2 y \:\sec \: x\)
Correct Ans: C
Solution:
\(\Rightarrow\) Impedance diagram can be drawn by representing each component is by its equivalent circuit. For example, the synchronous generator at the generating station by a voltage source in series with the resistance and reactance, the transmission line by a nominal \(\pi\) or \(T\) equivalent circuit. The loads are represented by a resistive and inductive reactance in the series, etc.
Correct Ans: B
Solution:
\(\Rightarrow\) From the truth table of the NAND gate it is evident that the output of the NAND gate is high when both the inputs are low.
\(\Rightarrow\) The same possibility exists in NOR and ExNOR gates also.
Correct Ans: D
Solution:
\(\Rightarrow\) Self loop is part of Signal Flow Graph representation.
\(\Rightarrow\) While Block Diagram representation is done with Blocks, Takeoff point, Summing Point.
Correct Ans: D
Solution:
\(\Rightarrow\) For givem characteristic equation,
No. of Poles = P = 4 and No. of Zeros = Z = 1
\(\Rightarrow \) The number of asymptotes of this control system\(=PZ=41=3\)
Correct Ans: A
Solution:
\(\Rightarrow \mathbf{BandLimited\:Interpolation:}\) It is a Interpotation using the impulse response of an ideal low pass filter for the reconstruction of the signal.
Correct Ans: D
Solution:
Given,
\(\rightarrow\) Let two numbers, \(X \: \& \: MX \),
\(\rightarrow X^2 + (MX)^2 = Z \)
For minima,
\(\frac {dZ}{dx} = 2x  2M + 2x = 4x  2M\)
\(\frac {dZ}{dx} = 0 \implies 4x  2M = 0 \implies x = \frac{1}{2}M\)
\(\frac {d^2Z}{dx^2} = 4 > 0\), (\(\because it \: has \: minimum \: value)\)
\(\Rightarrow \) Minima value of the sum of their squares at \(x = \frac{M}{2}\)
\(\implies Z = X^2 + (MX)^2 \) \(= \frac{1}{4} M^2 + \frac{1}{4} M^2 = \frac{1}{2} M^2 \)
Correct Ans: C
Solution:
\(\Rightarrow \)For Simple RC Low Pass Filter, the circuit will be:
\(\Rightarrow\) As per voltagedivider relationship,
\(V_{out} = \frac{jX_c}{RjX_c}\:V_{in}\)
\(\implies \) Transfer Function \(=H(j\omega) =\frac{V_{out}}{V_{in}}= \frac{jX_c}{RjX_c}\)
\(\quad \quad = \frac{\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}\)
\(\quad \quad = \frac{1}{1+j\omega RC}\)
Correct Ans: D
Solution:
\(\mathbf{Delta\: connected\: System:}\)
\(\Rightarrow\) \(V_L=V_{ph}\)
\(\Rightarrow\) \(I_L=\sqrt{3} I_{ph}\)
\(\Rightarrow\) It is a threewire system and does not contain a neutral point.
\(\mathbf{Note:}\)
\(\Rightarrow\) Delta connection is most suitable for rotary convertors.
\(\Rightarrow\) Most of the threephase induction motors are delta connected.
\(\Rightarrow\) Delta connection is most suitable for rotary convertors.
\(\Rightarrow\) Delta connection is highly reliable.
Correct Ans: A
Solution:
\(\Rightarrow\) Type A chopper \(\rightarrow\) First quadrant operation (+ve voltage and +ve current)
\(\Rightarrow\) Type B chopper \(\rightarrow\) Second quadrant operation (+ve voltage and ve current)
\(\Rightarrow\) Type C chopper \(\rightarrow\) First and second quadrant operation (+ve voltage and current in both direction)
\(\Rightarrow\) First and fourth quadrant operation (ve and +ve voltage and +ve current)
\(\Rightarrow\) Type E chopper \(\rightarrow\) Four quadrant operation
Correct Ans: D
Solution:
\(\Rightarrow\) The three most commonly used coordinate systems used in the study of electromagnetics are rectangular coordinates, cylindrical coordinates and spherical coordinates.
Correct Ans: D
Solution:
\(\mathbf{Capacitors\: in \:Series}\)
For all the capacitors connected in series, the charge on all of them is always same and the voltage across them is different.
\(\Rightarrow\) \(Q_1=Q_2=Q_3=Q=It\)
\(\Rightarrow\) \(\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)
\(\mathbf{Capacitors\: in \:Parallel:}\)
For all the capacitors connected in parallel, the charge on all of them is always different and the voltage across them is same.
\(\Rightarrow\) \(Q=Q_1+Q_2+Q_3\)
\(\Rightarrow\) \({C_{eq}}=C_1+C_2+C_3\)
Correct Ans: A
Solution:
\(\Rightarrow\) As per Time Differentiation Property of Fourier Transform,
\(\frac{d}{dt}x(t) \rightarrow j\omega X(\omega)\)
\(\Rightarrow\)Now \(\delta (t) \rightarrow 1 \implies \frac{d}{dt} \delta(t) \rightarrow j\omega \implies \delta^{'}(t) \rightarrow j\omega \)
Correct Ans: B
Solution:
Given,
\( f(x,y,z) =x^2 + 5y^2+5z^24x+40y40z+300 \)
For minimum value,
\(\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 0\)
\(\frac{\partial f}{\partial x} = 2x  4 =0 \)
\(\implies x = 2\)
\(\frac{\partial f}{\partial y} = 10 y + 40 =0\)
\(\implies y = 4\)
\(\frac{\partial f}{\partial z} = 10 z  40 =0 \)
\(\implies z = 4\)
\( f(x,y,z) =x^2 + 5y^2+5z^24x+40y40z+300 \)
\(\therefore f(2, 4, 4) = 2^2 + 5(4)^2+5(4)^24(2)+40(4)  40(4)+300 = 136 \)
Correct Ans: C
Solution:
\(\Rightarrow\) In given schematic diagram, there is a starting winding in which capacitor is connected to split phase.
\(\Rightarrow\) Also there is a centrifugal switch connected in starting winding which disconnect it when motor attain speed.
\(\Rightarrow\) So it is schematic diagram of capacitor start induction motor.
Correct Ans: D
Solution:
\(\Rightarrow\) In Power system network, the components may operate at different voltage and power levels. It will be convenient for analysis of power system if the voltage, power, current, impedance ratings of the components of the power system is expressed with reference to a common value called as base value.
Correct Ans: B
Solution:
\(\frac{C(s)}{R(s)} = \frac{Z_C}{Z_C+Z_R} = \frac{\frac{1}{sC}}{\frac{1}{sC}+R} = \frac{1}{sRC+1}\)
For impulse response, \(R(s)=1 \implies H(s) = \frac{1}{sRC+1}\)
\( \implies h(t) = \frac{1}{RC}e^{\frac{t}{RC}}, t \geq 0\)
Correct Ans: B
Solution:
\(\Rightarrow\) As shown in the figure, a full adder can be formed by two half adders and one OR gate.
Correct Ans: A
Solution:
Equivalent Inductance of series combination of inductors
\(L_{Eq}=L_1+L_2+2M\)
\(\mathbf{Dot \:conversion:}\)
\(\Rightarrow\) Dot conversion is very useful in deciding the sign of the coefficient of mutual inductance in formulating the current equilibrium equations.
\(\Rightarrow\) Two coils are mutually coupled, each coil will induce voltage in the other coils, the magnitude will depend upon the current flowing through the coil and the polarity and its relative direction of flow with respect to dot of the coil.
Correct Ans: B
Solution:
\(\Rightarrow\) Cylindrical coordinates \((\rho, \: \phi, \: z)\) corresponding to Cartesian coordinate \((x, \: y, \: z )\)
$$ \rho = \sqrt{x^2 +y^2} $$
$$ \phi = \tan^{1}\frac{y}{x} $$
$$ z=z $$
\(\Rightarrow\) Cartesian coordinate \((x, \: y, \: z )\) corresponding to Cylindrical coordinates \((\rho, \: \phi, \: z)\)
$$ x=\rho\cos \phi $$
$$ y=\rho \sin \phi $$
$$ z=z $$
Correct Ans: A
Solution:
\(\mathbf{Passive\:Network:}\)
A passive network does not contain any voltage or current source.
\(\mathbf{Active\:Network:}\)
Network that contains an active source is known as active network.
\(\mathbf{Note:}\)
\(\mathbf{Bilateral \:Network:}\)
\(\Rightarrow\) A network, whose properties or characteristics are the same in either direction is known as Bilateral Network.
\(\mathbf{Unilateral \:Network:}\)
\(\Rightarrow\) A Network, whose properties or characteristics are not the same in either direction is known as Unilateral Network.
Correct Ans: B
Solution:
Eddy current losses in a transformer is given by:
\(W_e = K_e\:{B_m}^2\:f^2\:t^2\)
Where
\(K_e =\) Eddy current constant
\(B_m=\) Maximum flux density
\(f=\) Frequency
\(t=\)Thickness of the core
Correct Ans: D
Solution:
\(\mathbf{Given \:data:}\)
An electric heater rating \(P=1\:kW=1000\:W,\:V=200\:V\)
\(P=\frac{V^2}{R}\implies R=\frac{V^2}{P}=\frac{(200)^2}{1000}=40\:\Omega\)
Correct Ans: D
Solution:
\(\Rightarrow\) An inhibit gate is a gate that gives a certain output no matter what is the input given.
Correct Ans: C
Solution:
\(\Rightarrow\) In a DC generator, as compared to shunt field winding, the series field winding consist of fewer turns of wire of large cross sectional area because it carries armature current through it.
\(\Rightarrow\)Field windings are meant to produce mmf (Ampere turns).
\(\Rightarrow\)Because series winding carries armature current which is very high and so it require less number of turns to produce required AT.
\(\Rightarrow\)To carry high armature current series winding should have higher cross sectional area.
Correct Ans: D
Solution:
\(\Rightarrow\) 120° mode of operation has poor utilisation of switches as compare to the 180° conduction mode.
\(\Rightarrow\) It provides six step waveform of phase voltages for deltaconnected load and hence it is preferred for the deltaconnected load.
\(\Rightarrow\) In this mode only two devices are in conduction state at any instance.
Correct Ans: D
Solution:
\(y(s)=\frac{2s}{(s^2a^2)}\)
Taking both side Inverse Laplace transform
\(L^{1}\lbrace y(s)\rbrace=L^{1}\lbrace \frac{2s}{(s^2a^2)}\rbrace\)
\(L^{1}\lbrace y(s)\rbrace=2L^{1}\lbrace \frac{s}{(s^2a^2)}\rbrace\)
\(L^{1}\lbrace y(s)\rbrace=2\cosh\:at\)
Correct Ans: A
Solution:
Correct Ans: D
Solution:
\(\int (\sqrt{1\sin 2x}) \: dx \)
\(=\int (\sqrt{\sin^2 x + \cos ^2 x  2 \sin x \cos x}) dx\)
\(=\int (\sqrt{(\sin x  \cos x )^2} dx \)
\(=\int (\sin x  \cos x ) dx \)
\(=\cos \: x  \sin \: x + C\)
\(=\sin x  \cos x +C\)
Correct Ans: A
Solution:
\(\Rightarrow\) Schering bridge can be used to measure capacitance, dielectric loss, relative permittivity and power factor.
Correct Ans: A
Solution:
\(\Rightarrow\) First of all the given circuit diagram (IC 74LS293) will work as a 4bit ripple counter if \(Q_A\) is connected to inputB.
\(\Rightarrow\) In this ripple counter when \(R_0(1) \:\&\: R_0(2)\) are high the counter will get reset and counting will restart.
\(\Rightarrow\) So when \(R_0(1) \:\&\: R_0(2)\) are connected to \(Q_C\: \&\: Q_D\), it will reset after counting 0 to 11.
\(\Rightarrow\) This is because \((11)_{10} \implies (Q_D =1, Q_C =1, Q_B =0, Q_A =0)_2\).
Correct Ans: D
Solution:
Correct Ans: D
Solution:
\(\Rightarrow\) The electric potential difference between two points P and Q at a distance \(R_p\) and \(R_q\) from the center is:
\(V_{pq}= \frac{Q}{4 \pi \varepsilon_0} \Big [\frac{1}{R_p}  \frac{1}{R_q} \Big] \:\:\:Volt \)
Correct Ans: D
Solution:
\(\Rightarrow\) In lap winding back pitch and front pitch should not be even.
Correct Ans: B
Solution:
\(\Rightarrow\) Kelvin's bridge is used for the measurement of very small resistance \((<1 \Omega)\).
Correct Ans: A
Solution:
Applying KVL to the given circuit
\(12=100i+400i+300i\)
\(12=800i\implies i=\frac{12}{800}=0.015\:A\)
Voltage \(V_2=i\times 400=0.015\times 400=6\:V\)
\(\mathbf{Note:}\)
\(\mathbf{Kirchhoff's \:Voltage \:Law (KVL):}\)
\(\Rightarrow\) KVL law states that the algebraic sum of all branch voltages around any closed loop in a circuit is zero at any instant of time.
$$OR$$
\(\Rightarrow\) KVL law states that the algebraic sum of e.m.f.s around a closed loop equals the algebraic sum of IR drops around the loop.
\(\Rightarrow\) This law represents conversation of energy.
\(\Rightarrow\) Kirchhoff's laws apply to circuits with linear, nonlinear, active, passive, timevarying as well as time invariant elements.
\(\Rightarrow\) Kirchhoff's law is applicable to both AC and DC circuits.
Correct Ans: B
Solution:
Short circuit capacity \(= V_o \times I_{f}\)
Where,
\(V_0\) = Prefault voltage
\(I_f\) = Fault current
Correct Ans: D
Solution:
\(\Rightarrow\) In case of propogation in dielectric, the attenuation factor of the medium is measured in neper/meter or dB/meter.
Correct Ans: B
Solution:
\(1\:Wb=10^8\:maxwells=10^8\:magnetic\:lines\)
\(5\:\mu Wb=5\times 10^{6}\times10^8\:maxwells=500 \:maxwells\)
\(\mathbf{Note:}\)
\(\Rightarrow\) The SI unit of magnetic flux is Weber(Wb).
\(\Rightarrow\) Magnetic flux is denoted by the symbol \(\phi\).
Correct Ans: D
Solution:
\(\Rightarrow\) Torque(T) developed in threephase induction motor is given by:
\(T = \frac{sE_2^2R_2}{\sqrt{R_2^2+sX_2^2}}\times\frac{3}{2\pi n_s}\)