Subject | Number of questions |
---|---|
POWER SYSTEM | 45 |
ELECTRICAL MACHINES | 22 |
ELECTRICAL ENERGY CONSERVATION AND AUDITING | 15 |
BASICS OF COMPUTER | 8 |
ELECTRICAL CIRCUIT ANALYSIS | 8 |
MEASUREMENT & INSTRUMENTATION | 2 |
Total | 100 |
Correct Ans: A
Correct Ans: C
Correct Ans: B
Correct Ans: A & B
Correct Ans: A
Correct Ans: D
Correct Ans: D
Correct Ans: C
Correct Ans: B
Correct Ans: D
Correct Ans: A
Correct Ans: D
Correct Ans: C
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: D
Correct Ans: C
Correct Ans: A
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: D
Correct Ans: A
Correct Ans: D
Correct Ans: D
Correct Ans: C
Correct Ans: D
Correct Ans: C
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: D
Correct Ans: C
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: C
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: C
Correct Ans: C
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: D
Correct Ans: D
Correct Ans: A
Correct Ans: B
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: D
Correct Ans: B
Correct Ans: A
Correct Ans: C
Correct Ans: C
Correct Ans: D
Correct Ans: D
Correct Ans: A
Correct Ans: D
Correct Ans: D
Correct Ans: D
Correct Ans: C
Correct Ans: D
Correct Ans: D
Correct Ans: A
Correct Ans: D
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: A
Subject | Number of questions |
---|---|
ELECTRICAL CIRCUIT ANALYSIS | 12 |
ENGLISH | 10 |
POWER SYSTEM | 65 |
MEASUREMENT & INSTRUMENTATION | 1 |
ELECTRICAL ENGINEERING MATERIALS | 1 |
UTILIZATION OF ELECTRICAL ENERGY | 1 |
GENERAL KNOWLEDGE | 10 |
Total | 100 |
Correct Ans: C
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: D
Correct Ans: A
Correct Ans: C
Correct Ans: A
Correct Ans: A
Correct Ans: B
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: D
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Correct Ans: A
Subject | Number of questions |
---|---|
BASICS OF COMPUTER | 10 |
ELECTRICAL CIRCUIT ANALYSIS | 13 |
CONTROL SYSTEM ENGINEERING | 7 |
ELECTRICAL MACHINES | 9 |
POWER ELECTRONICS & DRIVES | 4 |
ENGINEERING MATHEMATICS | 8 |
MEASUREMENT & INSTRUMENTATION | 3 |
DIGITAL ELECTRONICS | 3 |
POWER SYSTEM | 6 |
ELECTROMAGNETICS | 3 |
BASIC ELECTRONICS | 2 |
SIGNALS & SYSTEMS | 2 |
ENGLISH | 10 |
GENERAL KNOWLEDGE | 10 |
GUJARATI | 10 |
Total | 100 |
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
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: 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: 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: 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
Subject | Number of questions |
---|---|
GENERAL KNOWLEDGE | 10 |
GUJARATI | 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 |
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 |
---|---|
BASICS OF COMPUTER | 10 |
ENGINEERING MATHEMATICS | 6 |
ELECTRICAL CIRCUIT ANALYSIS | 9 |
ELECTRICAL MACHINES | 10 |
DIGITAL ELECTRONICS | 2 |
POWER SYSTEM | 4 |
ELECTROMAGNETICS | 6 |
MEASUREMENT & INSTRUMENTATION | 4 |
SIGNALS & SYSTEMS | 4 |
CONTROL SYSTEM ENGINEERING | 4 |
POWER ELECTRONICS & DRIVES | 8 |
BASIC ELECTRONICS | 3 |
ENGLISH | 10 |
GENERAL KNOWLEDGE | 10 |
GUJARATI | 10 |
Total | 100 |
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:
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 push-pull amplifier increases the output voltage and current delivering capability of the operational amplifier.
Correct Ans: A
Solution:
\(\mathbf{Given \:data:}\)
\(W_1=W_2=600\:W\)
\(\mathbf{ Measurement\:of\:Power\:and\:Power\:factor\:by\:Two\:Wattmeter\:Method\:in\:a\:balanced\:three\:phase\:load}\)
\(Wattmeter\: reading\: W_1=V_LI_L\cos(30^\circ-\phi)....(1)\)
\(Wattmeter\: reading \:W_2=V_LI_L\cos(30^\circ+\phi)....(2)\)
\(Using\: equation\:(1)\:and\:(2)\)
\(Total\:power\:absorbed\:in\:a\:three\:phase\:W=W_1+W_2=\sqrt3V_LI_L\cos\phi....(3)\)
\(W_1-W_2=V_LI_L\sin\phi....(4)\)
\(Using\: equation\:(3)\:and\:(4)\)
\(\frac{W_1-W_2}{W_1+W_2}=\frac{V_LI_L\sin\phi}{\sqrt3V_LI_L\cos\phi}\)
\(\tan\phi=\sqrt3[\frac{W_1-W_2}{W_1+W_2}]....(5)\)
\(\phi=\tan^{-1}[\frac{\sqrt3(W_1-W_2)}{W_1+W_2}]....(6)\)
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 3-phase 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 three-phase machine gives more output compared to a single-phase machine of the same size.
\(\mathbf{4.}\) A three-phase 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 Skew-symmetric matrix is \(A^T = -A\)
For matrix \(A =\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}\)
\(\implies A^T = \begin{bmatrix}0 & -5 \\5 & 0 \end{bmatrix} = (-1) \begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix} = -A \)
\(\therefore Skew-symmetric \: matrix
=\begin{bmatrix}0 & -5 \\5 & 0 \end{bmatrix}\)
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 cross-section 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 \(2^n-1\) counts.
\(\Rightarrow\) Here, the counter needs to count 3000. \(\quad \therefore 2^n-1 \geq 3000\).
\(\quad \therefore 2^n \geq 3001 \implies n = \frac{\log 3001}{\log 2} \approx 12.\)
Correct Ans: C
Solution:
Types of Shunt type of faults:
\(\rightarrow\) Line to Ground fault (L-G)
\(\rightarrow\) Line to Line fault (L-L)
\(\rightarrow\) Double Line to Ground fault (L-L-G)
\(\rightarrow\) 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 half-power points.
\(\Rightarrow\) \(f_1\) - Lower cut-off frequency (Lower half-power frequency)
\(\Rightarrow\) \(f_2\) - Upper cut-off frequency (Upper half-power frequency)
\(\Rightarrow\) \(f_r\) - Resonance frequency is the Geometric mean of the two half power frequencies.
\(\Rightarrow\) Bandwidth (B.W) \(=f_2-f_1\)
\(\Rightarrow\) 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 voltage-divider relationship,
\(V_{out} = \frac{R}{R-jX_c}\:V_{in}\)
\(\implies \) Transfer Function \(=H(j\omega) =\frac{V_{out}}{V_{in}}= \frac{R}{R-jX_c}\)
\(\quad \quad = \frac{R}{R+\frac{1}{j\omega C}}\)
\(\quad \quad = \frac{j\omega RC}{1+j\omega RC}\)
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 non-touching loops] + [Sum of gain products of all possible three non-touching 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,
\(\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}} \)
\(\Rightarrow\) \(\sqrt{1-x^2} \: dy = \sqrt{1-y^2} \: dx\)
\(\Rightarrow\) \(\frac{dy}{\sqrt{1-y^2}} = \frac{dx}{\sqrt{1-x^2}} \)
Apply Integration both side,
\(\Rightarrow\) \(\int \frac{dy}{\sqrt{1-y^2}} = \int \frac{dx}{\sqrt{1-x^2}} \)
\(\implies \sin^{-1} y = \sin^{-1} x + C\)
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 \(0-100\:V\) 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.