Question 1 |
In the circuit shown, the diodes are ideal, the inductance is small, and I_o \neq 0. Which one of the following statements is true?
D_1 conducts for greater than 180^\circ and D_2 conducts for greater than 180^\circ | |
D_2 conducts for more than 180^\circ and D_1 conducts for 180^\circ | |
D_1 conducts for 180^\circ and D_2 conducts for 180^\circ | |
D_1 conducts for more than 180^\circ and D_2 conducts for 180^\circ |
Question 1 Explanation:
Both diodes will conduct for more than 180^{\circ}.
Question 2 |
For a 3-input logic circuit shown below, the output Z can be expressed as
Q+\bar{R} | |
P\bar{Q}+R | |
\bar{Q}+R | |
P+\bar{Q}+R |
Question 2 Explanation:
Z=\overline{\overline{P\bar{Q}}\cdot Q\cdot \overline{Q\cdot R} }
\;\;=P\bar{Q}+\bar{Q}+Q R
\;\;=\bar{Q}(P+1)+QR
\;\;=\bar{Q}+QR
\;\;=(\bar{Q}+Q)(\bar{Q}+R)
\;\;=\bar{Q}+R
Question 3 |
An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and
discarded without noticing its colour. The probability to get a red ball in the second draw is
\frac{1}{2} | |
\frac{4}{9} | |
\frac{5}{9} | |
\frac{6}{9} |
Question 3 Explanation:
P(red)=\frac{5}{10}\frac{4}{9}+\frac{5}{10}\frac{5}{9}=\frac{45}{90}=\frac{1}{2}
Question 4 |
When a unit ramp input is applied to the unity feedback system having closed loop transfer
function
\frac{C(s)}{R(s)}=\frac{Ks+b}{s^{2}+as+b}(a \gt 0,b \gt 0,K\gt 0),
the steady state error will be
\frac{C(s)}{R(s)}=\frac{Ks+b}{s^{2}+as+b}(a \gt 0,b \gt 0,K\gt 0),
the steady state error will be
0 | |
\frac{a}{b} | |
\frac{a+K}{b} | |
\frac{a-K}{b} |
Question 4 Explanation:
Closed loop transfer function =\frac{Ks+b}{s^2+as+b}
Open loop transfer function = G(s)=\frac{Ks+b}{s^2+as+b-Ks-b}
G(s)=\frac{Ks+b}{s^2+as-Ks} =\frac{Ks+b}{s(s+a-K)}
Steady state error for ramp input given to type-1 system =1/K_V
where, velocity error coefficient,
K_V=\lim_{s \to 0}s\cdot \frac{Ks+b}{s(s+a-K)} =\frac{b}{a-K}
Steady state error,
e_{ss}=\frac{a-K}{b}
Open loop transfer function = G(s)=\frac{Ks+b}{s^2+as+b-Ks-b}
G(s)=\frac{Ks+b}{s^2+as-Ks} =\frac{Ks+b}{s(s+a-K)}
Steady state error for ramp input given to type-1 system =1/K_V
where, velocity error coefficient,
K_V=\lim_{s \to 0}s\cdot \frac{Ks+b}{s(s+a-K)} =\frac{b}{a-K}
Steady state error,
e_{ss}=\frac{a-K}{b}
Question 5 |
A three-phase voltage source inverter with ideal devices operating in 180^{\circ} conduction mode is
feeding a balanced star-connected resistive load. The DC voltage input is V_{dc}. The peak of the
fundamental component of the phase voltage is
\frac{V_{dc}}{\pi} | |
\frac{2V_{dc}}{\pi} | |
\frac{3V_{dc}}{\pi} | |
\frac{4V_{dc}}{\pi} |
Question 5 Explanation:
3-\phi \;\text{ VSI }180^{\circ} \text{ mode:}
\begin{aligned} V_{Rn}&=\frac{6 V_{dc}/3}{n \pi} \sin n\omega t\\ &= \frac{2 \times V_{dc}}{n \pi} \sin n\omega t \\ V_{Rn} &= \frac{2 V_{dc}}{ \pi} \sin \omega t \end{aligned}
\begin{aligned} V_{Rn}&=\frac{6 V_{dc}/3}{n \pi} \sin n\omega t\\ &= \frac{2 \times V_{dc}}{n \pi} \sin n\omega t \\ V_{Rn} &= \frac{2 V_{dc}}{ \pi} \sin \omega t \end{aligned}
Question 6 |
The figures show diagrammatic representations of vector fields \vec{X},\vec{Y} and \vec{Z} respectively.
Which one of the following choices is true?
\bigtriangledown \cdot \vec{X}=0, \bigtriangledown \times \vec{Y}\neq 0,\bigtriangledown \times \vec{Z}=0 | |
\bigtriangledown \cdot \vec{X} \neq 0, \bigtriangledown \times \vec{Y}=0, \bigtriangledown \times \vec{Z} \neq 0 | |
\bigtriangledown \cdot \vec{X}\neq 0, \bigtriangledown \times \vec{Y}\neq 0, \bigtriangledown \times \vec{Z}\neq 0 | |
\bigtriangledown \cdot \vec{X}=0, \bigtriangledown \times \vec{Y}= 0, \bigtriangledown \times \vec{Z}=0 |
Question 6 Explanation:
\vec{X} is going away so \vec{\triangledown } \cdot \vec{X}\neq 0
\vec{Y} is moving circulator direction so \vec{\triangledown } \cdot \vec{Y}\neq 0
\vec{Z} has circular rotation so \vec{\triangledown } \cdot \vec{Z}\neq 0
\vec{Y} is moving circulator direction so \vec{\triangledown } \cdot \vec{Y}\neq 0
\vec{Z} has circular rotation so \vec{\triangledown } \cdot \vec{Z}\neq 0
Question 7 |
Assume that in a traffic junction, the cycle of the traffic signal lights is 2 minutes of green
(vehicle does not stop) and 3 minutes of red (vehicle stops). Consider that the arrival time of
vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time
(in minutes) for the vehicle at the junction is ________.
0.4 | |
0.9 | |
1.5 | |
2.6 |
Question 7 Explanation:
t be arrival time of vehicles of the junction is uniformaly distributed in [0,5].
Let y be the waiting time of the junction.
\begin{aligned} \text{Then }y&=\left\{\begin{matrix} 0 & t \lt 2 \\ 5-t & 2\leq t \lt 5 \end{matrix}\right.\\ y\rightarrow &[0,5]\\ f(y)&=\frac{1}{5-0}=\frac{1}{5}\\ E(y)&=\int_{-\infty }^{0}y(y)dy=\int_{0}^{5}yf(y)dy\\ &=\int_{2}^{5}y\left ( \frac{1}{5} \right )dy=\frac{1}{5}\int_{2}^{5}(5-t)dt\\ &=\frac{1}{5}\left ( 5t-\frac{t^2}{2} \right )|_2^5\\ &=\frac{1}{5}\left [ \left ( 25-\frac{25}{2} \right )-\left ( 10-\frac{4}{2} \right ) \right ]\\ &=\frac{1}{5}\left ( \frac{25}{2}-8 \right )=\frac{1}{5}\frac{9}{2}=0.9 \end{aligned}
Let y be the waiting time of the junction.
\begin{aligned} \text{Then }y&=\left\{\begin{matrix} 0 & t \lt 2 \\ 5-t & 2\leq t \lt 5 \end{matrix}\right.\\ y\rightarrow &[0,5]\\ f(y)&=\frac{1}{5-0}=\frac{1}{5}\\ E(y)&=\int_{-\infty }^{0}y(y)dy=\int_{0}^{5}yf(y)dy\\ &=\int_{2}^{5}y\left ( \frac{1}{5} \right )dy=\frac{1}{5}\int_{2}^{5}(5-t)dt\\ &=\frac{1}{5}\left ( 5t-\frac{t^2}{2} \right )|_2^5\\ &=\frac{1}{5}\left [ \left ( 25-\frac{25}{2} \right )-\left ( 10-\frac{4}{2} \right ) \right ]\\ &=\frac{1}{5}\left ( \frac{25}{2}-8 \right )=\frac{1}{5}\frac{9}{2}=0.9 \end{aligned}
Question 8 |
Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million
electrons are added to this sphere, these electrons will be distributed.
uniformly over the entire volume of the sphere | |
uniformly over the outer surface of the sphere | |
concentrated around the centre of the sphere | |
along a straight line passing through the centre of the sphere |
Question 8 Explanation:
Added charge (one million electrons) to be solid spherical conductor is uniformly distributed over the outer surface of the sphere.
Question 9 |
The transfer function C(s) of a compensator is given below.
C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
0.1 \lt \omega \lt 1 | |
1 \lt \omega \lt 10 | |
10 \lt \omega \lt 100 | |
\omega \gt 100 |
Question 9 Explanation:
Pole zero diagram of compensator transfer function is shown below.
Maximum phase lead is between 0.1 and 1.
0.1 \lt \omega \lt 1
Maximum phase lead is between 0.1 and 1.
0.1 \lt \omega \lt 1
Question 10 |
The figure show the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where \alpha is a complex number with non-zero real and imaginary parts.
For the given circuit, Y_{bus}\; and \; Z_{bus} are bus admittance matrix and bus impedance matrix, respectively, each of size 2x2. Which one of the following statements is true?
For the given circuit, Y_{bus}\; and \; Z_{bus} are bus admittance matrix and bus impedance matrix, respectively, each of size 2x2. Which one of the following statements is true?
Both Y_{bus}\; and \; Z_{bus} are symmetric | |
Y_{bus} is symmetric and bus Z_{bus} is unsymmetric | |
Y_{bus} is unsymmetric and Z_{bus} is symmetric | |
Both Y_{bus}\; and \; Z_{bus} are unsymmetric |
Question 10 Explanation:
Both Y_{BUS} and Z_{BUS} are unsymmetrical with transformer.
There are 10 questions to complete.