GATE EE 2017 SET 2


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?
A
D_1 conducts for greater than 180^\circ and D_2 conducts for greater than 180^\circ
B
D_2 conducts for more than 180^\circ and D_1 conducts for 180^\circ
C
D_1 conducts for 180^\circ and D_2 conducts for 180^\circ
D
D_1 conducts for more than 180^\circ and D_2 conducts for 180^\circ
Power Electronics   Phase Controlled Rectifiers
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
A
Q+\bar{R}
B
P\bar{Q}+R
C
\bar{Q}+R
D
P+\bar{Q}+R
Digital Electronics   Logic Gates
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
A
\frac{1}{2}
B
\frac{4}{9}
C
\frac{5}{9}
D
\frac{6}{9}
Engineering Mathematics   Probability and Statistics
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
A
0
B
\frac{a}{b}
C
\frac{a+K}{b}
D
\frac{a-K}{b}
Control Systems   Time Response Analysis
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}
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
A
\frac{V_{dc}}{\pi}
B
\frac{2V_{dc}}{\pi}
C
\frac{3V_{dc}}{\pi}
D
\frac{4V_{dc}}{\pi}
Power Electronics   Inverters
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}




There are 5 questions to complete.