# 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,
$K_V=\lim_{s \to 0}s\cdot \frac{Ks+b}{s(s+a-K)} =\frac{b}{a-K}$
$e_{ss}=\frac{a-K}{b}$
 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.