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}
There are 5 questions to complete.