Question 1 |
Roots of the algebraic equation x^{3}+x^{2}+x+1=0 are
(+1, +j, -j) | |
(+1, -1, +1) | |
(0, 0, 0) | |
(-1, +j, -j) |
Question 1 Explanation:
-1 is one of the root since
(-1)^3+(-1)^2+(-1)+1=0
By polynomial division
\frac{x^3+x^2+x+1}{(x-(-1))}=x^2+1 \Rightarrow \;\; x^3+x^2+x+1=(x^2+1)(x+1)
So root are (-1,+j,-j)
(-1)^3+(-1)^2+(-1)+1=0
By polynomial division
\frac{x^3+x^2+x+1}{(x-(-1))}=x^2+1 \Rightarrow \;\; x^3+x^2+x+1=(x^2+1)(x+1)
So root are (-1,+j,-j)
Question 2 |
With K as a constant, the possible solution for the first order differential equation \frac{dy}{dx}=e^{-3x} is
-\frac{1}{3}e^{-3x}+K | |
-\frac{1}{3}e^{3x}+K | |
-\frac{1}{3}e^{-3x}+K | |
-3e^{-x}+K |
Question 2 Explanation:
\begin{aligned} \frac{dy}{dx}&=e^{-3x}\\ \int dy&=\int e^{-3x}dx\\ y&=\frac{e^{-3x}}{-3}+K\\ y&=-\frac{1}{3}e^{-3x}+K \end{aligned}
Question 3 |
The r.m.s value of the current i(t) in the circuit shown below is
\frac{1}{2}A | |
\frac{1}{\sqrt{2}}A | |
1A | |
\sqrt{2}A |
Question 3 Explanation:
V_s=1 \sin t\equiv V_m \sin \omega t
V_m=1 V and \omega =1 rad/sec
Impedance of the branch containing inductor and capacitor
Z=j(X_L-X_C)
\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )
\;\;=j\left ( 1 times 1 -\frac{1}{1 times 1} \right )=0
So, this branch is short circuit and the whole current flow through it
i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t
rms value of the current =\frac{1}{\sqrt{2}}A
V_m=1 V and \omega =1 rad/sec
Impedance of the branch containing inductor and capacitor
Z=j(X_L-X_C)
\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )
\;\;=j\left ( 1 times 1 -\frac{1}{1 times 1} \right )=0
So, this branch is short circuit and the whole current flow through it
i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t
rms value of the current =\frac{1}{\sqrt{2}}A
Question 4 |
The fourier series expansion f(t)=a_{0}+\sum_{n=1}^{\infty }a_{n}cosn\omega t+b_{n}sin n\omega t of the periodic signal shown below will contain the following nonzero terms
a_{0} \; and \; b_{n},n=1,3,5,...\infty | |
a_{0} \; and \; a_{n},n=1,2,3,...\infty | |
a_{0},a_{n} \; and \; b_{n},n=1,3,5,...\infty | |
a_{0} \; and \; a_{n},n=1,3,5,...\infty |
Question 4 Explanation:
Let, x(t)= Even and Hws
Fourier series expansion of x(t) contains cos terms with odd harmonics.
NOw, f(t)=1+x(t)
Fourier series of f(t) contains dc and cos terms with odd harmonics.
Fourier series expansion of x(t) contains cos terms with odd harmonics.
NOw, f(t)=1+x(t)
Fourier series of f(t) contains dc and cos terms with odd harmonics.
Question 5 |
A 4-point starter is used to start and control the speed of a
dc shunt motor with armature resistance control | |
dc shunt motor with field weakening control | |
dc series motor | |
dc compound motor |
Question 5 Explanation:
4-point starter is used to control the speed of shunt motor in field weakening region, i.e. above rates speeds.
In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.
In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.
There are 5 questions to complete.