Question 1 |
If E denotes expectation, the variance of a random variable X is given by
E[X^{2}]-E^{2}[X] | |
E[X^{2}]+E^{2}[X] | |
E[X^{2}] | |
E^{2}[X] |
Question 1 Explanation:
\begin{array}{l} \sigma_{X}^{2}=E\left[X^{2}\right]-E^{2}[X] \\ \text { AC. Power }=\text { Total power - DC power } \end{array}
Question 2 |
The following plot shows a function which varies linearly with x. The value of
the integral I=\int_{1}^{2} ydx is
1 | |
2.5 | |
4 | |
5 |
Question 2 Explanation:
\begin{aligned} y &=x+1 \\ I &=\int_{1}^{2} y d x \\ &=\int_{1}^{2}(x+1) d x=\left.\frac{(x+1)^{2}}{2}\right|_{1} \\ &=\frac{1}{2}(9-4)=2.5 \end{aligned}
Question 3 |
For |x| \lt \lt 1, \; coth (x) can be approximated as
x | |
x^{2} | |
\frac{1}{x} | |
\frac{1}{x^{2}} |
Question 3 Explanation:
\cot h x=\frac{\cos h x}{\sin h x}=\frac{1}{x}
Question 4 |
\lim_{\theta \rightarrow 0}\frac{sin(\frac{\theta }{2})}{\theta } is
0.5 | |
1 | |
2 | |
not defined |
Question 4 Explanation:
\lim _{\theta \rightarrow 0} \frac{\frac{1}{2} \times \sin \left(\frac{\theta}{2}\right)}{\theta \times \frac{1}{2}}=\frac{1}{2} \lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}}=\frac{1}{2}=0.5
Question 5 |
Which one of following functions is strictly bounded?
1/x^{2} | |
e^{x} | |
x^{2} | |
e^{-x^{2}} |
Question 5 Explanation:
y=\frac{1}{x^{2}}
y=\theta^{x}
y=x^{2}
y=e^{-x^{2}}
Question 6 |
For the function e^{-x}, the linear approximation around x=2 is
(3-x)e^{-2} | |
1-x | |
[3+2\sqrt{2}-(1+\sqrt{2})x]e^{2} | |
e^{-2} |
Question 7 |
An independent voltage source in series with an impedance Z_{s}=R_{s}+jX_{s}, delivers a maximum average power to a load impedance Z_{L} when
Z_{L}=R_{s}+jX_{s} | |
Z_{L}=R_{s} | |
Z_{L}=jX_{s} | |
Z_{L}=R_{s}-jX_{s} |
Question 7 Explanation:
Z_{L}=R_{s}-j X_{s}
For maximum power transfer
\begin{array}{l} Z_{L}=Z_{S}^{*} \\ Z_{L}=R_{s}-j X_{s} \end{array}
For maximum power transfer
\begin{array}{l} Z_{L}=Z_{S}^{*} \\ Z_{L}=R_{s}-j X_{s} \end{array}
Question 8 |
The RC circuit shown in the figure is
a low-pass filter | |
a high-pass filter | |
a band-pass filter | |
a band-reject filter |
Question 8 Explanation:
At \omega \rightarrow \infty , Capacitor \rightarrow short circuited
Circuit looks like,
at\omega \rightarrow 0, Capacitor \rightarrow open circuited
Circuit looks like
So frequency response of the circuit will be,
So the circuit is Band pass filter
Circuit looks like,
at\omega \rightarrow 0, Capacitor \rightarrow open circuited
Circuit looks like
So frequency response of the circuit will be,
So the circuit is Band pass filter
Question 9 |
The electron and hole concentrations in an intrinsic semiconductor are n_{i} per cm^{3} at 300 K. Now, if acceptor impurities are introduced with a concentration
of N_{A} per cm^{3} (where N_{A}\gt \gt n_{i} ), the electron concentration per cm^{3} at 300 K
will be
n_{i} | |
n_{i}+N_{A} | |
N_{A}-n_{i} | |
\frac{{n_{i}}^2}{N_{A}} |
Question 9 Explanation:
By the law of electrical neutrality
\begin{aligned} p+N_{0} &=n+N_{A} \\ \text{as}\quad\quad N_{0} &=0\\ N_{A}&>>n_{i} \cong 0 \quad p=N_{A}\\ \end{aligned}
using mass action law \mathrm{np}=n_{i}^{2}
So. \quad n=\frac{n_{i}^{2}}{p}=\frac{n_i^{2}}{N_{A}}
\begin{aligned} p+N_{0} &=n+N_{A} \\ \text{as}\quad\quad N_{0} &=0\\ N_{A}&>>n_{i} \cong 0 \quad p=N_{A}\\ \end{aligned}
using mass action law \mathrm{np}=n_{i}^{2}
So. \quad n=\frac{n_{i}^{2}}{p}=\frac{n_i^{2}}{N_{A}}
Question 10 |
In a p^{+}n junction diode under reverse biased the magnitude of electric field is
maximum at
the edge of the depletion region on the p-side | |
the edge of the depletion region on the n-side | |
the p^{+}n junction | |
the centre of the depletion region on the n-side |
Question 10 Explanation:
Electrical field is always maximum at the junction.
There are 10 questions to complete.