Question 1 |
The voltage e_o in figure is


2V | |
\frac{4}{3}V | |
4V | |
8V |
Question 1 Explanation:
Applying KCL,
\begin{aligned} \frac{e_{0}-12}{4}+\frac{e_{0}}{4}+\frac{e_{0}}{4}&=0 \\ \Rightarrow \quad 3 e_{0}&=12 \\ \therefore \quad e_{0}&=4 \mathrm{V} \end{aligned}
\begin{aligned} \frac{e_{0}-12}{4}+\frac{e_{0}}{4}+\frac{e_{0}}{4}&=0 \\ \Rightarrow \quad 3 e_{0}&=12 \\ \therefore \quad e_{0}&=4 \mathrm{V} \end{aligned}
Question 2 |
If each branch of a Delta circuit has impedance \sqrt{3}Z, then each branch of the
equivalent Wye circuit has impedance.
\frac{Z}{\sqrt{3}} | |
3Z | |
3\sqrt{3}Z | |
\frac{Z}{3} |
Question 2 Explanation:
\begin{aligned} Z_{\Delta} &=3 Z_{Y} \\ \Rightarrow \quad \sqrt{3} Z_{\Delta} &=3 Z_{Y} \\ Z_{Y} &=\frac{Z_{\Delta}}{\sqrt{3}} \end{aligned}
Question 3 |
The transfer function of a system is given by H(s)=\frac{1}{s^2(s-2)}. The impulse
response of the system is: (* denotes convolution, and U(t) is unit step function)
(t^2 * e^{-2t})U(t) | |
(t * e^{2t})U(t) | |
(te^{-2t})U(t) | |
(te^{-2}t)U(t) |
Question 3 Explanation:
Impulse response of system is L^{-1}[H(s)]
\frac{1}{s^{2}(s-2)}=\frac{1}{s^{2}} \times \frac{1}{s-2}=\left(t * e^{+2 t}\right) u(t)
\frac{1}{s^{2}(s-2)}=\frac{1}{s^{2}} \times \frac{1}{s-2}=\left(t * e^{+2 t}\right) u(t)
Question 4 |
The admittance parameter Y_{12} in the 2-port network in figure is


0.2 mho | |
0.1 mho | |
-0.05 mho | |
0.05 mho |
Question 4 Explanation:
\left[\begin{array}{cc} y_{1}+y_{3} & -y_{3} \\ -y_{3} & y_{2}+y_{3} \end{array}\right]=\left[\begin{array}{cc} y_{11} & y_{12} \\ y_{21} & y_{22} \end{array}\right]
y_{12}=-y_{3}

y_{12}=-\frac{1}{20}=-0.05 \mathrm{mho}
y_{12}=-y_{3}

y_{12}=-\frac{1}{20}=-0.05 \mathrm{mho}
Question 5 |
The region of convergence of the z-transform of a unit step function is
\mid z\mid \gt 1 | |
\mid z\mid \lt 1 | |
(real part of Z) \gt 0 | |
(real part of Z) \lt 0 |
Question 5 Explanation:
\begin{aligned} h(n)&=u(n) \\ H(z)&=\sum_{n=0}^{\infty} 1 . z^{-n} \end{aligned}
For the convergence of H(z)
\sum_{n=0}^{n}\left(z^{-1}\right)^{n}\lt \infty
\therefore ROC is the range of values of z for which
\left|z^{-1}\right| \lt 1\text{ or } |z|\gt |1|
For the convergence of H(z)
\sum_{n=0}^{n}\left(z^{-1}\right)^{n}\lt \infty
\therefore ROC is the range of values of z for which
\left|z^{-1}\right| \lt 1\text{ or } |z|\gt |1|
Question 6 |
The current gain of a BJT is
g_{m}r_{0} | |
\frac{g_{m}}{r_{o}} | |
g_{m}r_{\pi } | |
\frac{g_{m}}{r_{\pi }} |
Question 6 Explanation:
h_{f e}=g_{m} \cdot r_{\pi}
Question 7 |
MOSFET can be used as a
current controlled capacitor | |
voltage controlled capacitor | |
current controlled inductor | |
voltage controlled inductor |
Question 7 Explanation:
Voltage controlled capacitor.
Question 8 |
The effective channel length of a MOSFET in saturation decreases with increase in
gate voltage
| |
drain voltage | |
source voltage | |
body voltage |
Question 8 Explanation:
At the edge of saturation i.e. when drain to source voltage reaches V_{D_{sat}} the inversion layer charge at the drain end becomes zero (ideally). The channel is said to be pinched off at the drain end. If the drain to source voltage V_{DS} is increased even further beyond the saturation edge so that V_{DS} \gt V_{D_{sat}}, an even larger portion of the channel becomes pinched off and effective channel length is reduced.
Question 9 |
The ideal OP-AMP has the following characteristics.
R_{i}=\infty , A=\infty , R_{0}=0 | |
R_{i}=0, A=\infty , R_{0}=0 | |
R_{i}=\infty , A=\infty , R_{0}=\infty | |
R_{i}=0 , A=\infty , R_{0}=\infty |
Question 10 |
The 2's complement representation of -17 is
101110 | |
101111 | |
111110 | |
110001 |
Question 10 Explanation:
\begin{aligned} 17 &=010001 \\ -17 &=101111(2 \text { 's complement }) \end{aligned}
There are 10 questions to complete.