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|
There are 5 questions to complete.