Question 1 |
In the circuit shown below what is the output voltage (V_{out}) if a silicon transistor Q and an ideal op-amp are used?
-15 V | |
-0.7 V | |
+0.7 V | |
+15 V |
Question 1 Explanation:
Using the concept of vitual ground, V=0
V_{out}=-0.7V
Question 2 |
The transfer function \frac{V_{2}(s)}{V_{1}(s)} of the circuit shown below is
\frac{0.5s+1}{s+1} | |
\frac{3s+6}{s+2} | |
\frac{s+2}{s+1} | |
\frac{s+1}{s+2} |
Question 2 Explanation:
\frac{V_2(s)}{V_1(s)}=\frac{R+\frac{1}{Cs}}{\frac{1}{Cs}+R+\frac{1}{Cs}}
=\frac{1+RCs}{2+RCs}
=\frac{1+10 \times 10^3 \times 100 \times 10^{-6}s}{2+10 \times 10^3 \times 100 \times 10^{-6}s}
=\frac{s+1}{s+2}
Question 3 |
Assuming zero initial condition, the response y(t) of the system given below to a
unit step input u(t) is
u(t) | |
tu(t) | |
\frac{t^{2}}{2}u(t) | |
e^{-t}u(t) |
Question 3 Explanation:
\begin{aligned} Y(s)&=\frac{1}{s}U(s)=\frac{1}{s^2}\\ y(t)&=tu(t) \end{aligned}
Question 4 |
The impulse response of a system is h(t)=tu(t). For an input u(t-1), the output is
\frac{t^{2}}{2}u(t) | |
\frac{t(t-1)}{2}u(t-1) | |
\frac{(t-1)^{2}}{2}u(t-1) | |
\frac{t^{2}-1}{2}u(t-1) |
Question 4 Explanation:
\begin{aligned} h(t) &=tu(t) \\ H(s) &=\frac{1}{s^2} \\ \Rightarrow \; \frac{Y(s)}{U(s)} &=\frac{1}{s^2} \\ Y(s) &=\frac{1}{s^2} \frac{e^{-s}}{s}\\ \Rightarrow \; y(t) &=\frac{(t-1)^2}{2}u(t-1) \end{aligned}
Question 5 |
Which one of the following statements is NOT TRUE for a continuous time
causal and stable LTI system?
All the poles of the system must lie on the left side of the j \omega axis | |
Zeros of the system can lie anywhere in the s-plane | |
All the poles must lie within |s| = 1 | |
All the roots of the characteristic equation must be located on the left side
of the j \omega axis. |
Question 5 Explanation:
All poles must lie within |Z|=1
Question 6 |
Two systems with impulse responses h_{1}(t) \; and \; h_{2}(t) are connected in cascade. Then the overall impulse response of the cascaded system is given by
product of h_{1}(t) \; and \; h_{2}(t) | |
sum of h_{1}(t) \; and \; h_{2}(t) | |
convolution of h_{1}(t) \; and \; h_{2}(t) | |
subtraction of h_{1}(t) \; and \; h_{2}(t) |
Question 6 Explanation:
In cascade connection,
\begin{aligned} H(s) &=H_1(s)\cdot H_2(s)\\ \Rightarrow \; h(t)&=h_1(t) * h_2(t) \end{aligned}
\begin{aligned} H(s) &=H_1(s)\cdot H_2(s)\\ \Rightarrow \; h(t)&=h_1(t) * h_2(t) \end{aligned}
Question 7 |
A source v_{s}(t)=V \cos 100\pi t has an internal impedance of (4+j3)\Omega. If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in \Omega should be
3 | |
4 | |
5 | |
7 |
Question 7 Explanation:
Using maximum power transfer theorem,
R_L=|Z|=|4-j3|
\;\;=\sqrt{4^2+3^2}=5\Omega
R_L=|Z|=|4-j3|
\;\;=\sqrt{4^2+3^2}=5\Omega
Question 8 |
A single-phase load is supplied by a single-phase voltage source. If the current
flowing from the load to the source is 10\angle -150^{\circ}A and if the voltage at the load terminal is 100\angle 60^{\circ}V, then the
load absorbs real power and delivers reactive power | |
load absorbs real power and absorbs reactive power | |
load delivers real power and delivers reactive power | |
load delivers real power and absorbs reactive power |
Question 8 Explanation:
Complex power supplied by load =(100\angle 60^{\circ})(10\angle -150^{\circ})=-866.022-j500 VA
As supplied active and reactive power are negative.
Load absorbs active and reactive power both.
Question 9 |
A single-phase transformer has no-load loss of 64W, as obtained from an open
circuit test. When a short-circuit test is performed on it with 90% of the rated
currents flowing in its both LV and HV windings, he measured loss is 81 W. The
transformer has maximum efficiency when operated at
50.0% of the rated current | |
64.0% of the rated current | |
80.0% of the rated current | |
88.8% of the rated current |
Question 9 Explanation:
For 90% current,
\begin{aligned} P_{cu} &=(0.9)^2 P_{fl\;cu} \\ P_{fl\;cu} &= \frac{81}{0.81}=100W\\ x&=\sqrt{\frac{P_{core}}{P_{fl\;cu}}}=\sqrt{\frac{64}{100}}=0.8=80\% \end{aligned}
\begin{aligned} P_{cu} &=(0.9)^2 P_{fl\;cu} \\ P_{fl\;cu} &= \frac{81}{0.81}=100W\\ x&=\sqrt{\frac{P_{core}}{P_{fl\;cu}}}=\sqrt{\frac{64}{100}}=0.8=80\% \end{aligned}
Question 10 |
The flux density at a point in space is given by B=4xa_{x}+2kya_{y}+8a_{z} Wb/m^{2}. The value of constant k must be equal to
-2 | |
-0.5 | |
0.5 | |
2 |
Question 10 Explanation:
\begin{aligned} \bigtriangledown \cdot B &=0 \\ \left ( \frac{\partial }{\partial x}a_x +\frac{\partial }{\partial y}a_y + \frac{\partial }{\partial z}a_z\right )&(4xa_x+2kya_y+8a_z)=0 \\ 4+2k&=0\\ k&=-2 \end{aligned}
There are 10 questions to complete.