GATE EE 2013

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?
A
-15 V
B
-0.7 V
C
+0.7 V
D
+15 V
Analog Electronics   Operational Amplifiers
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
A
\frac{0.5s+1}{s+1}
B
\frac{3s+6}{s+2}
C
\frac{s+2}{s+1}
D
\frac{s+1}{s+2}
Control Systems   Mathematical Models of Physical Systems
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
A
u(t)
B
tu(t)
C
\frac{t^{2}}{2}u(t)
D
e^{-t}u(t)
Signals and Systems   Laplace Transform
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
A
\frac{t^{2}}{2}u(t)
B
\frac{t(t-1)}{2}u(t-1)
C
\frac{(t-1)^{2}}{2}u(t-1)
D
\frac{t^{2}-1}{2}u(t-1)
Signals and Systems   Linear Time Invariant Systems
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?
A
All the poles of the system must lie on the left side of the j \omega axis
B
Zeros of the system can lie anywhere in the s-plane
C
All the poles must lie within |s| = 1
D
All the roots of the characteristic equation must be located on the left side of the j \omega axis.
Signals and Systems   Laplace Transform
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
A
product of h_{1}(t) \; and \; h_{2}(t)
B
sum of h_{1}(t) \; and \; h_{2}(t)
C
convolution of h_{1}(t) \; and \; h_{2}(t)
D
subtraction of h_{1}(t) \; and \; h_{2}(t)
Signals and Systems   Linear Time Invariant Systems
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}
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
A
3
B
4
C
5
D
7
Electric Circuits   Network Theorems
Question 7 Explanation: 
Using maximum power transfer theorem,
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
A
load absorbs real power and delivers reactive power
B
load absorbs real power and absorbs reactive power
C
load delivers real power and delivers reactive power
D
load delivers real power and absorbs reactive power
Power Systems   Load Flow Studies
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
A
50.0% of the rated current
B
64.0% of the rated current
C
80.0% of the rated current
D
88.8% of the rated current
Electrical Machines   Transformers
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}
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
A
-2
B
-0.5
C
0.5
D
2
Electromagnetic Fields   Magnetostatic Fields
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.
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