EC GATE 2010 (Test 6)



Tag: ec gate 2010
Q.1
For the two-port network shown below, the short-circuit admittance parameter matrix is

A.
B.
C.
D.
Answer : Option A
Explaination / Solution:

Given circuit is as shown below

By writing node equation at input port

By writing node equation at output port

From (1) and (2), we have admittance matrix


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Q.2
For parallel RLC circuit, which one of the following statements is NOT correct ?
A. The bandwidth of the circuit decreases if R is increased
B. The bandwidth of the circuit remains same if L is increased
C. At resonance, input impedance is a real quantity
D. At resonance, the magnitude of input impedance attains its minimum values.
Answer : Option D
Explaination / Solution:

A parallel RLC circuit is shown below :
Input impedance 


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Q.3
In the circuit shown, the switch S is open for a long time and is closed at t = 0. The current i (t) for t  0+is

A. i (t) = 0.5 - 0.125e-1000tA
B. i (t) = 1.5 - 0.125e-1000tA
C. i (t) = 0.5 - 0.5e-1000tA
D. i (t) = 0.375e-1000tA
Answer : Option A
Explaination / Solution:


When the switch S is open for a long time before t < 0, the circuit is

At t = 0, inductor current does not change simultaneously, So the circuit is

Current is resistor (AB)
i(0) = 0.75/2 = 0.375 A
Similarly for steady state the circuit is as shown below


B = 0.375 - 0.5 =- 0.125


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Q.4
The current I in the circuit shown is

A. -j1A
B. j1A
C. 0 A
D. 20 A
Answer : Option A
Explaination / Solution:



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Q.5
In the circuit shown, the power supplied by the voltage source is

A. 0 W
B. 5 W
C. 10 W
D. 100 W
Answer : Option A
Explaination / Solution:

Applying nodal analysis

Current from voltage source is

Since current through voltage source is zero, therefore power delivered is zero.

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Q.6
The eigen values of a skew-symmetric matrix are
A. always zero
B. always pure imaginary
C. either zero or pure imaginary
D. always real
Answer : Option C
Explaination / Solution:

Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in conjugate pairs.

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Q.7
The trigonometric Fourier series for the waveform f (t) shown below contains

A. only cosine terms and zero values for the dc components
B. only cosine terms and a positive value for the dc components
C. only cosine terms and a negative value for the dc components
D. only sine terms and a negative value for the dc components
Answer : Option C
Explaination / Solution:

For a function x(t) trigonometric fourier series is

For an even function x(t),Bn = 0
Since given function is even function so coefficient Bn = 0, only cosine and constant
terms are present in its fourier series representation.
Constant term :


Constant term is negative.


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Q.8
A function n(x) satisfied the differential equation where L is a constant. The boundary conditions are :n(0) = K and n() = 0. The solution to this equation is
A. n(x) = Kexp(x/L)
B. n(x) = Kexp(- x/ L)
C. n(x) = K2 exp(- x/L)
D. n(x) = Kexp(- x/L)
Answer : Option D
Explaination / Solution:

Given differential equation


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Q.9
Consider the z -transform  The inverse z -transform x[n] is
A. 5𝛿[n + 2] + 3𝛿[n] + 4𝛿[n - 1]
B. 5𝛿[n - 2] + 3𝛿[n] + 4𝛿[n + 1]
C. 5u[n + 2] + 3u[n] + 4u[n - 1]
D. 5u[n - 2] + 3u[n] + 4u[n + 1]
Answer : Option A
Explaination / Solution:



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Q.10
If ey = x1/x , then y has a
A. maximum at x = e
B. minimum at x = e
C. maximum at x = e-1
D. minimum at x = e-1
Answer : Option A
Explaination / Solution:



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