# 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. 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 Workspace
Report
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.
Explaination / Solution:

A parallel RLC circuit is shown below : Input impedance Workspace
Report
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
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

Workspace
Report
Q.4
The current I in the circuit shown is A. -j1A
B. j1A
C. 0 A
D. 20 A
Explaination / Solution: Workspace
Report
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
Explaination / Solution:

Applying nodal analysis Current from voltage source is Since current through voltage source is zero, therefore power delivered is zero.

Workspace
Report
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
Explaination / Solution:

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

Workspace
Report
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
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.

Workspace
Report
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)
Explaination / Solution:

Given differential equation Workspace
Report
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]
Explaination / Solution: Workspace
Report
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 