Question 1 |

Given a system of equations

x + 2y + 2z = b_1

5x + y + 3z = b_2

What of the following is true regarding its solutions

x + 2y + 2z = b_1

5x + y + 3z = b_2

What of the following is true regarding its solutions

The system has a unique solution for any given b_1 \; and \; b_2 | |

The system will have infinitely many solutions for any given b_1 \; and \; b_2 | |

Whether or not a solution exists depends on the given b_1 \; and \; b_2 | |

The systems would have no solution for any values of b_1 \; and \; b_2 |

Question 2 |

Let f(x)=xe^{-x}. The maximum value of the function in the interval (0,\infty ) is

e^{-1} | |

e | |

1-e^{-1} | |

1+e^{-1} |

Question 2 Explanation:

\begin{aligned} f(x)&=xe^{-x}\\ f'(x)&=e^{-x}-xe^{-x}=0\\ e^{-x}(1-x)&=0\\ x&=1\\ f''(x)&=-e^{-x}-e^{-x}+xe^{-x}\\ &=e^{-x}(x-2)\\ f''(1)&=e^{-1}(-1)=-e^{-1} \lt 0 \end{aligned}

Hence f(x) has maximum value at x=1

f(1)=1\cdot e^{-1}=e^{-1}

Hence f(x) has maximum value at x=1

f(1)=1\cdot e^{-1}=e^{-1}

Question 3 |

The solution for the differential equation

\frac{d^{2}x}{dt^{2}}=-9x

with initial conditions x(0)=1 and \frac{dx}{dt}|_{t=0}=1, is

\frac{d^{2}x}{dt^{2}}=-9x

with initial conditions x(0)=1 and \frac{dx}{dt}|_{t=0}=1, is

t^{2}+t+1 | |

sin 3t+ \frac{1}{3} cos3t+\frac{2}{3} | |

\frac{1}{3} sin 3t+cos3t | |

cos3t+t |

Question 3 Explanation:

\begin{aligned} \frac{d^2x}{dt^2}&=-9x\\ \frac{d^2x}{dt^2}+9x&=0\\ (D^2+9)x &=0 \end{aligned}

Auxiliary equation is m^2+9=0

\begin{aligned} m&= \pm 3i\\ x&= C_1 \cos 3t+C_2 \sin 3t\;\;...(i)\\ x(0) &=1\;\;i.e.\;x\rightarrow 1\; when t\rightarrow 0 \\ 1&=C_1 \\ \frac{dx}{dt}&=-3C_1 \sin 3t+3C_2 \cos 3t\;\;...(ii) \\ x'(0)&=1\;\; i.e.\;x'\rightarrow 1\; when \; t\rightarrow 0 \\ 1&=3C_2 \\ C_2&=\frac{1}{3}\\ \therefore \;x&=\cos 3t+\frac{1}{3}\sin 3t \end{aligned}

Auxiliary equation is m^2+9=0

\begin{aligned} m&= \pm 3i\\ x&= C_1 \cos 3t+C_2 \sin 3t\;\;...(i)\\ x(0) &=1\;\;i.e.\;x\rightarrow 1\; when t\rightarrow 0 \\ 1&=C_1 \\ \frac{dx}{dt}&=-3C_1 \sin 3t+3C_2 \cos 3t\;\;...(ii) \\ x'(0)&=1\;\; i.e.\;x'\rightarrow 1\; when \; t\rightarrow 0 \\ 1&=3C_2 \\ C_2&=\frac{1}{3}\\ \therefore \;x&=\cos 3t+\frac{1}{3}\sin 3t \end{aligned}

Question 4 |

Let X(s)=\frac{3s+5}{s^{2}+10s+21} be the Laplace Transform of a signal x(t). Then, x(0^+) is

0 | |

3 | |

5 | |

21 |

Question 4 Explanation:

Given, X(s)=\left [ \frac{3s+5}{s^2+10s+21} \right ]

Using initial value theorem,

\begin{aligned} x(0^+) &= \lim_{s \to \infty }[sX(s)]\\ x(0^+) &= \lim_{s \to \infty } \left [ \frac{s(3s+5)}{s^2+10+21} \right ] \\ &= \lim_{s \to \infty }\left [ \frac{3+\frac{5}{s}}{1+\frac{10}{s}+\frac{21}{s^2}} \right ]=3 \end{aligned}

Using initial value theorem,

\begin{aligned} x(0^+) &= \lim_{s \to \infty }[sX(s)]\\ x(0^+) &= \lim_{s \to \infty } \left [ \frac{s(3s+5)}{s^2+10+21} \right ] \\ &= \lim_{s \to \infty }\left [ \frac{3+\frac{5}{s}}{1+\frac{10}{s}+\frac{21}{s^2}} \right ]=3 \end{aligned}

Question 5 |

Let S be the set of points in the complex plane corresponding to the unit circle. (That is, S={z:|z|=1}). Consider the function f(z)=zz* where z* denotes the complex conjugate of z. The f(z) maps S to which one of the following in the complex plane

unit circle | |

horizontal axis line segment from origin to (1, 0) | |

the point (1, 0) | |

the entire horizontal axis |

Question 5 Explanation:

\begin{aligned} z&=x+iy\\ z^*&=x-iy\\ zz^*&=(x+iy)(x-iy)\\ &=x^2+y^2 \end{aligned}

which is equal to (1) always as given

\begin{aligned} |z|&=1\\ zz^*&=x^2+y^2 \end{aligned}

Question 6 |

The three circuit elements shown in the figure are part of an electric circuit. The
total power absorbed by the three circuit elements in watts is _____.

110 | |

250 | |

330 | |

360 |

Question 6 Explanation:

Given electrical circuit is shown below:

Applying KCL at node, current through 15V voltage source =2 A

Power absorbed by 100 V voltage source = 10x100=1000 Watt.

Power absorbed by 80 V voltage source =-(80x8)=-640 Watts and power absorbed by 15 V voltage source = -(15x2)=-30 Watt.

Therefore, total power absorbed by the three circuit element=(100-640-30_ watts =330 Watts

Applying KCL at node, current through 15V voltage source =2 A

Power absorbed by 100 V voltage source = 10x100=1000 Watt.

Power absorbed by 80 V voltage source =-(80x8)=-640 Watts and power absorbed by 15 V voltage source = -(15x2)=-30 Watt.

Therefore, total power absorbed by the three circuit element=(100-640-30_ watts =330 Watts

Question 7 |

C_0 is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity \epsilon _r , the expression for the modified capacitance is

\frac{C_{0}}{2}(1+\epsilon _{r}) | |

(C_{0}+\epsilon _{r}) | |

\frac{C_{0}}{2}(\epsilon _{r}) | |

C_{0}(1+\epsilon _{r}) |

Question 7 Explanation:

Let A be the area of the parallel plate capacitor and d be the distance between the plates.

Let C_1 be the capacitance of half portion with air as dielectic medium and C_2 be capacitance with a dielectic of permitivity, \varepsilon _r.

Then, C_1=\frac{\varepsilon _0 \left ( \frac{A}{2} \right )}{d} (As area become half)

and C_2=\frac{\varepsilon _0 \varepsilon _r \left ( \frac{A}{2} \right )}{d} (As area becomes half)

Now, these two capacitance will be in parallel if a voltage is applied between the plates as same potential difference will be there between both the capacitance.

Equivalent capacitance is

\begin{aligned} C_{eqv} &= C_1+C_2\\ &=\frac{\varepsilon _0 A}{2d}+ \frac{\varepsilon _0 \varepsilon _r A}{2d}\\ &= \frac{\varepsilon _0 A}{2d}(1+ \varepsilon _r)=\frac{C_0}{2}(1+\varepsilon _r) \end{aligned}

Therefore, modified capacitance,

C_{eqv}=\frac{C_0}{2}(1+\varepsilon _r)

Let C_1 be the capacitance of half portion with air as dielectic medium and C_2 be capacitance with a dielectic of permitivity, \varepsilon _r.

Then, C_1=\frac{\varepsilon _0 \left ( \frac{A}{2} \right )}{d} (As area become half)

and C_2=\frac{\varepsilon _0 \varepsilon _r \left ( \frac{A}{2} \right )}{d} (As area becomes half)

Now, these two capacitance will be in parallel if a voltage is applied between the plates as same potential difference will be there between both the capacitance.

Equivalent capacitance is

\begin{aligned} C_{eqv} &= C_1+C_2\\ &=\frac{\varepsilon _0 A}{2d}+ \frac{\varepsilon _0 \varepsilon _r A}{2d}\\ &= \frac{\varepsilon _0 A}{2d}(1+ \varepsilon _r)=\frac{C_0}{2}(1+\varepsilon _r) \end{aligned}

Therefore, modified capacitance,

C_{eqv}=\frac{C_0}{2}(1+\varepsilon _r)

Question 8 |

A combination of 1 \muF capacitor with an initial voltage v_c(0)=-2V in series with a 100 \Omega resistor is connected to a 20 mA ideal dc current source by operating both switches at t=0s as shown. Which of the following graphs shown in the options approximates the voltage v_s across the current source over the next few seconds ?

A | |

B | |

C | |

D |

Question 8 Explanation:

Given C=1\mu F, V_c(0)=-2V, R=100\Omega , I=20mA. Circuit fot the given condition at time t \gt 0 is shown below:

Applying KVL, we have,V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )

\;\;=\frac{1}{s}\left [ -2+IR+\frac{I}{Cs} \right ]

\;\;=\frac{1}{s}\left [ (IR-2) +\frac{I}{Cs}\right ]

Putting values of R, C and I, we get,

V_s(s)=\frac{1}{s}\left [ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right ]

\;\;=\frac{1}{s}\left [ (2-2)+20 \times 10^3 \times \frac{1}{s} \right ]

\;\;=\frac{20 \times 10^3}{s^2}

\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}

V_s(t)=20000 t u(t)

\therefore \;\; V_s(s)=(20000)tu(t)

Which is equation of a straight line passing through origin. Hence option (C) is correct.

Applying KVL, we have,V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )

\;\;=\frac{1}{s}\left [ -2+IR+\frac{I}{Cs} \right ]

\;\;=\frac{1}{s}\left [ (IR-2) +\frac{I}{Cs}\right ]

Putting values of R, C and I, we get,

V_s(s)=\frac{1}{s}\left [ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right ]

\;\;=\frac{1}{s}\left [ (2-2)+20 \times 10^3 \times \frac{1}{s} \right ]

\;\;=\frac{20 \times 10^3}{s^2}

\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}

V_s(t)=20000 t u(t)

\therefore \;\; V_s(s)=(20000)tu(t)

Which is equation of a straight line passing through origin. Hence option (C) is correct.

Question 9 |

x(t) is nonzero only for T_x \lt t \lt T'_x, and similarly, y(t) is nonzero only for T_y \lt t \lt T'_y. Let z(t) be convolution of x(t) and y(t). Which one of the following statements is TRUE ?

z(t) can be nonzero over an unbounded interval. | |

z(t) is nonzero for t \lt T_{x}+T_{y} | |

z(t) is zero outside of T_{x}+T_{y} \lt t \lt T'_{x}+T'_{y} | |

z(t) is nonzero for t\gt T'_{x}+T'_{y} |

Question 9 Explanation:

x(t) is non-zero for T_x \lt t \lt T'_x

and y(t) is non-zero for T_y \lt t \lt T'_y

\because \;\; z(t)=x(t)\otimes y(t)

then the limits of the resultant signal is

T_x+T_y \lt t \lt T'_x+T'_y

i.e. z(t) is non-zero for T_x+T_y \lt t \lt T'_x+T'_y

and y(t) is non-zero for T_y \lt t \lt T'_y

\because \;\; z(t)=x(t)\otimes y(t)

then the limits of the resultant signal is

T_x+T_y \lt t \lt T'_x+T'_y

i.e. z(t) is non-zero for T_x+T_y \lt t \lt T'_x+T'_y

Question 10 |

For a periodic square wave, which one of the following statements is TRUE ?

The Fourier series coefficients do not exist | |

The Fourier series coefficients exist but the reconstruction converges at no point | |

The Fourier series coefficients exist but the reconstruction converges at most point | |

The Fourier series coefficients exist and the reconstruction converges at every
point. |

Question 10 Explanation:

Reconstruction of signal by its Fourier series coefficient is not possible at those points where signal is discontinuous.

In the above figure, at integer multiples of 'T/2', signal recovery is not possible by using its coefficient.

Therefore, reconstruction of x(t) by using its coefficient is possible at most of the points except those instants where x(t) is discontinous.

There are 10 questions to complete.