Question 1 |

Given f(z)=g(z)+h(z), where f, g, h are complex valued functions of a complex variable z. Which one of the following statements is TRUE?

If f(z) is differentiable at z_0, then g(z) and g(z) are also differentiable at z_0. | |

If g(z) and h(z) are differentiable at z_0, then f(z) is also differentiable at z_0. | |

If f(z) is continuous at z_0, then it is differentiable at z_0. | |

If f(z) is differentiable at z_0, then so are its real and imaginary parts. |

Question 2 |

We have a set of 3 linear equations in 3 unknowns. 'X \equiv Y' means X and Y are equivalent statements and 'X\not\equiv Y' means X and Y are not equivalent statements.

P: There is a unique solution.

Q: The equations are linearly independent.

R: All eigenvalues of the coefficient matrix are nonzero.

S: The determinant of the coefficient matrix is nonzero.

Which one of the following is TRUE?

P: There is a unique solution.

Q: The equations are linearly independent.

R: All eigenvalues of the coefficient matrix are nonzero.

S: The determinant of the coefficient matrix is nonzero.

Which one of the following is TRUE?

P\equiv Q\equiv R\equiv S | |

P\equiv R\not\equiv Q\equiv S | |

P\equiv Q\not\equiv R\equiv S | |

P\not\equiv Q\not\equiv R\not\equiv S |

Question 2 Explanation:

\begin{aligned} a_{11}x_1+a_{12}x_2+a_{13}x_3 &=b_1 \\ a_{21}x_1+a_{22}x_2+a_{23}x_3 &=b_2 \\ a_{31}x_1+a_{32}x_2+a_{33}x_3 &=b_3 \\ \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22}& a_{21}\\ a_{31}& a_{32} &a_{33} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}&= \begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix} \end{aligned}

if |A|\neq 0 then AX=B can be written as X=A^{-1}B

It leads unique solutions.

If |A|\neq 0 then \lambda _1 \cdot \lambda _2\cdot \lambda _3\neq 0 each \lambda _i is non-zero.

If |A|\neq 0 then all the row (column) vectors of A are linearly independent.

if |A|\neq 0 then AX=B can be written as X=A^{-1}B

It leads unique solutions.

If |A|\neq 0 then \lambda _1 \cdot \lambda _2\cdot \lambda _3\neq 0 each \lambda _i is non-zero.

If |A|\neq 0 then all the row (column) vectors of A are linearly independent.

Question 3 |

Match the following.

P-2 Q-1 R-4 S-3 | |

P-4 Q-1 R-3 S-2 | |

P-4 Q-3 R-1 S-2 | |

P-3 Q-4 R-2 S-1 |

Question 3 Explanation:

Stokes theorem \oint \vec{A}\cdot dl=\int \int (\triangledown \times A)\cdot \hat{n}ds

Gauss theorem \int \int D\cdot ds=Q

Divergence theorem \oint \oint A\cdot \hat{n}ds=\int \int \int \triangledown \cdot \bar{A}dV

Cauchy integral theorem \oint _cf(z)dz=0

Gauss theorem \int \int D\cdot ds=Q

Divergence theorem \oint \oint A\cdot \hat{n}ds=\int \int \int \triangledown \cdot \bar{A}dV

Cauchy integral theorem \oint _cf(z)dz=0

Question 4 |

The Laplace transform of f(t)=2\sqrt{t/\pi } is s^{-3/2}. The Laplace transform of g(t)=\sqrt{1/\pi t} is

3s^{-5/2}/2 | |

s^{-1/2} | |

s^{1/2} | |

s^{3/2} |

Question 4 Explanation:

Given that,

f(t)=2\sqrt{\frac{t}{\pi}}\rightleftharpoons F(s)=s^{-3/2}

By using property of differentiation In time,

\begin{aligned} \frac{df(t)}{dt}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{2}{\sqrt{\pi}}\cdot \frac{1}{2}t^{-1/2}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s\cdot s^{-3/2} \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s^{-1/2} \end{aligned}

f(t)=2\sqrt{\frac{t}{\pi}}\rightleftharpoons F(s)=s^{-3/2}

By using property of differentiation In time,

\begin{aligned} \frac{df(t)}{dt}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{2}{\sqrt{\pi}}\cdot \frac{1}{2}t^{-1/2}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s\cdot s^{-3/2} \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s^{-1/2} \end{aligned}

Question 5 |

Match the following.

P-1 Q-2 R-1 S-3 | |

P-1 Q-2 R-1 S-3 | |

P-1 Q-2 R-3 S-3 | |

P-3 Q-1 R-2 S-1 |

Question 6 |

A 3-phase balanced load which has a power factor of 0.707 is connected to a balanced supply. The power consumed by the load is 5 kW. The power is measured by the two-wattmeter method. The readings of the two wattmeters are

3.94 kW and 1.06 kW | |

2.50 kW and 2.50 kW | |

5.00 kW and 0.00 kW | |

2.96 kW and 2.04 kW |

Question 6 Explanation:

\begin{aligned} p.f. &= 0.707\\ \phi &= cos^{-1}(0.707)=45^{\circ}\\ P&= 5kW\\ W_1 +W_2&= 5\;\;...(i)\\ \tan \phi &=\frac{\sqrt{3(W_1-W_2)}}{(W-1+W_2)} \\ (W-1+W_2) \times 1 &= \sqrt{3}(W_1-W_2) \\ W_1-W_2 &= \frac{5}{\sqrt{3}} \;\;...(ii)\\ \text{From equation} & \text{ (i) and (ii),} \\ W-1&= 3.94kW\\ W_2&= 1.06kW \end{aligned}

Question 7 |

A capacitive voltage divider is used to measure the bus voltage V_{bus} in a high-voltage 50 Hz AC system as shown in the figure. The measurement capacitors C_1 \; and \; C_2 have tolerances of \pm10% on their nominal capacitance values. If the bus voltage V_{bus} is 100 kV rms, the maximum rms output voltage V_{out} (in kV), considering the capacitor tolerances, is __________.

8.52 | |

11.95 | |

16.35 | |

22.25 |

Question 7 Explanation:

V_{BUS} \text{is} 100 kV_{rms}

C_1=1\mu F \pm 10\%

C_2=9\mu F \pm 10\%

To get maximum output voltage we need minimum C_2 and maximum C_1,

So, C_2=8.1\mu F and C_1=1.1\mu F

So, V_{out\;rms}=\left ( \frac{C_1}{C_1+C_2} \right )V_{BUS_{rms}}

\;\;\;=11.95kV

Question 8 |

In the following circuit, the input voltage V_{in} is 100sin(100\pit). For 100\piRC=50, the average voltage across R (in Volts) under steady-state is nearest to

100 | |

31.8 | |

200 | |

63.6 |

Question 8 Explanation:

Given,

\begin{aligned} 100 \pi RC&=50\\ 100 \pi\cdot \frac{RC}{2}&=25\\ \omega \tau &=25 \end{aligned}

For one cycle \omega \tau =2 \pi \; rad=6.28

[\omega \tau =25] \gt \gt [\omega \tau =2 \pi \; rad=6.28]

Each capacitor voltage is approximately equal to

\begin{aligned} V_m&=100V \\ V_0&=2V_m\simeq 200V \end{aligned}

\begin{aligned} 100 \pi RC&=50\\ 100 \pi\cdot \frac{RC}{2}&=25\\ \omega \tau &=25 \end{aligned}

For one cycle \omega \tau =2 \pi \; rad=6.28

[\omega \tau =25] \gt \gt [\omega \tau =2 \pi \; rad=6.28]

Each capacitor voltage is approximately equal to

\begin{aligned} V_m&=100V \\ V_0&=2V_m\simeq 200V \end{aligned}

Question 9 |

Two semi-infinite dielectric regions are separated by a plane boundary at y = 0. The dielectric constants of region 1 (y\lt0) and region 2 (y\gt0) are 2 and 5, respectively. Region 1 has uniform electric field \vec{E}=3\hat{a}_{x}+4\hat{a}_{y}+2\hat{a}_{z}, where \hat{a}_{x},\hat{a}_{y} and \hat{a}_{z} are unit vectors along the x, y and z axes, respectively. The electric field in region 2 is

3\hat{a}_{x} + 1.6\hat{a}_{y} + 2\hat{a}_{z} | |

1.2\hat{a}_{x} + 4\hat{a}_{y} + 2\hat{a}_{z} | |

1.2 \hat{a}_{x} + 4\hat{a}_{y} + 0.8\hat{a}_{z} | |

3\hat{a}_{x} + 10\hat{a}_{y} + 0.8\hat{a}_{z} |

Question 9 Explanation:

Given that,

at the interface (y=0) there is no surface charge.

Region 1 (y \lt 0) has \varepsilon _r=2

Region 2 (y \gt 0) has \varepsilon _r=5

Electric field in region 1 is 3a_x+4a_y+2a_z

Normal component of electric field is 4a_y tangential component of electric field is 3a_x+2a_y

Now, E_1=E_{1t}+E_{1n}

The tangentail component of E_1= tangentail component of E_2

E_{1t}=E_{2t}=3a_x+2a_z

\begin{aligned} \varepsilon _1 E_{1n}&=\varepsilon _2 E_{2n}\\ E_{2n}&=\frac{\varepsilon _1 \times 4a_y}{\varepsilon _2} \\ &= \frac{2 \times 4a_y}{5}=1.6a_y\\ E_2&=E_{2t}+E_{2n}\\ &=3a_x+1.6a_y+2a_z \end{aligned}

at the interface (y=0) there is no surface charge.

Region 1 (y \lt 0) has \varepsilon _r=2

Region 2 (y \gt 0) has \varepsilon _r=5

Electric field in region 1 is 3a_x+4a_y+2a_z

Normal component of electric field is 4a_y tangential component of electric field is 3a_x+2a_y

Now, E_1=E_{1t}+E_{1n}

The tangentail component of E_1= tangentail component of E_2

E_{1t}=E_{2t}=3a_x+2a_z

\begin{aligned} \varepsilon _1 E_{1n}&=\varepsilon _2 E_{2n}\\ E_{2n}&=\frac{\varepsilon _1 \times 4a_y}{\varepsilon _2} \\ &= \frac{2 \times 4a_y}{5}=1.6a_y\\ E_2&=E_{2t}+E_{2n}\\ &=3a_x+1.6a_y+2a_z \end{aligned}

Question 10 |

A circular turn of radius 1 m revolves at 60 rpm about its diameter aligned with the x-axis as shown in the figure. The value of \mu _{0} is 4\pi \times 10^{-7} in SI unit. If a uniform magnetic field intensity \vec{H}=10^{7}\hat{z} A/m is applied, then the peak value of the induced voltage, V_{turn} ( in Volts), is _________.

150.35 | |

200.25 | |

248.05 | |

300.54 |

Question 10 Explanation:

The circular turn rotate with 60 rpm, let the angle made by ring w.r.t. x-axis \theta

and \;\;\; \theta =\omega _0 t

the turn rotate at 60 rpm,

so, \;\;\; \omega _0=2 \pi

So, the flux flowing through the circular turn wil be

\begin{aligned} \Psi &=(\mu _0 H_z \times \text{Area of turn} \times \cos \omega _0 t) \\ \Psi &= 4 \pi \times 10^{-7} \times 10^7\; A/m \times \pi \times 1^2 \times \cos \omega _0 t\\ &\text{Maximum voltage induced is } \\ \\ \left. \begin{matrix} \frac{d\Psi}{dt} \end{matrix}\right|_{max}&=(\omega _0 \times 4 \pi \times \pi \sin \omega _0 t)_{max}\\ V_{max}&=(4 \pi^2 \times 2 \pi)=248.05 volts \end{aligned}

There are 10 questions to complete.