# GATE EE 2015 SET 2

 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?
 A If f(z) is differentiable at $z_0$, then g(z) and g(z) are also differentiable at $z_0$. B If g(z) and h(z) are differentiable at $z_0$, then f(z) is also differentiable at $z_0$. C If f(z) is continuous at $z_0$, then it is differentiable at $z_0$. D If f(z) is differentiable at $z_0$, then so are its real and imaginary parts.
Engineering Mathematics   Complex Variables
 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?
 A $P\equiv Q\equiv R\equiv S$ B $P\equiv R\not\equiv Q\equiv S$ C $P\equiv Q\not\equiv R\equiv S$ D $P\not\equiv Q\not\equiv R\not\equiv S$
Engineering Mathematics   Linear Algebra
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 $X=A^{-1}B$
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.
 A P-2 Q-1 R-4 S-3 B P-4 Q-1 R-3 S-2 C P-4 Q-3 R-1 S-2 D P-3 Q-4 R-2 S-1
Electromagnetic Fields   Coordinate Systems and Vector Calculus
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 $\oint _cf(z)dz=0$
 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
 A $3s^{-5/2}/2$ B $s^{-1/2}$ C $s^{1/2}$ D $s^{3/2}$
Signals and Systems   Laplace Transform
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}
 Question 5
Match the following.
 A P-1 Q-2 R-1 S-3 B P-1 Q-2 R-1 S-3 C P-1 Q-2 R-3 S-3 D P-3 Q-1 R-2 S-1
Electrical and Electronic Measurements   Galvanometers, Voltmeters and Ammeters
 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
 A 3.94 kW and 1.06 kW B 2.50 kW and 2.50 kW C 5.00 kW and 0.00 kW D 2.96 kW and 2.04 kW
Electrical and Electronic Measurements   Measurement of Energy and Power
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 $\pm$10% 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 __________.
 A 8.52 B 11.95 C 16.35 D 22.25
Electrical and Electronic Measurements   Galvanometers, Voltmeters and Ammeters
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$\pi$t). For 100$\pi$RC=50, the average voltage across R (in Volts) under steady-state is nearest to
 A 100 B 31.8 C 200 D 63.6
Power Electronics   Phase Controlled Rectifiers
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}
 Question 9
Two semi-infinite dielectric regions are separated by a plane boundary at y = 0. The dielectric constants of region 1 (y$\lt$0) and region 2 (y$\gt$0) 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
 A $3\hat{a}_{x} + 1.6\hat{a}_{y} + 2\hat{a}_{z}$ B $1.2\hat{a}_{x} + 4\hat{a}_{y} + 2\hat{a}_{z}$ C $1.2 \hat{a}_{x} + 4\hat{a}_{y} + 0.8\hat{a}_{z}$ D $3\hat{a}_{x} + 10\hat{a}_{y} + 0.8\hat{a}_{z}$
Electromagnetic Fields   Electrostatic Fields
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}
 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 _________.
 A 150.35 B 200.25 C 248.05 D 300.54
Electromagnetic Fields   Time Varying Fields
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.