GATE EE 2017 SET 1

Given that, g(t)=\left\{\begin{matrix} t-\left \lfloor t \right \rfloor, & t \geq 0\\ t-\left \lceil t \right \rceil & \text{otherwise} \end{matrix}\right.
where,
\left \lfloor t \right \rfloor= greatest integer less than or equal to 't'.
\left \lceil t \right \rceil= smallest integer greater than or equal to 't'.
Now,



Since, g(t) is nonperiodic. So, there is no Fourier series expansion of this signal and hence no need to calculate harmonic here.

\begin{aligned} V_{an}&=220 \sin (100 \pi t)V\\ i_a&=10 \sin (100 \pi t)A\\ V_{bn}&=220 \cos (100 \pi t)V\\ i_b&=10 \cos (100 \pi t)A\\ p&=V_{an}i_a+V_{bn}i_b\\ &=2200 W \end{aligned}

For synchronous motor,
\bar{V_t}=\bar{E_f}+j\bar{I_a}X_s
As currrent is drawn at unity power factors.
Therefore,
\begin{aligned} E_f \cos \delta &=V_t \\ E_f&= \frac{V_t}{\cos \delta }=\frac{6.6kV}{\sqrt{3}} \times \frac{1}{\cos 30^{\circ}}\\ E_f&= \frac{6.6 \times 10^3}{\sqrt{3} \times \sqrt{3}} \times 2V\\ V_t &=\frac{6.6 \times 10^3}{\sqrt{3}} \\ \therefore \;\;P_e &=\frac{3V_{ph}E_{ph}}{X_s} \sin \delta \\ &= \frac{3 \times \frac{6.6 \times 10^3}{\sqrt{3}} \times \frac{6.6 \times 10^3}{\sqrt{3} \times \sqrt{3}} \times 2 }{30} \times \sin 30^{\circ}\\ &=838.3kW \end{aligned}

\begin{aligned} \lim_{z \to i}&=\frac{z^2+1}{z^3+2z-i(z^2+2)}\\ \lim_{z \to i}&=\frac{2z}{3z^2+2-i(2z)}\\ &=\frac{2i}{3i^2+2-i(2i)}\\ &=\frac{2i}{-3+2+2}\\ &=\frac{2i}{-3+4}=2i \end{aligned}

Given that electron, neutron and proton are in straight line.

The electron will move towards proton and proton will move towards electron and force will be same F=\frac{q_1q_2}{4 \pi \in _0 R^2}. But acceleration of electron will be more than proton as mass of electron \lt mass of proton. Since neutron are neutral they will not move. Thus electron will hit neutron first.

Time scaling property of convolution.
If, x(t)*y(t)=z(t)
Then, x(ct)*y(ct)=\frac{1}{c} z(ct)
z(ct)=c \times x(ct) * y(ct)

Given,
\begin{aligned} Y_{11} &=-j12 \; p.u. \\ Y_{22} &=-j15 \; p.u.\\ Y_{33} &=-j7 \; p.u. \\ &\text{We know that,} \\ Y_{11} &=y_{12}+y_{13}=-j12 \; p.u. \;\;...(i) \\ Y_{22} &=y_{12}+y_{23}=-j15 \; p.u. \;\;...(ii) \\ Y_{33} &=y_{13}+y_{23}=-j7 \; p.u. \;\;...(iii) \\ &\text{From eq. (i) and (ii)} \\ y_{13}-y_{23} &=j3\; p.u. \\ y_{13}+y_{23} &=-j7 \; p.u. \\ y_{13}&= -j2\; p.u.\\ y_{23}&= -j5\; p.u.\\ y_{12}&= -j10\; p.u. \end{aligned}
The p.u. values of line reactances p, q and r are
\begin{aligned} jr&=\frac{1}{-j2}=j0.5\; p.u.\\ jp&=\frac{1}{-j5}=j0.2\; p.u.\\ jq&=\frac{1}{-j10}=j0.1\; p.u.\\ \therefore \; p&=0.2, q=0.1, r=0.5 \end{aligned}

Consider the following circuit diagram,

After rearrangement we get

Now, R_{AB}=1+\frac{6}{5}+0.8=3\Omega

BC+A\bar{C}

\begin{aligned} V &= 220 \pm 1\% \\ I&= 5 \pm 1 \%\\ W &=555 \pm 2 \% \\ W&= VI \cos (\phi )\\ p.f. &=\cos (\phi )=\frac{W}{VI} \\ &= \frac{555 \pm 2 \%}{(220 \pm 1\%)(5 \pm 1 \%)}\\ &= \frac{555}{220 \times 5} \pm 4\%\\ p.f.&= 0.5 \pm 4\% \end{aligned}