# GATE EE 2003

 Question 1
Figure Shows the waveform of the current passing through an inductor of resistance $1\Omega$ and inductance 2 H. The energy absorbed by the inductor in the first four seconds is A 144J B 98J C 132J D 168J
Electric Circuits   Basics
Question 1 Explanation:
For $0 \lt t \lt 2s$ current varies linearly with time and given as $i(t)=3t$ and for $2s \lt t \lt 4s$ current is constant, $i(t)=6A$.
The energy absorbed by the inductor (Resistance neglected) in the first 2 sec,
$E_L=\int_{0}^{T}Li\frac{di}{dt}=E_{L_1}+E_{L_2}$
$E_{L_1}=\int_{0}^{2}Li \left (\frac{di}{dt} \right )dt$
$\;\;=\int_{0}^{2} 2 \times 3t \times 3\; dt$
$\;\;=18\int_{0}^{2}t\; dt$
$\;\;=18 \times \left.\begin{matrix} \frac{t^2}{2} \end{matrix}\right|^2_0$
$\;\;= 18 \times \left [ \frac{4}{2}-0 \right ]=36J$
The energy absorbed by the inductor in ($2\rightarrow 4$) second
$E_{L_2}=\int_{2}^{4}Li\left ( \frac{di}{dt}dt \right )$
$\;\;=\frac{2}{4}2 \times 6 \times 0 \;dt=0 J$
A pure indictor does not dissipate enegy but only stores it. Due to resistance, some energy is dissipated in the resistor. Therefore, total energy stored in the inductor and the energy dissipated in the resistor.
The energy dissipated by the resistance in 4 sec.
$E_R=\int_{0}^{T}i^2 R\;dt$
$\;\;=\int_{0}^{2}(3t)^2 \times 1 \;dt+\int_{2}^{4}6^2 \times 1 \; dt$
$\;\;=9\int_{0}^{2}t^2\;dt +36\int_{2}^{4}1 \; dt$
$\;\;=9 \times \left.\begin{matrix} \frac{t^3}{3} \end{matrix}\right|_0^2+36t|_2^4$
$\;\;=9 \times \left ( \frac{8}{3} \right )+36 \times 2$
$\;\;=24+72=96J$
The total energy absorbed by the inductor in 4 sec
$=96 J +36 J =132 J$
 Question 2
A segment of a circuit is shown in figure. $V_{R}$ = 5V, $V_{C}$ = $4 \sin 2t$. The voltage $V_{L}$ is given by A $3-8 \cos 2t$ B $32 \sin 2t$ C $16 \sin 2t$ D $16 \cos 2t$
Electric Circuits   Basics
Question 2 Explanation:
By KCL,
$I_P+I_Q+I_C+I_L=0$
$2+2+I_C+I_L=0$
But, $I_C=C \times \frac{dv}{dt}$
$\;\; =1 \times \frac{d}{dt}(4 \sin 2t)$
$\;\;=(8 \cos 2t)$
$\therefore \;\; I_L=-(2+1+8 \cos 2t)$
$\;\; \;\;=-3-8 \cos 2t$
$\therefore \;\; V_L=L\left ( \frac{di}{dt} \right )$
$\;\;\;=2 \times 2 \times 8 \sin 2t$
$\;\;\;=32 \sin 2t$
NOTE: KCL is based on the law of conservation of charges.

 Question 3
In the figure,
$Z_{1}=10\angle -60 ^{\circ} , Z_{2}=10\angle 60^{\circ} , Z_{3}=50\angle 53.13^{\circ}$.
The venin impedance seen form X-Y is A $56.66\angle 45^{\circ}$ B $60\angle 30^{\circ}$ C $70\angle 30^{\circ}$ D $34.4\angle 65^{\circ}$
Electric Circuits   Network Theorems
Question 3 Explanation:
By Thevenin's theorem $Z_{Th}=Z_{X-Y}=Z_1||Z_2+Z_3$
$\;\;=\frac{Z_1 \times Z_2}{Z_1+Z_2}+Z_3$
$\;\;=\frac{10\angle -60 \times 10\angle 60}{10\angle -60 + 10\angle 60}+50\angle 53.13$
$\;\;=56.66\angle 45^{\circ}\Omega$
 Question 4
Two conductors are carrying forward and return current of +I and -I as shown in figure. The magnetic field intensity H at point P is A $\frac{I}{\pi d}\vec{Y}$ B $\frac{I}{\pi d}\vec{X}$ C $\frac{I}{2 \pi d}\vec{Y}$ D $\frac{I}{2 \pi d}\vec{X}$
Electromagnetic Fields   Magnetostatic Fields
Question 4 Explanation:
$\bar{H}=\bar{H_1}+\bar{H_2}=\frac{I}{2 \pi d}\vec{y}+\frac{I}{2 \pi d}\vec{y}=\frac{I}{ \pi d}\vec{y}$ Question 5
Two infinite strips of width w m in x -direction as shown in figure, are carrying forward and return currents of +I and -I in the z - direction. The strips are separated by distance of x m. The inductance per unit length of the configuration is measured to be L H/m. If the distance of separation between the strips in snow reduced to x/2 m, the inductance per unit length of the configuration is A 2L H/m B L/4 H/m C L/2 H/m D 4L H/m
Electromagnetic Fields   Magnetostatic Fields
Question 5 Explanation:
Inductance is proportional to separation between the strips so when separation is reduced to x/2 then inductance will become L/2 H/m.

There are 5 questions to complete.