GATE EC 2002


Question 1
The dependent current source shown in Figure
A
delivers 80 W
B
absorbs 80 W
C
delivers 40 W
D
absorbs 40 W
Network Theory   Basics of Network Analysis
Question 1 Explanation: 
Applying KVL, 20-5 I-5\left(I+\frac{V_{1}}{5}\right)=0
20-10 I-20=0
\Rightarrow \quad I=0
\therefore Only dependent source acts.
\frac{V_{1}}{5}=4 \mathrm{A}
Power delivered =I^{2} R=16 \times 5=80 \mathrm{W}
Question 2
In figure, the switch was closed for a long time before opening at t = 0. the voltage V_x at t = 0^{+} is
A
25 V
B
50 V
C
-50 V
D
0 V
Network Theory   Transient Analysis
Question 2 Explanation: 
When switch was closed circuit was in steady state,


i_{L}(O-) = 2.5 A
At, t = 0^{+}


\begin{aligned} \Rightarrow & & V=I R \\ & &=2.5 \times 20=50 \mathrm{V} \end{aligned}
\therefore \quad V_{x}=-50 \mathrm{V}
(Polarity of V_{x} is given reverse of n)


Question 3
Convolution of x(t+5) with impulse function \delta(t-7) is equal to
A
x(t - 12)
B
x(t + 12)
C
x(t - 2)
D
x(t + 2)
Signals and Systems   LTI Systems Continuous and Discrete
Question 3 Explanation: 
\begin{aligned} x(t+5) * \delta(t-7) &=x(t+5-7) \\ &=x(t-2) \end{aligned}
Question 4
Which of the following cannot be the Fourier series expansion of a periodic signal?
A
x(t) = 2cos t + 3 cos 3t
B
x(t) = 2cos \pit + 7 cos t
C
x(t) = cos t + 0.5
D
x(t) = 2cos 1.5\pit + sin 3.5 \pit
Signals and Systems   Fourier Series
Question 4 Explanation: 
\begin{aligned} x(t) &=2 \cos \pi t+7 \cos t \\ T_{1} &=\frac{2 \pi}{(0)}=2 \\ T_{2} &=\frac{2 \pi}{1}=2 \pi \\ \frac{T_{1}}{T_{2}} &=\frac{1}{\pi}=\text { irrational } \end{aligned}
\therefore x(t) is not periodic and does not satisfy Dirichlet condition.
Question 5
In Figure shown below, a silicon is carrying a constant current of 1 mA. When the temperature of the diode is 20^{\circ}C, V_{D} is found to be 700 mV. If the temperature rises to 40^{\circ}C, V_{D} becomes approximately equal to
A
740 mV
B
660 mV
C
680 mV
D
700 mV
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 5 Explanation: 
For either Si or Ge\frac{d V}{d T} \simeq-2 \mathrm{mV} /^{\circ} \mathrm{C}
in order to maintain a constant value of I.
\begin{aligned} T_{2}-T_{1}&=40-20=20^{\circ} \mathrm{C} \\ -2 \times 20 \mathrm{mV}&=40 \mathrm{mV} \\ \text { Therefore, } \quad V_{D}&=700-40=660 \mathrm{mV} \end{aligned}


There are 5 questions to complete.