# 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 $\pi$t + 7 cos t C x(t) = cos t + 0.5 D x(t) = 2cos 1.5$\pi$t + sin 3.5 $\pi$t
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}
 Question 6
In a negative feedback amplifier using voltage-series (i.e. voltage-sampling series mixing) feedback. (Ri and Ro denote the input and output resistance respectively)
 A Ri decreases and Ro decreases B Ri decreases and Ro increases C Ri increases and Ro decreases D Ri increases and Ro increases
Analog Circuits   Feedback Amplifiers
Question 6 Explanation:
$R_{i}$ increases by factor of $1+A \beta$ and $R_{0}$ decreases by $1+A \beta$
 Question 7
A 741-type op-amp has a gain-bandwidth product of 1 MHz. A non-inverting amplifier using this op-amp and having a voltage gain of 20 dB will exhibit a 3-dB bandwidth of
 A 50KHz B 100KHz C 1000/17KHz D 1000/7.07KHz
Analog Circuits   Operational Amplifiers
Question 7 Explanation:
\begin{aligned} 20 \log x &=20 \mathrm{dB} \\ x &=10 \\ \text{Gain} \times B W &=1 \times 10^{6} \\ B W &=\frac{1 \times 10^{6}}{\text { Gain }}=\frac{10^{6}}{10}=100 \mathrm{k}=10^{5} \end{aligned}
 Question 8
Three identical RC-coupled transistor amplifiers are cascaded. If each of the amplifiers has a frequency response as shown in Figure, the overall frequency response is as given in

 A A B B C C D D
Analog Circuits   BJT Analysis
Question 8 Explanation:
$\begin{array}{l} f_{L}=20 \mathrm{Hz} \\ f_{H}=1 \mathrm{kHz} \text { for single stage } \end{array}$
$\begin{array}{l} f_{L}^{*}=\frac{f_{L}}{\sqrt{2^{1 / n}-1}}=\frac{20}{\sqrt{2^{1 / 3}-1}}=39.2 \mathrm{H}_{2} \\ f_{H}^{*}=f_{H} \sqrt{2^{1 / n}-1}=0.5 \mathrm{k} \Omega \end{array}$
 Question 9
4-bit 2's complement representation of a decimal number is 1000. The number is
 A 8 B 0 C -7 D -8
Digital Circuits   Number Systems
Question 9 Explanation:
1000
MSB is 1 so, -ve number
Take 2's complement for magnitude.
$+\begin{array}{cccc} 0&1&1&1\\ &&&1\\ \hline1&0&0&0 \end{array}=8$
 Question 10
If the input to the digital circuit (shown in figure ) consisting of a cascade of 20 XOR gates is X, then the output Y is equal to
 A 0 B 1 C $\overline{X}$ D X
Digital Circuits   Logic Gates
Question 10 Explanation:
$\begin{array}{l} \text { Output of } 1 \mathrm{st} \text { XOR gate }=\bar{X} \\ \text { Output of } 2 \text { nd } \mathrm{XOR} \text { gate }=\bar{X} \oplus X \\ =(\bar{X}) \cdot X+\bar{X} \cdot \bar{X}=X+\bar{X}=1 \end{array}$
Output of 20 XOR gates is 1
There are 10 questions to complete.