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}
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}
For cascaded stage
\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.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.