| Author: smslca 23 Sep 2009 | Member Level: Bronze | Rating:  Points: 4 |
Excellent idea dude If Im not wrong this question is from kanodia book....
Here my question
If a unit step function is passed through a capacitor what will be the voltage across the capacitor?
a)0
b)a step function
c)a ramp function
d)an impulse function
Answer:
a ramp function(c)
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| Author: GATE 2010 Aspirant 23 Sep 2009 | Member Level: Diamond | Rating: Points: 3 |
Hi
Its really interesting yaar.
First Question 2's complement of -7 is 1001. How 11001 and 111001 are 2's complement of -7.
Thanks and Regards,
Sundar.
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| Author: Ghulam 23 Sep 2009 | Member Level: Silver | Rating: Points: 3 |
Hi Sundar,
Take out 2's complement of 11001:
ie.
00110
+1
00111=(7)
Now, Take out 2's complement of 111001:
ie.
000110
+1
000111=(7)
The (-) sign appears in both the above 7's because the USB (Upper Significant bit) of both the binary forms are 1 and 1 at USB indicates a negative number.
I think You got it now.
If not, then please let me know.
Thanks.
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| Author: Ghulam 24 Sep 2009 | Member Level: Silver | Rating: Points: 3 |
Hi Smslsca,
Sorry yaar I dont know your name.
Please provide a brief description of your solution with your answer. It would be useful for everyone of us.
Really saying a nice question of yours.
If I am not wrong the Capacitor plays the role of a differentiator when step Input passes through it converts that step I/P to Ramp O/P.
Thanks.
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| Author: Ghulam 24 Sep 2009 | Member Level: Silver | Rating: Points: 4 |
Hi all,
QUESTION OF THE DAY- 24th Sep, 2009
What is the capacitive reactance of a circuit comprised of a 65.0-µF capacitor and a 50.0-Hz generator?
a. 49.0 ohms
b. 72.5 ohms
c. 97.6 ohms
d. 145 ohms
e. 308 ohms
Ans: 49.0 ohms (Capacitive reactance= 1/(2*pi*Fc*C); Fc=cut off frequency and C is the capacitance of the capacitor)
Thanks.
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| Author: smslca 24 Sep 2009 | Member Level: Bronze | Rating: Points: 2 |
Capacitor acts as an integrator so, for step signal the output is ramp.i think it clears all the doubts......
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| Author: Ghulam 25 Sep 2009 | Member Level: Silver | Rating: Points: 5 |
QUESTION OF THE DAY (E&TCE)
SEMICONDUCTORS:
Given Parameters of a Junction Diode are:
Cjo=20pF
Vb=1V
m=0.5
Calculate change in Junction Capacitance if the reverse bias Voltage on the diode changes from 5V to 10V?
A)2.2 microFarad
B)2.2 picoFarad
C)1.1 microFarad
D)1.1 picoFarad
Ans) B(2.2 picoFarad)
Expalnation:
Junction Capacitance Cj is given by
Cj = Cjo/[(1-Vd/Vb)^m]
calculate Cj for Vd1=5V and Vd2=10V and find the change.
Thanks.
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| Author: Ghulam 25 Sep 2009 | Member Level: Silver | Rating: Points: 4 |
Hi all,
I wouldn't be able to post questions from tomorrrow onwards as I have an operation on 28th.
But my request to you all, dont stop this process.
Continue it to its success.
I request SMSLCA and Sundar to continue this process.
I will Take care of it after 10th October.
All the best.
Thanks.
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| Author: smslca 25 Sep 2009 | Member Level: Bronze | Rating: Points: 3 |
Today's question
Relaxation oscillator is one which
a) has 2 stable states
b) oscillates continuously
c) relaxes indefinitely
d) produces non-sinusoidal output
Answer :
D(produces non-sinusoidal output)
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| Author: smslca 25 Sep 2009 | Member Level: Bronze | Rating: Points: 2 |
Hi all
I am also not available for next one week so please try to continue the process i will be available from 5th october.....
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| Author: Ghulam 25 Sep 2009 | Member Level: Silver | Rating: Points: 5 |
Hi,
Notes on Relaxation Oscillator:
A relaxation oscillator is an oscillator in which a capacitor is charged gradually and then discharged rapidly. It is usually implemented with a resistor or current source, a capacitor, and a threshold device such as a neon lamp, thyratron, diac, unijunction transistor, or Gunn diode.
The output of relaxation Oscillator is generally a Sawtooth Wave.
Thanks.
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| Author: smslca 07 Oct 2009 | Member Level: Bronze | Rating: Points: 3 |
Find the closed loop gain of an Op-amp if its open loop gain is 1000 and Rf(feed back resistance) is 100k and r1(resistance connected to inverting input) is 1k
A)90.83
B)99.0
C)96.78
D)89.98
Answer is "A"
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| Author: smslca 07 Oct 2009 | Member Level: Bronze | Rating: Points: 1 |
Closed loop gain G= (Rf/r1)/(1+((1+Rf/r1)/A))
Where A is open loop gain
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| Author: Priyanka 20 Oct 2009 | Member Level: Bronze | Rating: Points: 3 |
Hi all,
Could anyone of u plzzz help me out by suggesting me d best available book for clearing d basic concepts in Electronics- mainly for network analysis and basic electronics part..
I wud be really thankful....
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| Author: Ghulam 20 Oct 2009 | Member Level: Silver | Rating: Points: 6 |
Hi Priyanka,
There are lot many good books you can follow for clearing your basic concepts:
Network Analysis:
* Network Analysis VanValkenburg
* Engg Circuit Analysis Hayt & Kemmerly
Basic Electronics Part:
* Micro Electronic Circuits by Sedra& Smith
* Electronic Principals by Malvino
* Electronic Devices And Circuits by Boylestead
But from my opinion try not to follow all these books.
Pick up any one from these get thorough through the topics.
Any One book from each section is enough.
Meanwhile I am trying to post some soft copies related to electronics in the File Manager tab. Try going through those. It will be of great help!!
Thanks.
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| Author: Ghulam 21 Oct 2009 | Member Level: Silver | Rating: Points: 5 |
Hi all,
Here's the Question of the Day from SEMICONDUCTORS:
Q: The thermal-equilibrium concentration of hole Po in silicon at T=300 K is 10^15 cm-3. The value of No is:
A.3.8*10^8 cm-3
B.4.4*10^4 cm-3
C.2.6*10^4 cm-3
D.4.3*10^8 cm-3
Ans: (B)4.4*10^4 cm-3
Sol: Po= Nv exp -{(Ef-Ev)/kT}
=> (Ef-Ev)= kT ln (Nv/Po)
At 300K, Nv= 10^19 cm-3
=> Ef-Ev = 0.0259 ln (10^19/10^15) = 0.239eV
No= Nc exp{(Ec-Ef)/kT}
At 300K, Nc=2.8*10^19 cm-3
=> Ec-Ef=1.12-0.239= 0.881eV
=> No=4.4*10^4 cm-3 (Ans)
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| Author: swazz 24 Oct 2009 | Member Level: Bronze | Rating: Points: 0 |
thanks for suggest this books
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| Author: harshi 24 Oct 2009 | Member Level: Bronze | Rating: Points: 1 |
how many hours do u study for this gate exam??
this especially for "ghulam"!!
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| Author: Ghulam 26 Oct 2009 | Member Level: Silver | Rating: Points: 4 |
Hi Harshi,
No need to burden youself too much preparing for GATE.
It hardly needs an hour or two everyday to clear your concepts topic by topic if you sit with a cool mind with all your devotions.
GATE is after all " A test to test your concepts".
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| Author: Ghulam 26 Oct 2009 | Member Level: Silver | Rating: Points: 5 |
Question of the Day on TRANSMISSION LINES:
A telephone line has the following parameters:
R=60 ohm/mt, G=600 microSiemens/mt, L=0.3 microHenry/mt,
C=0.75 nanoFarad/mt
If the line operate at 10 MHz, the characteristic impedance Zo is:
A.21.0 - j17.6 ohms
B.29.6 - j21.4 ohms
C.10.8 - j7.9 ohms
D.14.0 - j42 ohms
Ans) B)29.6 - j21.4 ohms
Expln:
R+jwL= 60 + j*2*pi*(10*10^6)(3*10^-7) [Since, w=2*pi*f]
= |62| <17.44deg
G+jwC= [600*10^-6] + j*2pi*(10*10^6)*(0.75*10^-9)
= |0.047| <89.27deg
Zo=(R+jwL)/(G+jwC)
= 29.6 - j21.4 (Ans)
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| Author: Ghulam 27 Oct 2009 | Member Level: Silver | Rating: Points: 4 |
Question of the Day on AMPLITUDE MODULATION:
Q.A 2MHz carrier is amplitude modulated by a 500Hz modulating signal to a depth of 60%. If the unmodulated carrier power is 1 kW, the power of the modulated signal is
(A)1.83 kW
(B)1.36 kW
(C)1.18 kW
(D)1.26 kW
Ans. (C)1.18 kW
Expln:
Pc=1 kW, m=60%=0.6
Pt=Pc[1+{(m^2)/2}]^0.5
=1[1+{(0.6*0.6)/2}]^0.5
=1.18 kW (Ans)
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| Author: Ghulam 28 Oct 2009 | Member Level: Silver | Rating: Points: 6 |
Question of the Day on TRANSISTORS:
Q)Two Nmos Transistors are connected in series and the gates of both the transistors are connected to 5V. One end of
the Transistor is connected to a 10V supply and the Vth=1V. What is the voltage at the other end?
A.3 Volts
B.4 Volts
C.5 Volts
D.6 Volts
Ans) B)4 Volts
Expln:
Consider a single NMOS as a switch. The max voltage at the other end can reach max of VG - Vt, after that NMOS will be off.
So if the voltage at one end is less than Vg-Vt it passes that value to the other end, but if it is more, it reaches Vg-Vt at the end and stops there bcoz after that the MOS switch will be off. So in this case, first NMOS which has 12v, at the input, gives 4v out at its source, the other Transistor which has 4v at the input transmits samething to the other end as it is.
So final voltage is 4V.
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