Electronics Engineering

Qam only

A capacitor is used to restore the DC charge for shifting the DC level of the AC signal. ie. it helps to maintain the level depending on the its time constant.

An oscilloscope with 2 vertical inputs that are multiplexed to the CRT, either by chopping them together on every sweep or alternating between them on every other sweep. Some oscilloscopes also support a differential mode where the 2 inputs are subtracted to produce a single combined signal.

binary coded decimal counter with carry propagated bit to bit by ripple method instead of carry lookahead combinatorial logic method. this is easy to build but has long settling time with invalid codes occurring before it settles. for example these counters will generate a 1010 code between 1001 and 0000.

9, 8 to shift the bits in & 1 to move the byte into another register.

power supplied=sum of powers delivered to individual elements

two crosscoupled NAND or NOR gates.

Photo transistor always work with solar power

No

There is no 'equivalent resistance' for three resistors connected in star.

By changing the length of wire, say reducing it, the resistance will drop and that will increase current flow but the voltage is less likely to change V=IR.

no i dont think so

forward drop is the same as any other silicon diode, about 0.7V

transformers cannot amplify!

Since we know that the amplifier gain is given by

A=Output voltage/input voltage (where A is the amplifier gain)

So, it can be written as output voltage=A*input voltage,

so when the output part increases gain increases but when input part increases gain decreases

· Gruvlok Coupling

Hatdog

Cheese dog

Ga kalog kalog

Imong itlog

for a two input gate to represent as an n-input gate excatly n-1 two input gates are required. this implies that for a two input OR gate to represent a four input OR gate exactly three two input OR gates are required

let F is =a+b+c+d

=(((a+b)+c)+d)

=((a+b)+(c+d))

in both the above cases + is used three times so three two input OR gates make a four input OR gates.

This discussion doesnot hold good for NAND gates an example can illlustrate the reson:-

take F=(a.b.c.d)'=a'+b'+c'+d' --------------------------->(1)

(this is obtained by a four input NAND gate)

let us take this in the manner we did it for an OR gate and we will then verify the result.

=((a.b)'(c.d)')'

=((a'+b').(c'+d'))'

=(a'+b')'+(c'+d')'

=ab+cd <------------------------(2)

(1)is not equal to (2) so we can say that a NAND gate cannot be replaced in the manner as OR gate is replaced

by using voltmeter

no

the internal RAM is for data storage and for many applications is all that is needed.

well to get the answer first know the principle by which a capacitor(consider a capacitor without dielectric) gets charged .let a capacitor with plates p1 and p2 and resistor in parlell are connected to a DC source , when the switch is closed the circuit is in ON state and current starts flowing ,assume the flow of current as the movement of negative charges then the concept would be much clear.

assume that p1 is connected with positive terminal of the battery and p2 with the negative terminal ,now as the switch is closed the negative charge on p1 is attracted by the positive terminal of the battery and is driven to the other plate p2 of capacitor.as this process continues charge seperation increases and potential difference starts getting developed and after a very long time the potential diff. across capacitor becomes equal to the applied voltage V.

during discharging first of all we short the battery , this mplies that we are short circuiting plates p1 and p2 so again charge redistribution takes place and after a longtime both the plates become electrically nuetral