i jus know da answer.... nto how 2 write coz i m not CBSE.....
Q2 **mera ans jaldbazi mei wrong tha... pl. refer SIR'S answer**
Q1 b) (A' + B')' = A.B
Q1 a) [(A + B)']' = (A'.B')' = A + B
kal paper hai aur mujhe logic gates ko chor kar sab kuch aata hai.......plzzzzz meri help karo aur cbse type answer do..........
all are ncert problems............mere paas koi help book nahin hai........[2][2]
Q1
Q2
i jus know da answer.... nto how 2 write coz i m not CBSE.....
Q2 **mera ans jaldbazi mei wrong tha... pl. refer SIR'S answer**
Q1 b) (A' + B')' = A.B
Q1 a) [(A + B)']' = (A'.B')' = A + B
1 a)
{(A+B)c}c
=A+B
1 b) (Ac+Bc)c = A.B
2 (X.X)c=Y
thus Xc=Y
so this is complement operator or not operator.
dude one Q . atleast you know the basic gates of AND OR and NOT no ? only asking . If yes then you should have no problem here nevertheless I'll post the answers .
@ sri no dude............i dont know anything...........i just know that basic truth table for every gate.......but nothing for combo..........i cant understand these formulae even.......
plz explain sir
hey dude...one silly doubt nand gate mein to ek hi semi circle part hota hai naa par isme uske peeche box kyun bana hai??
The box is not there... It is two wires from the same source.. So it is looking like a box!
1 a)
From the question, {(A+B)c}c
=A+B (bexaus (X')'=X
1 b) (Ac+Bc)c = A.B (DeMovier's theorem)
2 (X.X)c=Y
thus Xc=Y (Using X.X=X)
so this is complement operator or not operator.
Not gate complement deta hai
so if the input to a nor gate is X then the output is X'
Yahan pe input is (A+B)' so the output will be {(A+B)'}'
Eureka.. sorry I gave a differetn answer! :(
Eureka... jaroor CS nahi liya hoga.. ;P
Phy ed.. ya something like that... :)