Practice portal › Semiconductor Electronics › Digital Electronics and Logic Gates
Asked in NEET 2010 · Gate combinations and truth tables
Idea: reduce the two gates to one expression before testing any of the four sets.
The OR gate takes A and B and gives A+B.
The AND gate takes that signal together with C, so Y=(A+B)· C.
Y=1 needs both factors high: C=1, and at least one of A and B equal to 1.
A=1, B=0, C=1 gives (1+0)·1=1.
Each of the other three sets fails one of the two conditions, so this is the only one that works.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer