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To get an output Y = 1 from the circuit shown below, the input must be
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A
A = 1, B = 0, C = 0
B
A = 0, B = 1, C = 0
C
A = 0, B = 0, C = 1
D
A = 1, B = 0, C = 1
Detailed Solution
In the circuit, inputs A and B go to an OR gate, and the output of this OR gate together with C goes to an AND gate.
Boolean expression: $Y = (A + B)\cdot C$
For an AND gate the output is 1 only when both of its inputs are 1, so we need $A + B = 1$ and $C = 1$.
A = 1, B = 0, C = 0: $Y = (1 + 0)\cdot0 = 0$
A = 0, B = 1, C = 0: $Y = (0 + 1)\cdot0 = 0$
A = 0, B = 0, C = 1: $Y = (0 + 0)\cdot1 = 0$
A = 1, B = 0, C = 1: $Y = (1 + 0)\cdot1 = 1$
Hence the input must be A = 1, B = 0, C = 1.
Boolean expression: $Y = (A + B)\cdot C$
For an AND gate the output is 1 only when both of its inputs are 1, so we need $A + B = 1$ and $C = 1$.
A = 1, B = 0, C = 0: $Y = (1 + 0)\cdot0 = 0$
A = 0, B = 1, C = 0: $Y = (0 + 1)\cdot0 = 0$
A = 0, B = 0, C = 1: $Y = (0 + 0)\cdot1 = 0$
A = 1, B = 0, C = 1: $Y = (1 + 0)\cdot1 = 1$
Hence the input must be A = 1, B = 0, C = 1.