2023 A/L Physics Past Paper · Question 46 (English Medium)

Answer: (3) Circuit (3)

What the question asks

We want: the circuit that gives the table.

The inputs are $A$ and $B$, each $0$ or $1$. The output is $F$.

Read the table: $F = 1$ when $B = 0$, and $F = 0$ when $B = 1$. The value of $A$ does not change $F$.

So the table says $F = \bar{B}$.

Relevant theory

F = Ā Where Ā is the inversion of A

The logic operation of the NAND gate is the inverse of that of the AND gate. It will be

The first quote says that a NOT gate gives the inversion of its input: $F = \bar{A}$, so $0$ becomes $1$ and $1$ becomes $0$.

The second quote says that a NAND gate does the inverse of an AND gate. An AND gate gives $1$ when both inputs are $1$, so a NAND gate gives $0$ when both inputs are $1$, and $1$ in the other three cases.

Each circuit here has two NAND gates feeding a third NAND gate. The last gate gives $F = 1$ when at least one of the first two gates gives $0$, that is, when both inputs of the top gate are $1$, or both inputs of the bottom gate are $1$.

For example, think of a doorbell with two buttons, one at the front door and one at the back. It rings when someone presses the front button, or when someone presses the back button.

The last NAND gate works like that bell: it gives $1$ when either of the two front gates "presses" by giving $0$.

Reading the options

Read each circuit. Follow the wires and write the two inputs of the top gate and of the bottom gate. A small triangle with a circle on a wire is a NOT gate, so it turns $A$ into $\bar{A}$ or $B$ into $\bar{B}$.

Circuit (1): top $A, B$. Bottom $\bar{A}, B$. So $F = 1$ when $B = 1$: $F = B$.

Circuit (2): top $A, B$. Bottom $\bar{A}, \bar{B}$. So $F = 1$ when $A = B$.

Circuit (3): top $\bar{A}, \bar{B}$. Bottom $A, \bar{B}$. So $F = 1$ when $B = 0$: $F = \bar{B}$.

Circuit (4): top $\bar{A}, \bar{B}$. Bottom $\bar{A}, B$. So $F = 1$ when $A = 0$: $F = \bar{A}$.

Circuit (5): top $A, B$. Bottom $A, \bar{B}$. So $F = 1$ when $A = 1$: $F = A$.

The four rows match the table: $1, 0, 1, 0$.

$A = 0, B = 0$

$F = 1$

Top gate gets $1, 1$ and gives $0$, so $F = 1$.

$A = 0, B = 1$

$F = 0$

Top gate gets $1, 0$ and bottom gets $0, 0$. Both give $1$, so $F = 0$.

$A = 1, B = 0$

Bottom gate gets $1, 1$ and gives $0$, so $F = 1$.

$A = 1, B = 1$

Top gate gets $0, 0$ and bottom gets $1, 0$. Both give $1$, so $F = 0$.

Now match it with the table.

The table is $F = \bar{B}$. Circuit (3) gives $F = \bar{B}$, and its four rows are $1, 0, 1, 0$, the same as the table. It matches.

Circuits (1), (2), (4) and (5) give $B$, $A = B$, $\bar{A}$ and $A$. None of these is $\bar{B}$.

Why the other options fail

Circuit (1) gives $F = B$: at $A = 0, B = 0$ it gives $0$, but the table has $1$.

Circuit (2) gives $1$ when $A = B$: at $A = 1, B = 0$ it gives $0$, but the table has $1$.

Circuit (4) gives $F = \bar{A}$: at $A = 0, B = 1$ it gives $1$, but the table has $0$.

Circuit (5) gives $F = A$: at $A = 0, B = 0$ it gives $0$, but the table has $1$.

Conclusion

$F = \bar{B}$

The table is $F = \bar{B}$. Circuit (3) has $\bar{A}, \bar{B}$ on the top NAND gate and $A, \bar{B}$ on the bottom one, so it gives $F = \bar{B}$.

The answer is (3).

The quotes give the NOT gate and the NAND gate. The rule that two NAND gates into a third NAND gate give $1$ when either first gate has both inputs at $1$ is our reasoning from the NAND quote, checked row by row above.

Teacher's note

First read the table as a short rule, here $F = \bar{B}$. Then, for each circuit, write the two inputs of each first NAND gate. The output is $1$ when one of those gates has both inputs at $1$.

Exam tip