ExpertMedium3 marksStructured
AQA GCSE · Question 03.2 · Computer systems
A game uses three sensors. A red light (R) in the game switches on if all of the following conditions are true:
- sensor D is off
- sensor L is on
- sensor W is on.
Complete the logic circuit for this game. You must use the correct symbols for the logic gates.
A game uses three sensors. A red light (R) in the game switches on if all of the following conditions are true:
- sensor D is off
- sensor L is on
- sensor W is on.
Complete the logic circuit for this game. You must use the correct symbols for the logic gates.
How to approach this question
1. **Analyze the conditions:**
* "sensor D is off": This means we need the logical opposite of D, which is NOT D. This requires a NOT gate for input D.
* "sensor L is on" and "sensor W is on": These inputs are used as they are.
* "**all** of the following conditions are true": The word "all" implies an AND condition. We need a gate that outputs true only when all its inputs are true. This is an AND gate.
2. **Construct the circuit:**
* Draw an input line for D and connect it to a NOT gate.
* Draw input lines for L and W.
* Take the output from the NOT gate (which is NOT D), the input line for L, and the input line for W, and connect all three to the inputs of a single 3-input AND gate.
* Connect the output of the AND gate to the output R.
Full Answer
The correct circuit diagram shows:
1. Input D connected to a NOT gate.
2. The output of the NOT gate, along with inputs L and W, are all connected to a 3-input AND gate.
3. The output of the AND gate is connected to R.
The problem states that the red light R turns on if (D is off) AND (L is on) AND (W is on).
- **"D is off"** translates to the Boolean expression `NOT D`. In a circuit diagram, this is represented by a NOT gate.
- **"L is on"** and **"W is on"** are represented by the inputs L and W directly.
- **"all ... are true"** translates to the Boolean operator `AND`. This is represented by an AND gate. Since there are three conditions to be met simultaneously, we need a 3-input AND gate.
The final logic circuit is constructed as follows:
1. The input D goes into a NOT gate.
2. The output of this NOT gate, along with the inputs L and W, all feed into a 3-input AND gate.
3. The output of the AND gate is R.
The Boolean expression for this circuit is `R = (NOT D) AND L AND W`.
<svg width="700" height="700" xmlns="http://www.w3.org/2000/svg"><style>text{font-family: Arial, sans-serif; font-size: 24px; text-anchor: middle; dominant-baseline: middle;}</style><text x="100" y="200">D</text><line x1="120" y1="200" x2="200" y2="200" stroke="black" stroke-width="2"/><path d="M 200 200 L 250 180 L 250 220 Z" stroke="black" stroke-width="2" fill="none"/><circle cx="260" cy="200" r="5" stroke="black" stroke-width="2" fill="none"/><line x1="265" y1="200" x2="350" y2="300" stroke="black" stroke-width="2"/><text x="100" y="400">L</text><line x1="120" y1="400" x2="350" y2="400" stroke="black" stroke-width="2"/><text x="100" y="600">W</text><line x1="120" y1="600" x2="350" y2="500" stroke="black" stroke-width="2"/><path d="M 350 275 L 425 275 L 425 525 L 350 525 L 350 400 A 125 125 0 0 1 350 275 Z" stroke="black" stroke-width="2" fill="none"/><line x1="425" y1="400" x2="580" y2="400" stroke="black" stroke-width="2"/><text x="600" y="400">R</text></svg>
Common mistakes
✗ Using the wrong logic gate symbols (e.g., American vs IEC standard if not specified, but consistency is key).
✗ Using an OR gate instead of an AND gate.
✗ Forgetting the NOT gate for input D.
✗ Using three separate 2-input AND gates instead of one 3-input AND gate.
Practice the full AQA GCSE Computer Science Paper 2
46 questions · hints · full answers · grading
More questions from this exam
Q01.1Convert the binary number 11010100 into decimal.EasyQ01.2Convert the binary number 10111001 into hexadecimal. You should show your working.MediumQ01.3State the largest decimal number that can be represented using 6 bits.EasyQ02.1Add together the following three binary numbers and give your answer in binary.
00110110
1001...MediumQ02.2Apply a binary shift three places to the right on the bit pattern 10101000. Give the result using...Easy