Study the following circuit:

Definitions

S1 = 0 / S2 = 0 means the button is pressed down

S1 = 1 / S2 = 1 means the button is not pressed down

Question

Suppose S1 = 0

Suppose S2 = 0

void setup() {
Serial.begin(9600);
pinMode(13, OUTPUT);
pinMode(2, INPUT);
pinMode(3, INPUT);
}

void loop() {
delay(2000);
int button1_val = !digitalRead(2); // button1_val = 1 when pressed, 0 when not pressed
int button2_val = !digitalRead(3); // button2_val = 1 when pressed, 0 when not pressed

if (button1_val == 1 && button2_val == 1) { // AND = && (both of them)
digitalWrite(13, 1); // turn on
}
else {
digitalWrite(13, 0); // turn off
}
}

Use the following codes to describe the state of the LED, and values of the variables.

Your answer should either be 0, 1, 2, 3, 4, 5, 6, or 7. If your answer is 0, Canvas might flag the question as unanswered. It is safe to ignore this message.

CODE LED button1_val button2_val
0 OFF 0 0
1 OFF 0 1
2 OFF 1 0
3 OFF 1 1
4 ON 0 0
5 ON 0 1
6 ON 1 0
7 ON 1 1

Question

Study the following circuit:

Definitions

S1 = 0 / S2 = 0 means the button is pressed down

S1 = 1 / S2 = 1 means the button is not pressed down

Question

Suppose S1 = 0

Suppose S2 = 0

void setup() {
  Serial.begin(9600);
  pinMode(13, OUTPUT);
  pinMode(2, INPUT);
  pinMode(3, INPUT);
}

void loop() {
  delay(2000);
  int button1_val = !digitalRead(2); // button1_val = 1 when pressed, 0 when not pressed
  int button2_val = !digitalRead(3); // button2_val = 1 when pressed, 0 when not pressed
  
  if (button1_val == 1 && button2_val == 1) { // AND = && (both of them)
    digitalWrite(13, 1); // turn on
  }
  else {
    digitalWrite(13, 0); // turn off
  }
}

Use the following codes to describe the state of the LED, and values of the variables.

Your answer should either be 0, 1, 2, 3, 4, 5, 6, or 7. If your answer is 0, Canvas might flag the question as unanswered. It is safe to ignore this message.

CODE	LED	button1_val	button2_val
0	OFF	0	0
1	OFF	0	1
2	OFF	1	0
3	OFF	1	1
4	ON	0	0
5	ON	0	1
6	ON	1	0
7	ON	1	1

BlackTom AI · Accepted Answer

Because both buttons are pressed (S1 = 0 and S2 = 0) the input pins will read LOW (0) and the code negates those readings so each button value becomes 1.

1) Relevant concepts
- digitalRead(pin) returns 0 when the pin is pulled to ground, 1 when pulled to Vcc. The circuit uses pull-up resistors to 5V and the switches connect the pin to ground when pressed, so a pressed button gives digitalRead = 0.  
- The code uses the logical NOT operator (!) so button_val = !digitalRead(pin). That converts a physical pressed state (digitalRead = 0) into button_val = 1.  
- The if condition requires both button1_val == 1 && button2_val == 1 to turn the LED on via digitalWrite(13, 1).

2) Step-by-step reasoning with intermediate values
- Given S1 = 0 (pressed) → digitalRead(2) = 0 → button1_val = !0 = 1.  
- Given S2 = 0 (pressed) → digitalRead(3) = 0 → button2_val = !0 = 1.  
- Evaluate if condition: (button1_val == 1 && button2_val == 1) → (1 && 1) is true.  
- Because condition is true, the code executes digitalWrite(13, 1) which switches the LED ON.

3) Why this answer is correct
- The code logic explicitly turns the LED only when both inverted readings are 1. With both physical buttons pressed, both inverted values are 1, satisfying the AND condition and turning the LED on.

4) Why other options are incorrect
- Codes 0–3 list LED = OFF; they cannot be correct because the AND condition is satisfied and the LED is turned ON.  
- Codes 4–6 show LED = ON but have at least one button_val equal to 0; they do not match the actual inverted readings (both should be 1). Only the entry with LED = ON and button1_val = 1 and button2_val = 1 matches the actual state.

5) Conclusion
- The correct code is 7, which corresponds to LED = ON, button1_val = 1, button2_val = 1. This matches the provided answer: 7.

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