delay() function behavior in Arduino - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to pause the program for 1000 milliseconds.

Arduino

delay([1]);

Drag options to blanks, or click blank then click option'

A1000

B10

C5000

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 will not pause the program.

Using 10 will pause only 10 milliseconds, which is very short.

The delay() function pauses the program for the number of milliseconds given as the argument. Here, 1000 means 1 second.

2fill in blank

medium

Complete the code to pause the program for half a second.

Arduino

delay([1]);

Drag options to blanks, or click blank then click option'

A1000

B50

C1500

D500

Attempts:

3 left

💡 Hint

Common Mistakes

Using 50 pauses only 0.05 seconds, which is too short.

Using 1500 pauses 1.5 seconds, which is too long.

500 milliseconds is half a second, so delay(500); pauses the program for 0.5 seconds.

3fill in blank

hard

Fix the error in the code to correctly pause for 2 seconds.

Arduino

delay([1]);

Drag options to blanks, or click blank then click option'

A2000

B200

C20

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 2 pauses only 2 milliseconds, which is too short.

Using 200 pauses 0.2 seconds, which is less than 2 seconds.

The delay() function expects milliseconds. 2000 milliseconds equals 2 seconds.

4fill in blank

hard

Fill both blanks to create a delay of 3 seconds and then 500 milliseconds.

Arduino

delay([1]);
delay([2]);

Drag options to blanks, or click blank then click option'

A3000

B500

C1500

D1000

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1500 instead of 3000 for 3 seconds.

Mixing up the order of delays.

The first delay(3000); pauses for 3 seconds, and the second delay(500); pauses for half a second.

5fill in blank

hard

Fill all three blanks to create a sequence of delays: 1 second, 2 seconds, and 750 milliseconds.

Arduino

delay([1]);
delay([2]);
delay([3]);

Drag options to blanks, or click blank then click option'

A1000

B2000

C750

D500

Attempts:

3 left

💡 Hint

Common Mistakes

Using 500 instead of 750 for the last delay.

Confusing the order of delays.

The delays are 1 second (1000 ms), 2 seconds (2000 ms), and 750 milliseconds respectively.

Practice

1. What does the delay(1000); function do in an Arduino program?

easy

A. Pauses the program for 1000 milliseconds (1 second)

B. Stops the program permanently

C. Speeds up the program by 1000 times

D. Restarts the Arduino board

Solution

1. Step 1: Understand the delay() parameter

   The number inside delay() is the time in milliseconds to pause the program.

2. Step 2: Interpret delay(1000)

   1000 milliseconds equals 1 second, so the program pauses for 1 second.

3. Final Answer:

   Pauses the program for 1000 milliseconds (1 second) -> Option A

4. Quick Check:

   delay(1000) = 1 second pause [OK]

Hint: delay(x) pauses for x milliseconds, 1000ms = 1 second [OK]

Common Mistakes:

• Thinking delay stops the program forever
• Confusing milliseconds with seconds
• Assuming delay speeds up the program

2. Which of the following is the correct syntax to pause an Arduino program for half a second?

easy

A. delay(0.5);

B. delay = 500;

C. delay(500s);

D. delay(500);

Solution

1. Step 1: Check correct function usage

   delay() takes an integer number of milliseconds inside parentheses.

2. Step 2: Validate each option

   A uses delay(0.5); which is a float and incorrect. B assigns delay which is invalid. C uses delay(500); which is correct for 500 milliseconds. D uses '500s' which is invalid syntax.

3. Final Answer:

   delay(500); -> Option D

4. Quick Check:

   delay(500) is correct syntax [OK]

Hint: Use delay(milliseconds); with integer inside parentheses [OK]

Common Mistakes:

• Using decimal numbers instead of integers
• Assigning delay like a variable
• Adding units like 's' inside delay()

3. What will be the output on the Serial Monitor after running this Arduino code?

void setup() {
  Serial.begin(9600);
  Serial.println("Start");
  delay(2000);
  Serial.println("End");
}
void loop() {}

medium

A. Start and End printed immediately together

B. Only Start is printed, End never prints

C. Start immediately, then End after 2 seconds

D. No output because delay stops Serial

Solution

1. Step 1: Analyze Serial prints and delay

   "Start" prints immediately, then delay(2000) pauses 2 seconds before next line.

2. Step 2: Understand delay effect on output

   After 2 seconds pause, "End" prints. Both lines appear but with 2 seconds gap.

3. Final Answer:

   Start immediately, then End after 2 seconds -> Option C

4. Quick Check:

   delay pauses program, output delayed [OK]

Hint: delay pauses code, so output after delay appears later [OK]

Common Mistakes:

• Thinking delay stops Serial output completely
• Assuming both prints happen instantly
• Believing delay affects only loop(), not setup()

4. Identify the problem in this Arduino code snippet:

void setup() {
  pinMode(13, OUTPUT);
}
void loop() {
  digitalWrite(13, HIGH);
  delay(1000)
  digitalWrite(13, LOW);
  delay(1000);
}

medium

A. pinMode should be in loop()

B. Missing semicolon after delay(1000)

C. digitalWrite needs delay before it

D. delay() cannot be used in loop()

Solution

1. Step 1: Check syntax line by line

   delay(1000) is missing a semicolon at the end, causing a syntax error.

2. Step 2: Validate other statements

   pinMode is correctly in setup(), digitalWrite and delay usage is correct except missing semicolon.

3. Final Answer:

   Missing semicolon after delay(1000) -> Option B

4. Quick Check:

   Every statement must end with ; [OK]

Hint: Check for missing semicolons after delay() calls [OK]

Common Mistakes:

• Putting pinMode inside loop unnecessarily
• Thinking delay() can't be in loop
• Forgetting semicolons after statements

5. You want to blink an LED connected to pin 9 exactly 3 times with 0.5 second ON and 0.5 second OFF, then stop. Which code snippet correctly uses delay() to do this?

hard

A. for(int i=0; i<3; i++) { digitalWrite(9, HIGH); delay(500); digitalWrite(9, LOW); delay(500); }

B. while(true) { digitalWrite(9, HIGH); delay(500); digitalWrite(9, LOW); delay(500); }

C. digitalWrite(9, HIGH); delay(1500); digitalWrite(9, LOW); delay(1500);

D. for(int i=0; i<=3; i++) { digitalWrite(9, HIGH); delay(1000); digitalWrite(9, LOW); delay(1000); }

Solution

1. Step 1: Understand blinking requirements

   LED must turn ON and OFF 3 times, each ON and OFF lasting 0.5 seconds (500 ms).

2. Step 2: Analyze each option

   A uses a for loop 3 times with 500ms delay ON and OFF, matching requirements. B loops forever, no stop. C delays 1500ms which is too long and only blinks once. D loops 4 times (i<=3) with 1000ms delays, wrong timing and count.

3. Final Answer:

   for(int i=0; i<3; i++) { digitalWrite(9, HIGH); delay(500); digitalWrite(9, LOW); delay(500); } -> Option A

4. Quick Check:

   Loop 3 times with 500ms delay ON/OFF [OK]

Hint: Use for loop with delay(500) ON and OFF, repeat 3 times [OK]

Common Mistakes:

• Using infinite loops instead of fixed count
• Wrong delay times for ON/OFF
• Looping one extra time with <= instead of <