Displaying sensor data on screen in Arduino - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to read the sensor value.

Arduino

int sensorValue = analogRead([1]);

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

AA0

B5

C13

DLED_BUILTIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using a digital pin number instead of an analog pin.

Using LED_BUILTIN instead of the sensor pin.

The sensor is connected to analog pin A0, so we use analogRead(A0) to read its value.

2fill in blank

medium

Complete the code to initialize the serial communication.

Arduino

Serial.begin([1]);

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A19200

B4800

C115200

D9600

Attempts:

3 left

💡 Hint

Common Mistakes

Using a baud rate that does not match the serial monitor setting.

Choosing a very low or very high baud rate without reason.

9600 baud rate is the standard speed for serial communication in many Arduino examples.

3fill in blank

hard

Fix the error in printing the sensor value to the serial monitor.

Arduino

Serial.println([1]);

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

AanalogRead

BsensorValue

Cvalue

Dsensor

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that was never declared.

Trying to print the function name instead of the variable.

The variable holding the sensor reading is sensorValue, so we print that.

4fill in blank

hard

Fill both blanks to create a dictionary-like structure to store sensor names and values.

Arduino

String sensorData = "[1]: " + String([2]);

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ATemperature

BsensorValue

CHumidity

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name instead of a string for the sensor name.

Not converting the sensor value to a string before concatenation.

We label the data as "Temperature" and convert the sensor value to a string to combine them.

5fill in blank

hard

Fill all three blanks to read sensor data, convert it, and print with a label.

Arduino

int [1] = analogRead(A0);
float [2] = [3] * (5.0 / 1023.0);
Serial.println("Voltage: " + String([2]));

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

AsensorRaw

BsensorVoltage

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same variable name for both raw and converted values.

Not converting the raw value correctly to voltage.

We read raw sensor data into sensorRaw, convert it to voltage in sensorVoltage, and print the voltage.

Practice

1. What is the main purpose of displaying sensor data on a screen in an Arduino project?

easy

A. To store data permanently

B. To see real-time measurements easily

C. To power the sensor

D. To connect to the internet

Solution

1. Step 1: Understand sensor data usage

   Sensor data is collected to monitor or measure something in real time.

2. Step 2: Purpose of displaying data

   Showing data on a screen helps users see these measurements immediately.

3. Final Answer:

   To see real-time measurements easily -> Option B

4. Quick Check:

   Displaying sensor data = real-time viewing [OK]

Hint: Displaying sensor data means showing live values on screen [OK]

Common Mistakes:

• Confusing display with data storage
• Thinking display powers the sensor
• Assuming display connects to internet

2. Which Arduino command correctly clears the screen before showing new sensor data?

easy

A. lcd.reset();

B. lcd.clean();

C. lcd.erase();

D. lcd.clear();

Solution

1. Step 1: Identify the correct LCD clear command

   The standard command to clear an LCD screen in Arduino is lcd.clear();.

2. Step 2: Check other options

   Commands like reset(), erase(), or clean() are not valid for LCD clearing.

3. Final Answer:

   lcd.clear(); -> Option D

4. Quick Check:

   Clear screen command = lcd.clear(); [OK]

Hint: Use lcd.clear() to refresh display before new data [OK]

Common Mistakes:

• Using non-existent lcd.reset() or lcd.erase()
• Forgetting parentheses in lcd.clear()
• Confusing clear with reset functions

3. What will be displayed on the LCD if the following code runs and the sensor value is 512?

int sensorValue = analogRead(A0);
lcd.clear();
lcd.setCursor(0, 0);
lcd.print("Value: ");
lcd.print(sensorValue);

medium

A. Value: 512

B. sensorValue

C. Value: A0

D. 512

Solution

1. Step 1: Understand variable assignment

   The variable sensorValue stores the analog reading from pin A0, which is 512.

2. Step 2: Analyze LCD print commands

   The code clears the screen, sets cursor to start, prints "Value: " text, then prints the number 512.

3. Final Answer:

   Value: 512 -> Option A

4. Quick Check:

   lcd.print("Value: ") + lcd.print(sensorValue) = Value: 512 [OK]

Hint: lcd.print(variable) shows its value, not the name [OK]

Common Mistakes:

• Printing variable name instead of value
• Confusing pin name with value
• Missing lcd.clear() causing old data

4. Identify the error in this code snippet that tries to display sensor data:

int sensorValue = analogRead(A0);
lcd.setCursor(0, 0);
lcd.print("Temp: ");
lcd.print(sensorValue);
lcd.clear();

medium

A. lcd.clear() should be before lcd.print()

B. analogRead() needs a semicolon

C. lcd.setCursor() is missing parameters

D. sensorValue must be declared as float

Solution

1. Step 1: Check lcd.clear() placement

   lcd.clear() clears the screen, so it must be called before printing new data to avoid erasing it immediately.

2. Step 2: Verify other code parts

   analogRead() has semicolon, setCursor() has correct parameters, and sensorValue as int is fine for analog values.

3. Final Answer:

   lcd.clear() should be before lcd.print() -> Option A

4. Quick Check:

   Clear screen before printing = correct order [OK]

Hint: Always clear screen before printing new data [OK]

Common Mistakes:

• Clearing screen after printing erases output
• Misplacing lcd.setCursor() parameters
• Wrong variable type assumptions

5. You want to display temperature in Celsius on an LCD using a sensor that outputs voltage. The sensor voltage is read by analogRead(A1) and converted by tempC = (sensorValue * 5.0 / 1023) * 100;. Which code snippet correctly displays the temperature with one decimal place?

hard

A. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(int(tempC)); lcd.print(" C");

B. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC); lcd.print(" Celsius");

C. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC, 1); lcd.print(" C");

D. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC + " C");

Solution

1. Step 1: Understand temperature conversion

   tempC is a float value representing temperature in Celsius with decimals.

2. Step 2: Display with one decimal place

   lcd.print(value, 1) prints float with 1 decimal. Adding " C" after shows units clearly.

3. Step 3: Check other options

   lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC); lcd.print(" Celsius"); prints full float but no decimal control; lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(int(tempC)); lcd.print(" C"); truncates decimals; lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC + " C"); tries to add string to float incorrectly.

4. Final Answer:

   lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC, 1); lcd.print(" C"); -> Option C

5. Quick Check:

   lcd.print(float, 1) = one decimal place [OK]

Hint: Use lcd.print(value, 1) to show one decimal place [OK]

Common Mistakes:

• Not controlling decimal places
• Trying to add string directly to float
• Using int() truncates decimals