Displaying sensor data on screen in Arduino - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output on the LCD screen?

Given the Arduino code below, what will be displayed on the LCD screen after running?

Arduino

#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);
int sensorValue = 512;
void setup() {
  lcd.begin(16, 2);
  lcd.print("Sensor:");
  lcd.setCursor(0, 1);
  lcd.print(sensorValue);
}
void loop() {
  // no changes
}

A

Sensor:
1023

B Sensor: 512

C

Sensor:
512

D Sensor: 1023

💡 Hint

Look at how the cursor is set before printing the sensor value.

The lcd.print("Sensor:") writes on the first line. lcd.setCursor(0, 1) moves to the start of the second line, then lcd.print(sensorValue) prints 512 there.

🧠 Conceptual

intermediate

Why use lcd.setCursor() before printing sensor data?

In Arduino LCD programming, why do we use lcd.setCursor() before printing sensor data?

A To move the cursor to the desired position on the screen before printing

B To clear the screen before printing new data

C To change the font size of the printed text

D To turn off the LCD backlight

💡 Hint

Think about where the text appears on the LCD.

lcd.setCursor(column, row) moves the cursor to a specific position so the next printed text appears there.

❓ Predict Output

advanced

What will the serial monitor show?

Consider this Arduino code snippet reading a sensor and printing to Serial Monitor. What is the output?

Arduino

void setup() {
  Serial.begin(9600);
}
void loop() {
  int sensorValue = analogRead(A0);
  Serial.print("Value: ");
  Serial.println(sensorValue);
  delay(1000);
}

A No output, code has error

B Value: 0 to 1023 (changing every second)

C Value: 1023 only

D Value: 0 only

💡 Hint

analogRead reads values from 0 to 1023 depending on sensor input.

analogRead(A0) returns a value between 0 and 1023 depending on the sensor voltage. The code prints this value every second.

🔧 Debug

advanced

Identify the error in this LCD display code

What error will this Arduino code cause when trying to display sensor data on LCD?

Arduino

#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);
int sensorValue;
void setup() {
  lcd.begin(16, 2);
  sensorValue = analogRead(A0);
  lcd.print("Value: ");
  lcd.print(sensorValue);
}
void loop() {
  // empty
}

A analogRead(A0) cannot be called in setup() causing runtime error

B lcd.begin() is missing parameters causing compile error

C lcd.print() requires a delay after each call or it crashes

D lcd.print(sensorValue) prints a number but cursor is not moved, so text overlaps

💡 Hint

Think about how lcd.print works when printing multiple things in a row.

lcd.print("Value: ") prints text starting at cursor 0,0. Then lcd.print(sensorValue) continues printing immediately after. This can cause overlapping if not managed carefully.

🚀 Application

expert

How to display sensor data updating every 2 seconds on LCD?

Which code snippet correctly reads a sensor and updates the LCD display every 2 seconds, clearing old data before printing new?

A

#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);
void setup() {
  lcd.begin(16, 2);
}
void loop() {
  int val = analogRead(A0);
  lcd.clear();
  lcd.setCursor(0, 0);
  lcd.print("Sensor:");
  lcd.setCursor(0, 1);
  lcd.print(val);
  delay(2000);
}

B

#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);
void setup() {
  lcd.begin(16, 2);
}
void loop() {
  int val = analogRead(A0);
  lcd.setCursor(0, 0);
  lcd.print("Sensor:");
  lcd.setCursor(0, 1);
  lcd.print(val);
  delay(2000);
}

C

#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);
void setup() {
  lcd.begin(16, 2);
  lcd.clear();
}
void loop() {
  int val = analogRead(A0);
  lcd.setCursor(0, 0);
  lcd.print("Sensor:");
  lcd.setCursor(0, 1);
  lcd.print(val);
  delay(2000);
}

D

#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);
void setup() {
  lcd.begin(16, 2);
}
void loop() {
  int val = analogRead(A0);
  lcd.clear();
  lcd.print("Sensor:");
  lcd.setCursor(0, 1);
  lcd.print(val);
  delay(2000);
}

💡 Hint

Clearing the screen before printing new data avoids leftover characters.

Option A clears the LCD each loop, sets cursor positions explicitly, and prints the label and value properly, updating every 2 seconds.

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