Why power matters for battery projects in Arduino - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set the LED pin as output.

Arduino

void setup() {
  pinMode([1], OUTPUT);
}

void loop() {
  // LED control code
}

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

A7

B13

CA0

D12

Attempts:

3 left

💡 Hint

Common Mistakes

Using analog pins like A0 instead of digital pins.

Choosing pins not connected to the LED.

The built-in LED on most Arduino boards is connected to pin 13, so we set pin 13 as output.

2fill in blank

medium

Complete the code to turn the LED on.

Arduino

void loop() {
  digitalWrite(13, [1]);
  delay(1000);
  digitalWrite(13, LOW);
  delay(1000);
}

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

ALOW

BINPUT

CHIGH

DOUTPUT

Attempts:

3 left

💡 Hint

Common Mistakes

Using LOW instead of HIGH to turn on the LED.

Confusing pin modes with output values.

To turn the LED on, we set the pin to HIGH.

3fill in blank

hard

Fix the error in the code to correctly read battery voltage on analog pin A0.

Arduino

int sensorValue = analogRead([1]);
float voltage = sensorValue * (5.0 / 1023.0);

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

A13

BA0

CA1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'A0' as a string instead of 0.

Using digital pin numbers for analogRead.

analogRead expects the analog pin number as an integer, so A0 is pin 0.

4fill in blank

hard

Fill both blanks to calculate battery voltage with a voltage divider.

Arduino

float batteryVoltage = sensorValue * ([1] / 1023.0) * ([2] + 10.0) / 10.0;

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

A5.0

B3.3

C20.0

D15.0

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong reference voltage.

Incorrect resistor values in formula.

The Arduino uses 5V reference, and the voltage divider resistors are 10k and 20k ohms, so we multiply by (20+10)/10.

5fill in blank

hard

Fill all three blanks to print battery voltage with 2 decimal places.

Arduino

Serial.print("Battery Voltage: ");
Serial.print([1], [2]);
Serial.println([3]);

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

AbatteryVoltage

B2

C" V"

Dvoltage

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable name.

Forgetting to print the unit.

We print the variable batteryVoltage with 2 decimals, then print the unit " V" on a new line.

Practice

1. Why is it important to know the power (Voltage x Current) your Arduino project uses when choosing a battery?

easy

A. To select a battery that can supply enough energy without running out quickly

B. To make the project run faster

C. To reduce the size of the Arduino board

D. To change the color of the LEDs automatically

Solution

1. Step 1: Understand power in battery projects

   Power is the product of voltage and current, showing how much energy the project uses.

2. Step 2: Connect power to battery choice

   Knowing power helps pick a battery that can supply enough energy for the project to run well and last long.

3. Final Answer:

   To select a battery that can supply enough energy without running out quickly -> Option A

4. Quick Check:

   Power knowledge = correct battery choice [OK]

Hint: Power tells how much energy your battery must supply [OK]

Common Mistakes:

• Thinking power affects Arduino speed
• Confusing power with physical size
• Assuming power changes LED colors

2. Which of the following Arduino code snippets correctly calculates power given voltage and current variables?

easy

A. float power = voltage - current;

B. float power = voltage + current;

C. float power = voltage / current;

D. float power = voltage * current;

Solution

1. Step 1: Recall power formula

   Power equals voltage multiplied by current (Power = Voltage x Current).

2. Step 2: Check code syntax

   The code must multiply voltage and current variables to get power, which is done by '*'.

3. Final Answer:

   float power = voltage * current; -> Option D

4. Quick Check:

   Power = voltage * current [OK]

Hint: Use * operator to multiply voltage and current [OK]

Common Mistakes:

• Using + instead of *
• Dividing voltage by current
• Subtracting current from voltage

3. Given the Arduino code below, what will be printed to the Serial Monitor?

float voltage = 5.0;
float current = 0.2;
float power = voltage * current;
Serial.println(power);

medium

A. 5.0

B. 1.0

C. 0.2

D. 25.0

Solution

1. Step 1: Calculate power value

   Power = 5.0 (voltage) x 0.2 (current) = 1.0

2. Step 2: Understand Serial.println output

   The code prints the value of power, which is 1.0, to the Serial Monitor.

3. Final Answer:

   1.0 -> Option B

4. Quick Check:

   5.0 * 0.2 = 1.0 [OK]

Hint: Multiply voltage and current to find printed power [OK]

Common Mistakes:

• Printing current or voltage instead of power
• Multiplying incorrectly
• Confusing output with variable names

4. The following Arduino code is meant to calculate power but does not work as expected. What is the error?

float voltage = 9.0;
float current = 0.5;
float power = voltage + current;
Serial.println(power);

medium

A. Incorrect variable types for voltage and current

B. Missing semicolon after variable declarations

C. Using '+' instead of '*' to calculate power

D. Serial.println cannot print float values

Solution

1. Step 1: Identify the power calculation mistake

   Power should be voltage multiplied by current, but the code uses addition (+) instead of multiplication (*).

2. Step 2: Confirm other syntax is correct

   Semicolons are present, variable types are correct, and Serial.println can print floats.

3. Final Answer:

   Using '+' instead of '*' to calculate power -> Option C

4. Quick Check:

   Power = voltage * current, not voltage + current [OK]

Hint: Power needs multiplication, not addition [OK]

Common Mistakes:

• Adding voltage and current instead of multiplying
• Thinking semicolons are missing
• Believing Serial.println can't print floats

5. You want your Arduino project to run on a 3.7V battery with a current draw of 0.3A. To estimate how long a 1500mAh battery will last, which calculation is correct?

hard

A. Battery life (hours) = 1500mAh / 300mA

B. Battery life (hours) = 1500mAh x 0.3A

C. Battery life (hours) = (3.7V x 0.3A) / 1500mAh

D. Battery life (hours) = 1500mAh / 0.3A

Solution

1. Step 1: Convert units for current and capacity

   Battery capacity is 1500mAh, current draw is 0.3A = 300mA.

2. Step 2: Calculate battery life

   Battery life (hours) = capacity (mAh) / current draw (mA) = 1500mAh / 300mA = 5 hours.

3. Final Answer:

   Battery life (hours) = 1500mAh / 300mA -> Option A

4. Quick Check:

   Battery life = capacity / current [OK]

Hint: Divide battery mAh by current in mA for hours [OK]

Common Mistakes:

• Mixing amps and milliamps without conversion
• Multiplying capacity by current
• Using voltage in battery life calculation