Why Does result Become None?

def double_value(value):
    print(value * 2)

result = double_value(5)
print("result:", result)

Before running:

1. Predict both displayed lines.
2. Identify which line displays 10.
3. Explain why result is not 10.

Task 1: Repair and Test

Open starter/return_vs_print_bug.py.

• Change the function so it returns the doubled value.
• Keep the final print() in the main code.
• Test with 5, 0, and -3.
• Add one comment explaining the difference between displaying and returning.

Expected results from the repaired function:

Task 2: Calibrate One Reading

Create a function with this contract:

def calibrate_voltage(raw_voltage, offset=0.05):
    """Return raw_voltage corrected by offset."""
    # your code

Test these calls:

print(calibrate_voltage(4.92))
print(calibrate_voltage(4.92, 0.10))
print(calibrate_voltage(0.0, -0.02))

Record the three expected values before running the program.

Check the Default Argument

def calibrate_voltage(raw_voltage, offset=0.05):
    return raw_voltage + offset

Answer in a comment:

• Which call uses the default?
• Which call overrides it?
• Why should raw_voltage not have a default?

Task 3: Return a Classification

Implement:

def voltage_status(voltage):
    """Return Invalid, Low, Normal, or High."""
    # Invalid: voltage < 0
    # Low:     0 <= voltage < 4.75
    # Normal:  4.75 <= voltage <= 5.25
    # High:    voltage > 5.25

Your function must return exactly one string and must not call input().

Test the Boundaries

Use a separate call for every test.

print(voltage_status(-0.1))
print(voltage_status(4.74))
print(voltage_status(4.75))
print(voltage_status(5.25))
print(voltage_status(5.26))

Create a short test table in a comment:

Correct the function if any expected and actual values differ.

Task 4: Package an Accumulator

Implement:

def sum_multiples(limit, divisor):
    """Return the sum of positive multiples up to limit."""
    total = 0
    # add a for loop
    return total

Required tests:

print(sum_multiples(12, 3))   # 30
print(sum_multiples(10, 2))   # 30
print(sum_multiples(4, 5))    # 0

Explain why the last call returns 0.

Check-Off A: Functions in One File

Show the instructor or teaching assistant:

• calibrate_voltage() with a default argument
• voltage_status() passing all five boundary tests
• sum_multiples() passing all three tests
• one sentence explaining why input() belongs in the main code rather than these calculation functions

Do not continue until your functions return the expected values.

Task 5: Use the Standard math Module

For a series circuit, calculate impedance magnitude:

|Z| = square root of (resistance² + reactance²)

import math

def impedance_magnitude(resistance, reactance):
    # return the magnitude using math.sqrt
    pass

Test:

print(impedance_magnitude(3, 4))   # 5.0
print(impedance_magnitude(5, 12))  # 13.0

Two Files, Two Responsibilities

Download or open sensor_tools.py and main.py. Keep them in the same folder.

Task 6: Complete sensor_tools.py

def calibrate_voltage(raw_voltage, offset=0.05):
    """Return raw_voltage corrected by offset."""
    pass

def voltage_status(voltage):
    """Return Invalid, Low, Normal, or High."""
    pass

Replace every pass. Keep the function names and parameters unchanged so that main.py can call them.

Complete the Main Program

main.py

import sensor_tools

raw_voltage = 4.72

calibrated = sensor_tools.calibrate_voltage(raw_voltage)
status = sensor_tools.voltage_status(calibrated)

print("Raw:", raw_voltage)
print("Calibrated:", calibrated)
print("Status:", status)

Run main.py, then test at least three raw readings that produce different statuses.

Final Check-Off

Demonstrate a working two-file program and explain:

1. Which file defines the reusable functions?
2. Why do calls use the prefix sensor_tools.?
3. What value does calibrate_voltage() return?
4. Which boundary inputs did you test?
5. What would happen if sensor_tools.py were in a different folder?

Submit or retain the completed files according to the instructor’s directions.

Extension: Debug a Mathematical Expression

The intended expression is:

√2 × sin(π/4)

import math

wrong = math.sqrt(2) * 0.25 * math.pi
correct = math.sqrt(2) * math.sin(0.25 * math.pi)

Explain why the two expressions differ and verify that correct is approximately 1.0.