Estimate lamp resistance using Ohm’s law: A flashlight bulb draws 40 mA from a 4.5 V battery pack. What is the approximate resistance of the bulb filament?

Introduction / Context:
Batteries and small lamps are classic examples for applying Ohm’s law. Determining filament resistance helps with estimating power and verifying that the battery can supply the load without excessive voltage sag.

Given Data / Assumptions:

• Total battery voltage V = 4.5 V.
• Lamp current I = 40 mA = 0.04 A.
• Assume steady state at operating temperature; use linear Ohm’s law model.

Concept / Approach:
Compute resistance from R = V / I after converting current to amperes. The numerical accuracy is set by the significant figures of the given data, so a rounded answer is acceptable.

Step-by-Step Solution:

Verification / Alternative check:
Check power: P = V * I = 4.5 * 0.04 = 0.18 W. Also P = I^2 * R ≈ 0.0016 * 112.5 ≈ 0.18 W, confirming internal consistency.

Why Other Options Are Wrong:

• 11.2 Ω or 1.2 Ω: Would imply currents in amperes, not tens of milliamperes.
• 18 Ω: Would give I = 4.5 / 18 = 0.25 A = 250 mA, far larger than 40 mA.

Common Pitfalls:

• Forgetting to convert mA to A before dividing.
• Rounding too early and selecting a value off by a factor of 10.

Final Answer:
112 Ω