Power dissipation from voltage and resistance: What power is dissipated by a 1.2 kΩ resistor when 12 V is applied across it? (Assume a purely resistive load.)

Introduction / Context:
Computing resistor power is essential for selecting safe component wattage ratings and avoiding overheating. Using Ohm’s law with power relations ensures correct sizing in practical circuits like voltage dividers and bias networks.

Given Data / Assumptions:

• Resistance R = 1.2 kΩ = 1200 Ω.
• Applied voltage V = 12 V (DC or RMS AC for a resistor).
• Ideal resistor, no reactive effects.

Concept / Approach:

The relevant formula is P = V^2 / R for a fixed resistance with known voltage. Alternative equivalent forms include P = I^2 * R and P = V * I after computing current I via I = V / R.

Step-by-Step Solution:

Verification / Alternative check:

Compute current first: I = V / R = 12 / 1200 = 0.01 A = 10 mA. Then P = V * I = 12 V * 0.01 A = 0.12 W, confirming the result.

Why Other Options Are Wrong:

• 12 W / 1.2 W / 0.48 W: too large given the modest current of 10 mA.
• 12 mW: too small by a factor of 10.

Common Pitfalls:

• Forgetting to convert kilohms to ohms.
• Mixing up V^2/R with I^2*R leading to arithmetic slips; both yield the same when applied correctly.

Final Answer:

0.12 W