Problem Statement
A resistor connected to $12\,\text{V}$ passes a current of $0.5\,\text{A}$. Find (a) its resistance and (b) the power dissipated.
Given Information
- $V = 12\,\text{V}$
- $I = 0.5\,\text{A}$
Physical Concepts & Formulas
Ohm’s law relates voltage, current, and resistance: $V = IR$. Power dissipated is $P = VI = I^2 R = V^2/R$ — the Joule heating rate.
- $R = V/I$ — Ohm’s law
- $P = VI = I^2R$ — power dissipated
Step-by-Step Solution
Step 1 — Resistance: $$R = V/I = 12/0.5 = 24.00\,\Omega$$
Step 2 — Power: $$P = VI = 12\times0.5 = 6.0\,\text{W}$$
Step 3 — Check: $P = I^2 R = 0.5^2\times24.00 = 6.0\,\text{W}$ ✓
Worked Calculation
$$R = 24.00\,\Omega;\quad P = 6.0\,\text{W}$$
Answer
$$\boxed{R = 24.00\,\Omega,\quad P = 6.0\,\text{W}}$$
The 24.00 Ω resistor dissipates 6.0 W at 12 V. This power appears as heat, warming the resistor at a rate determined by its thermal mass and cooling.
Physical Interpretation
The 24.00 Ω resistor dissipates 6.0 W at 12 V. This power appears as heat, warming the resistor at a rate determined by its thermal mass and cooling.
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