Problem Statement
A resistor connected to $6\,\text{V}$ passes a current of $2.0\,\text{A}$. Find (a) its resistance and (b) the power dissipated.
Given Information
- $V = 6\,\text{V}$
- $I = 2.0\,\text{A}$
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: $$R = V/I = 6/2.0 = 3.00\,\Omega$$
Step 2 — Power: $$P = VI = 6\times2.0 = 12.0\,\text{W}$$
Step 3 — Check: $P = I^2 R = 2.0^2\times3.00 = 12.0\,\text{W}$ ✓
Worked Calculation
$$R = 3.00\,\Omega;\quad P = 12.0\,\text{W}$$
Answer
$$\boxed{R = 3.00\,\Omega,\quad P = 12.0\,\text{W}}$$
The 3.00 Ω resistor dissipates 12.0 W at 6 V. This power appears as heat, warming the resistor at a rate determined by its thermal mass and cooling.
Physical Interpretation
The 3.00 Ω resistor dissipates 12.0 W at 6 V. This power appears as heat, warming the resistor at a rate determined by its thermal mass and cooling.
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