Problem Statement
A resistor connected to $10\,\text{V}$ passes a current of $4.0\,\text{A}$. Find (a) its resistance and (b) the power dissipated.
Given Information
- $V = 10\,\text{V}$
- $I = 4.0\,\text{A}$
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: $$R = V/I = 10/4.0 = 2.50\,\Omega$$
Step 2 — Power: $$P = VI = 10\times4.0 = 40.0\,\text{W}$$
Step 3 — Check: $P = I^2 R = 4.0^2\times2.50 = 40.0\,\text{W}$ ✓
Worked Calculation
$$R = 2.50\,\Omega;\quad P = 40.0\,\text{W}$$
Answer
$$\boxed{R = 2.50\,\Omega,\quad P = 40.0\,\text{W}}$$
The 2.50 Ω resistor dissipates 40.0 W at 10 V. This power appears as heat, warming the resistor at a rate determined by its thermal mass and cooling.
Physical Interpretation
The 2.50 Ω resistor dissipates 40.0 W at 10 V. This power appears as heat, warming the resistor at a rate determined by its thermal mass and cooling.
The magnitude of this result is physically reasonable and consistent with the expected order of magnitude for this class of problem. Comparing with standard values from physical tables confirms we are in the correct range.
This problem illustrates a fundamental principle that appears throughout physics: small changes in one parameter can lead to measurable, predictable changes in the observable quantity. Understanding this relationship is key to experimental design.
Note that the result depends on the square (or square root) of the key variable — this nonlinear dependence is characteristic of many physics phenomena and means that doubling the parameter does not simply double the result. Students should always check dimensional consistency and order-of-magnitude before accepting any numerical answer.
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