Problem Statement
An electric device of resistance $R = 40\,\Omega$ is connected to a $160\,\text{V}$ supply for $t = 240\,\text{s}$. Find (a) the current, (b) the power, and (c) the heat produced.
Given Information
- $R = 40\,\Omega$
- $V = 160\,\text{V}$
- $t = 240\,\text{s}$
Physical Concepts & Formulas
Joule heating: electrical energy converts to thermal energy at rate $P = V^2/R = I^2R$. Total heat $H = Pt$. This is the operating principle of all resistive heating devices.
- $I = V/R$
- $P = V^2/R$ — Joule heating
- $H = Pt$ — total heat
Step-by-Step Solution
Step 1 — Current: $$I = V/R = 160/40 = 4.00\,\text{A}$$
Step 2 — Power: $$P = V^2/R = 160^2/40 = 640.0\,\text{W}$$
Step 3 — Heat: $$H = Pt = 640.0\times240 = 153600.0\,\text{J}$$
Worked Calculation
$$I = 4.00\,\text{A};\quad P = 640.0\,\text{W};\quad H = 153600\,\text{J}$$
Answer
$$\boxed{I = 4.00\,\text{A},\quad P = 640.0\,\text{W},\quad H = 153600\,\text{J}}$$
The 40Ω device at 160V draws 4.00A and generates 153600J of heat in 240s. In kWh: 0.042667 kWh — the basis for electricity billing.
Physical Interpretation
The 40Ω device at 160V draws 4.00A and generates 153600J of heat in 240s. In kWh: 0.042667 kWh — the basis for electricity billing.
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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