Problem Statement
An electric device of resistance $R = 90\,\Omega$ is connected to a $160\,\text{V}$ supply for $t = 300\,\text{s}$. Find (a) the current, (b) the power, and (c) the heat produced.
Given Information
- $R = 90\,\Omega$
- $V = 160\,\text{V}$
- $t = 300\,\text{s}$
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: $$I = V/R = 160/90 = 1.78\,\text{A}$$
Step 2 — Power: $$P = V^2/R = 160^2/90 = 284.4\,\text{W}$$
Step 3 — Heat: $$H = Pt = 284.4\times300 = 85333.3\,\text{J}$$
Worked Calculation
$$I = 1.78\,\text{A};\quad P = 284.4\,\text{W};\quad H = 85333\,\text{J}$$
Answer
$$\boxed{I = 1.78\,\text{A},\quad P = 284.4\,\text{W},\quad H = 85333\,\text{J}}$$
The 90Ω device at 160V draws 1.78A and generates 85333J of heat in 300s. In kWh: 0.023704 kWh — the basis for electricity billing.
Physical Interpretation
The 90Ω device at 160V draws 1.78A and generates 85333J of heat in 300s. In kWh: 0.023704 kWh — the basis for electricity billing.
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