A 500W electric oven plugged into a 220 V source will consume an electric current of

Answer Details

Electrical power delivered to a device is the product of the potential difference across it and the current through it: \[P = IV.\] The rating on an appliance states the power it consumes at its working voltage, so the current follows by rearranging: \[I = \frac{P}{V} = \frac{500}{220} = 2.27\,\text{A}\ (3\ \text{s.f.}).\] The oven therefore draws about \(2.27\,\text{A}\).

It is worth seeing where the other numbers could come from, because each represents a specific error. Dividing the voltage by the power, \(220/500\), gives \(0.44\), while using a mains value of \(110\,\text{V}\) instead of \(220\,\text{V}\) would double the answer to \(4.55\,\text{A}\). Only the direct substitution into \(I = P/V\) with the values actually given is defensible.

Two related results are often needed in the same question and follow from the same data: the resistance of the heating element at working temperature is \[R = \frac{V^2}{P} = \frac{220^2}{500} = 96.8\,\Omega,\] and the energy consumed in, for example, half an hour is \[E = Pt = 500 \times 1800 = 9.0 \times 10^{5}\,\text{J} = 0.25\,\text{kWh}.\] Keep the three forms \(P = IV = I^2R = \dfrac{V^2}{R}\) at hand, and choose the one whose quantities are actually given rather than working through an intermediate you do not need.