Circuit energy cost: current to power to kilowatt-hours
SPH3U Grade 11 Physics · Electricity and Magnetism
A 120 V shop heater carries two nichrome coils, 15.0 Ω and 9.0 Ω, and its LOW setting wires them in series. A student leaves it on LOW for 6.00 h in an unheated garage. Find the total resistance of the series pair, the current the heater draws, the power it dissipates, and the electrical energy it consumes over the run.
On LOW the coils sit end to end, so their resistances add. Nothing else about the circuit can be computed until this total is known.
Carried onward at full precision, not this rounded figure.
The outlet holds 120 V across the pair. Ohm's law turns that into the single current every part of a series loop shares.
Carried onward at full precision, not this rounded figure.
Power from the current and the resistance — the first step that draws on two earlier answers at once.
Carried onward at full precision, not this rounded figure.
Power sustained over time is energy. The answer lands in joules; the utility meters the same quantity in kilowatt-hours.
Carried onward at full precision, not this rounded figure.
Why this order
Every electricity bill is this chain. The supply voltage is fixed by the utility and the resistance is fixed by the appliance, so current is the first thing you can actually compute — nothing downstream is knowable until Ohm's law hands it over. Step 3 is where the chain stops being a straight line: P = I²R reaches back for the current from step 2 and the same total resistance from step 1. P = VI would have given the identical 600 W, and picking between them is really a question of which two quantities you measured rather than inferred.
The trap is the series setting. Two coils end to end share one current and their resistances add to 24 Ω — more opposition than either coil alone — so the heater draws less power on LOW, not more, which is exactly the point of the switch. Students routinely reason that two elements must mean more heat. The second trap is units: joules are the SI answer, but nobody bills in them. 12.96 MJ is 3.60 kWh, and at 12.5 cents a kilowatt-hour that garage cost about 45 cents to warm — the kind of arithmetic that turns an efficiency argument from a slogan into a number.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.