The rating on a device gives you two numbers: the supply voltage V and the power P. From them you can find the current and, once you know how long it runs, the energy transferred. This appears in short calculation questions and as the first step of longer cost or fuse questions.
It builds on potential difference, resistance and circuits and on power from work, energy and efficiency.
Which equations connect the rating to the energy?
Power is energy transferred per second, so E = P × t, with E in joules, P in watts and t in seconds.
Electrical power is also P = I × V, with I in amperes and V in volts. Combining them gives E = I × V × t.
If the question gives P and V, rearrange to I = P ÷ V. A sensible habit is to write the units next to each number before you multiply.
Worked example
A kettle is marked 230 V, 2300 W. It runs for 2 minutes 30 seconds to boil the water. The figures are invented for practice.
Step 1, current: I = P ÷ V = 2300 ÷ 230 = 10 A.
Step 2, time in seconds: 2 min 30 s = 120 + 30 = 150 s.
Step 3, energy: E = P × t = 2300 × 150 = 345 000 J.
Step 4, check with the other route: E = I × V × t = 10 × 230 × 150 = 345 000 J. The two agree.
Answer: current 10 A, energy 345 000 J (345 kJ).
The answer is large because a kettle has a high power. A quick plausibility check: 2300 J each second for 150 seconds must be a few hundred thousand joules.
The mistake to watch for
A common slip is to put the time in minutes straight into E = P × t.
Mistaken working: E = 2300 × 2.5 = 5750 J
The student converted 2 min 30 s to 2.5 minutes, then used minutes in an equation that needs seconds.
The correction is to ask “what does a watt mean?” A watt is one joule per second, so the time must be in seconds. 2.5 min = 150 s, and 2300 × 150 = 345 000 J. The wrong answer is 60 times too small, which is the clue that a minute-to-second conversion has been missed.
Check yourself
Try these on paper, then open each answer.
1. A lamp is marked 12 V, 36 W. Find the current, and the energy transferred in 10 minutes.
Show answer
I = 36 ÷ 12 = 3 A. t = 10 × 60 = 600 s. E = 36 × 600 = 21 600 J.
2. A device draws 5.0 A from a 230 V supply for 2 minutes. Find its power and the energy transferred.
Show answer
P = IV = 5.0 × 230 = 1150 W. t = 120 s. E = 1150 × 120 = 138 000 J. Check: 5.0 × 230 × 120 = 138 000 J.
3. A phone charger delivers 5 V at 2 A for 1 hour. How much energy is transferred in joules?
Show answer
P = 5 × 2 = 10 W. t = 3600 s. E = 10 × 3600 = 36 000 J (36 kJ).
Where this leads next
Next, see why a device also needs protection: compare a fuse purpose with a switch purpose. When energy is asked for in kilowatt-hours, continue to check a unit conversion in an energy-use calculation. The full module is at electrical energy and safe interpretation.
Some students get each formula right but choose the wrong one when the question gives unfamiliar wording. A teacher on online one-to-one Physics tuition can set fresh rating questions until that choice is automatic.