Lesson 3 of 4
Electrical power and energy
I can recall and use P = VI, P = I^2 R and P = V^2/R for electrical power.Syllabus
Syllabus 8867, 8(d). Recall and use P = VI, P = I^2 R and P = V^2/R for electrical power.
Electrical power is the rate of energy transfer. Multiply energy per charge by charge per second, using the voltage and current of the same component.
For steady p.d. V and current I, start with work W = VQ and power P = W/t:
For a resistor at an operating state described by V = IR, substitute that relation in two ways:
P = VI = V(V/R) = V2/R
Use the form that matches the given quantities. V is across the resistor and I is through it. The local resistor relation does not mean that a motor's useful mechanical output equals I2R: its terminal input is VI, and the useful output needs its own measurement.
State what stays fixed
- At fixed resistance, doubling voltage quadruples power: P is proportional to V2.
- At fixed current, increasing resistance increases power: P = I2R.
- At fixed voltage, increasing resistance decreases power: P = V2/R.
These comparisons hold different quantities constant. There is no contradiction between them.
Worked steady resistor
Check the result by all three forms
A few volts at a few amperes suggests power of order 10 W and a few hundred joules over tens of seconds. Now use the precise supplied model: 6.00 V across a constant 3.00 Ω resistor for 40.0 s.
P = VI = 6.00 × 2.00 = 12.0 W
P = I2R = 2.002 × 3.00 = 12.0 W
P = V2/R = 6.002/3.00 = 12.0 W
At this steady operating state:
The current and power are substantial for a small resistor. This is a supplied calculation model; a physical setup at these values would need suitable current and power ratings. The measurement example uses a separate, lower-power load.
Try it step by step
Work it out in steps: Current I = P / V in A, then Resistance R = V / I in ohm, then Energy transferred in 10 minutes E = Pt in J.
Show the answer
Problem 1.
Current I = P / V in A: I = 60 / 240 = 0.25 A
Resistance R = V / I in ohm: R = 240 / 0.25 = 960 ohm
Energy transferred in 10 minutes E = Pt in J: E = 60 x 600 = 36000 J (2 sf)
Problem 2.
Current I = P / V in A: I = 1500 / 230 = 6.52 A
Resistance R = V / I in ohm: R = 230 / 6.52 = 35.3 ohm
Energy transferred in 10 minutes E = Pt in J: E = 1500 x 600 = 900000 J, which is 900 kJ (3 sf)
Problem 3.
Current I = P / V in A: I = 11 / 230 = 0.0478 A
Resistance R = V / I in ohm: R = 230 / 0.0478 = 4810 ohm (3 sf)
Energy transferred in 10 minutes E = Pt in J: E = 11 x 600 = 6600 J (2 sf)
Energy over time
For constant power, the area under a power-time graph is a rectangle, Pt. For changing power, the total area still gives transferred energy; the average power multiplied by the full interval gives the same energy. Preserve the actual time widths when estimating an area from samples.
A kilowatt-hour is another energy unit:
480 J = 480/(3.6 × 106) kWh
= 1.33 × 10-4 kWh
kW is power; kWh is energy. In W = Pt, the quantity W denotes work or transferred energy in joules, whereas the upright unit W after a number denotes watts.
For a passive resistor, I2R is non-negative even when a chosen current reference makes I negative. With current defined into the terminal used as positive for the voltage, VI also describes the power absorbed by the load.
Check your understanding
A constant 3.00 ohm resistor has 6.00 V across it for 40.0 s. What electrical energy is transferred to it?
Show the answer
480 J
P = V2/R = 6.002/3.00 = 12.0 W. At this steady operating state, W = Pt = 12.0 x 40.0 = 480 J.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
A heater is rated 1.0 kW, 230 V. Calculate its resistance at normal operation. The supply voltage drops to 210 V. Assuming the resistance does not change, calculate the new power. [3]
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