Electrical Power & Energy
Tesla Supercharger V3 (Fremont, California, 2019): 250 kW at 400 V DC — peak current I = P/V = 625 A. Charges a 75 kWh battery from 10% to 80% (52.5 kWh) in 12.6 minutes; session cost $15.85 at US rates. This is P = IV and E = Pt made real at automotive scale.
A 1200 W toaster is plugged into 230 V mains for 3 minutes.
Predict 1: What current does the toaster draw?
Predict 2: How much electrical energy (in joules) does the toaster convert in those 3 minutes?
Electrical power is defined as:
Know
- $P = IV = I^2 R = V^2/R$ (watts)
- $E = Pt$ (joules)
- 1 kilowatt-hour (kWh) = 3.6 MJ
- Power unit: watt (W) = joule per second (J s⁻¹)
Understand
- Why $P \propto I^2$ — doubling $I$ quadruples power
- The distinction between power (rate) and energy (total)
- How to choose the correct power formula given available data
Can Do
- Calculate power and energy for any circuit component
- Verify total power in a circuit equals supply power
- Convert between joules and kilowatt-hours
Core Content
Hold a light globe in one hand for a minute — it gets warm. The filament is converting electrical energy to heat and light faster than the glass can radiate it away. A 100 W globe does this at the rate of 100 joules every second; a 10 W LED does the same job using only 10 joules per second. "Power" is the word for that rate — and three different formulae let you calculate it depending on which quantities you know.
$P = IV = I^2 R = \dfrac{V^2}{R}$
All three forms come from combining $P = IV$ with $V = IR$. Use whichever two quantities you know.
$E = Pt = VIt = I^2 Rt$ (joules, when $t$ is in seconds)
1 kWh = $1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}$
A 20 Ω resistor carries a current of 3.0 A. Find power dissipated and energy in 2 minutes.
- $P = I^2 R = (3.0)^2 \times 20 = 9.0 \times 20 = 180\ \text{W}$
- $E = Pt = 180 \times (2 \times 60) = 180 \times 120 = 21{,}600\ \text{J} = 21.6\ \text{kJ}$
- Check with $V = IR = 3.0 \times 20 = 60\ \text{V}$: $P = IV = 3.0 \times 60 = 180\ \text{W}$ ✓
Electrical power is the rate of energy conversion: $P = IV = I^2R = V^2/R$ (W). Choose the formula depending on which two of $I$, $V$, $R$ you know. Energy used: $E = Pt$ (J); $1\ \text{kWh} = 3.6 \times 10^6\ \text{J}$. Note: $P \propto I^2$, so doubling current quadruples power.
Pause — copy the highlighted formulas into your book before moving on.
Doubling the current through a resistor quadruples the power dissipated.
1 kilowatt-hour is equal to 1000 joules.
A hairdryer is rated at 1800 W and operates at 230 V.
- Calculate the current drawn by the hairdryer.
- Calculate the resistance of the heating element.
- How much energy (in kJ) does the hairdryer convert in 5 minutes?
A resistor dissipates 50 W when the current through it is 2.0 A. The resistance is:
A 2400 W electric kettle boils water in 3 minutes 30 seconds.
- Calculate the energy used in joules.
- Express the energy in kWh.
- If electricity costs 32 cents per kWh, what does one boil cost (in cents)?
Three of these correctly express electrical power. Pick the odd one out.
A 100 W lamp operates for 36,000 seconds. The energy it converts is _____ J.
A 40 Ω resistor is connected to a 20 V supply. Its power dissipation is:
ApplyBand 3(3 marks) 3. An electric heater has a resistance of 46 Ω and is connected to a 230 V supply. Calculate (a) the current drawn, (b) the power rating of the heater, and (c) the energy converted in 20 minutes.
AnalyseBand 4(3 marks) 4. Two resistors, $R_1 = 10\ \Omega$ and $R_2 = 40\ \Omega$, are connected in parallel to a 20 V supply. (a) Calculate the power dissipated by each resistor. (b) Find the total power and verify it equals $P_{\text{supply}} = V \times I_{\text{total}}$.
EvaluateBand 6(4 marks) 5. Power transmission lines carry electricity at high voltage (e.g. 330 kV) and low current. Using the formula $P_{\text{loss}} = I^2 R$, evaluate why transmission at high voltage rather than high current significantly reduces power losses in the line. Include a numerical example with a line resistance of 10 Ω.
Show all answers
Activity 1 — Model Answers
- $I = P/V = 1800/230 \approx 7.83\ \text{A}$
- $R = V^2/P = 230^2/1800 = 52{,}900/1800 \approx 29.4\ \Omega$
- $t = 5 \times 60 = 300\ \text{s}$; $E = 1800 \times 300 = 540{,}000\ \text{J} = 540\ \text{kJ}$
Activity 2 — Model Answers
- $t = 3 \times 60 + 30 = 210\ \text{s}$; $E = 2400 \times 210 = 504{,}000\ \text{J}$
- $E = 504{,}000/(3.6 \times 10^6) \approx 0.14\ \text{kWh}$
- Cost = $0.14 \times 32 \approx 4.5\ \text{cents}$
Short Answer — Model Answers
Q3 (3 marks): (a) $I = V/R = 230/46 = 5.0\ \text{A}$. (b) $P = IV = 5.0 \times 230 = 1150\ \text{W}$. (c) $t = 20 \times 60 = 1200\ \text{s}$; $E = 1150 \times 1200 = 1{,}380{,}000\ \text{J} = 1.38\ \text{MJ}$.
Q4 (3 marks): (a) $P_1 = V^2/R_1 = 400/10 = 40\ \text{W}$; $P_2 = V^2/R_2 = 400/40 = 10\ \text{W}$. (b) $P_{\text{total}} = 50\ \text{W}$. $I_{\text{total}} = I_1 + I_2 = 2.0 + 0.50 = 2.5\ \text{A}$; $P_{\text{supply}} = 20 \times 2.5 = 50\ \text{W}$ ✓.
Q5 (4 marks): From $P = IV$, the same power can be transmitted at high $V$ and low $I$, or low $V$ and high $I$. Example: transmit 1 MW over a 10 Ω line. Option A (high $I$): $V = 1000\ \text{V}$, $I = 1000\ \text{A}$. $P_{\text{loss}} = I^2 R = 1000^2 \times 10 = 10{,}000{,}000\ \text{W}$ — all power is lost! Option B (high $V$): $V = 100{,}000\ \text{V}$, $I = 10\ \text{A}$. $P_{\text{loss}} = 10^2 \times 10 = 1000\ \text{W}$ — only 0.1% lost. Because $P_{\text{loss}} = I^2 R \propto I^2$, reducing current by a factor of 100 reduces losses by a factor of 10,000. High-voltage transmission is therefore essential to minimise resistive heating losses.
Five timed questions on electrical power and energy.
⚔ Enter the arenaThe Tesla Supercharger V3 (Fremont, California, 2019) delivers 250 kW at 400 V — a peak current of 625 A — and charges a 75 kWh battery from 10% to 80% in 12.6 minutes for $15.85. That is E = Pt = 250,000 W × 756 s = 189 MJ = 52.5 kWh transferred in one session. Power and energy are related by time: power is the rate, energy is the total.
Now check your Think First answers: the 1200 W toaster draws I = P/V = 1200 ÷ 230 ≈ 5.2 A. Energy in 3 min: E = 1200 × 180 = 216,000 J = 216 kJ.