Year 11 Physics Module 4 ⏱ ~35 min 5 MC · 3 Short Answer Lesson 9 of 17

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.

Today's hook: The Tesla Supercharger V3, launched in Fremont, California in 2019, delivers 250 kW at 400 V DC, a peak current of 625 A. It charges a 75 kWh battery from 10% to 80% in just 12.6 minutes for about $15.85. That is P = IV and E = Pt applied at scale. Why does doubling the current quadruple the power dissipated in a resistor?
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Warm up first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Before you read, predict

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:

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Learning Intentions

Learning Intentions

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
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Key Terms

Key Terms
Power ($P$)Rate of energy conversion. $P = E/t$, measured in watts (W). 1 W = 1 J s⁻¹.
Electrical energy ($E$)Total energy converted by a device. $E = Pt$ (joules) or $E = VIt$ (joules). Also measured in kilowatt-hours for billing.
Kilowatt-hour (kWh)Energy used by a 1 kW device for 1 hour. 1 kWh = 3.6 × 10⁶ J = 3.6 MJ.
Power dissipationThe conversion of electrical energy to thermal (or other) energy in a resistor. $P = I^2 R$.
Cross-lesson links: L05–L08 established V, I, and R and circuit analysis. L09 introduces the rate at which electrical energy is converted, power. P = IV connects directly to Ohm's Law: combining with V = IR gives P = I²R and P = V²/R. The Tesla Supercharger example shows why P = IV matters beyond the textbook.
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Power Formulae

Core Content

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Power Formulae
+5 XP

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.

Electrical power

$P = IV = I^2 R = \dfrac{V^2}{R}$

All three forms come from combining $P = IV$ with $V = IR$. Use whichever two quantities you know.

Electrical energy

$E = Pt = VIt = I^2 Rt$ (joules, when $t$ is in seconds)

1 kWh = $1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}$

Worked example, power in a resistor

A 20 Ω resistor carries a current of 3.0 A. Find power dissipated and energy in 2 minutes.

  1. $P = I^2 R = (3.0)^2 \times 20 = 9.0 \times 20 = 180\ \text{W}$
  2. $E = Pt = 180 \times (2 \times 60) = 180 \times 120 = 21{,}600\ \text{J} = 21.6\ \text{kJ}$
  3. 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.

Electrical power summary showing power formulas, resistor calculations, energy conversion, kilowatt-hours and cost
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Activity 1, Power Calculations

Activity 1, Power Calculations
ApplyBand 3

A hairdryer is rated at 1800 W and operates at 230 V.

  1. Calculate the current drawn by the hairdryer.
  2. Calculate the resistance of the heating element.
  3. 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:

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Activity 2, Energy and kWh

Activity 2, Energy and kWh
ApplyBand 3

A 2400 W electric kettle boils water in 3 minutes 30 seconds.

  1. Calculate the energy used in joules.
  2. Express the energy in kWh.
  3. 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:

✓

Show what you have learned

Quick recall, electrical power
+5 XP
Short Answer, 10 marks
+5 XP

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

  1. $I = P/V = 1800/230 \approx 7.83\ \text{A}$
  2. $R = V^2/P = 230^2/1800 = 52{,}900/1800 \approx 29.4\ \Omega$
  3. $t = 5 \times 60 = 300\ \text{s}$; $E = 1800 \times 300 = 540{,}000\ \text{J} = 540\ \text{kJ}$

Activity 2, Model Answers

  1. $t = 3 \times 60 + 30 = 210\ \text{s}$; $E = 2400 \times 210 = 504{,}000\ \text{J}$
  2. $E = 504{,}000/(3.6 \times 10^6) \approx 0.14\ \text{kWh}$
  3. 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.

✓

Retrieve, reflect and finish

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How did your thinking change?

The 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.

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Module overview · Physics · Checkpoint 2 · Module quiz