Electrical Energy and Power

Understanding how electrical energy is transferred and how power is calculated is essential for circuit analysis. These concepts link the fundamental quantities of current, voltage, and resistance to the real-world behaviour of devices.

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Electrical Power

Power is the rate at which energy is transferred. In an electrical circuit:

P = IV

where:

• P = power in watts (W)
• I = current in amperes (A)
• V = potential difference in volts (V)

One watt means one joule of energy is transferred per second.

Deriving Alternative Forms

By substituting Ohm's law (V = IR) into P = IV, we get two additional useful expressions:

P = I²R (substitute V = IR into P = IV)

P = V²/R (substitute I = V/R into P = IV)

These three forms are all equivalent and are used in different situations:

• Use P = IV when you know both current and voltage.
• Use P = I²R when you know current and resistance (but not voltage).
• Use P = V²/R when you know voltage and resistance (but not current).

When to Use Which Power Formula

Known Quantities	Best Formula	Example Situation
I and V	P = IV	Ammeter and voltmeter readings available
I and R	P = I²R	Series circuit: same I through each R
V and R	P = V²/R	Parallel circuit: same V across each R

Worked Example 1

A kettle draws 10 A from a 230 V mains supply. What power does it consume?

Solution: P = IV = 10 × 230 = 2300 W (or 2.3 kW)

Worked Example 2

A 47 Ω resistor carries a current of 0.20 A. What power is dissipated in the resistor?

Solution: P = I²R = (0.20)² × 47 = 0.040 × 47 = 1.9 W

Worked Example 3

A 12 V supply is connected across a 24 Ω resistor. What power is dissipated?

Solution: P = V²/R = 12²/24 = 144/24 = 6.0 W

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Electrical Energy

Energy is the total amount of work done over a period of time:

E = Pt

Combining with P = IV:

E = IVt

where:

• E = energy in joules (J)
• I = current (A)
• V = potential difference (V)
• t = time (s)

Similarly:

• E = I²Rt (from P = I²R)
• E = V²t/R (from P = V²/R)

Worked Example 4

A 100 W lamp is left on for 8.0 hours. How much energy does it use?

Solution: E = Pt = 100 × (8.0 × 3600) = 100 × 28800 = 2,880,000 J (or 2.88 MJ)

Worked Example 5: Energy in a Resistor Network

Two resistors (3.0 Ω and 6.0 Ω) are connected in parallel across a 12 V supply for 5.0 minutes. Calculate the total energy dissipated.

Solution:

• R_total = (3.0 × 6.0)/(3.0 + 6.0) = 2.0 Ω
• Total current = V/R = 12/2.0 = 6.0 A
• Total power = IV = 6.0 × 12 = 72 W
• Energy = Pt = 72 × (5.0 × 60) = 72 × 300 = 21,600 J (or 21.6 kJ)

Alternatively, per resistor:

• 3.0 Ω: P = V²/R = 144/3.0 = 48 W → E = 48 × 300 = 14,400 J
• 6.0 Ω: P = V²/R = 144/6.0 = 24 W → E = 24 × 300 = 7,200 J
• Total = 14,400 + 7,200 = 21,600 J ✓

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The Kilowatt-Hour

The joule is a very small unit for domestic energy consumption. Electricity companies use the kilowatt-hour (kWh) instead.

1 kWh = 1 kW × 1 hour = 1000 W × 3600 s = 3,600,000 J = 3.6 MJ

To calculate energy in kWh: multiply the power in kW by the time in hours.

Worked Example 6

A 2.0 kW heater is used for 3.0 hours. Electricity costs 28p per kWh. Calculate the cost.

Solution:

• Energy = 2.0 × 3.0 = 6.0 kWh
• Cost = 6.0 × 28 = 168 p (or £1.68)

Worked Example 7: Comparing Appliance Costs

Appliance	Power (kW)	Time Used	Energy (kWh)	Cost at 30p/kWh
LED bulb	0.010	8 h	0.080	2.4p
Filament bulb	0.060	8 h	0.48	14.4p
Kettle	3.0	5 min (0.083 h)	0.25	7.5p
Electric shower	9.5	10 min (0.167 h)	1.58	47.5p
Washing machine	0.50	1.5 h	0.75	22.5p

This comparison illustrates why high-power appliances used for short periods can cost more than low-power appliances used for long periods.

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Efficiency

No energy conversion is 100% efficient. Some energy is always dissipated as heat. The efficiency of a device is:

Efficiency = useful output power / total input power × 100%

Or equivalently:

Efficiency = useful output energy / total input energy × 100%

Worked Example 8

An electric motor draws 500 W from the supply and does 350 W of useful mechanical work. What is its efficiency?

Solution: Efficiency = 350/500 × 100% = 70%

The remaining 150 W is dissipated as heat in the motor's windings and bearings.

Worked Example 9: Efficiency with Circuit Context

A battery (ε = 6.0 V, r = 0.50 Ω) drives a current through a 5.5 Ω motor. The motor has a mechanical efficiency of 80%. Calculate (a) the current, (b) the electrical power delivered to the motor, (c) the useful mechanical power output.

Solution:

• (a) I = ε/(R + r) = 6.0/(5.5 + 0.50) = 6.0/6.0 = 1.0 A
• (b) P_motor = I²R = (1.0)² × 5.5 = 5.5 W
• (c) Useful P = 0.80 × 5.5 = 4.4 W
• Total power from battery = εI = 6.0 × 1.0 = 6.0 W
• Power wasted in battery = I²r = 1.0² × 0.50 = 0.50 W
• Power wasted as heat in motor = 5.5 - 4.4 = 1.1 W
• Check: 4.4 + 1.1 + 0.50 = 6.0 W ✓

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Power Dissipation in Circuit Components

Every component with resistance dissipates energy as heat. This is important for practical circuit design:

• Resistors must have an adequate power rating. A 0.25 W resistor cannot safely dissipate 1 W of power — it will overheat and potentially fail.

• In series circuits, the component with the largest resistance dissipates the most power (since P = I²R and I is the same for all).

• In parallel circuits, the component with the smallest resistance dissipates the most power (since P = V²/R and V is the same for all).

This is counterintuitive for parallel circuits. Students often assume the bigger resistor dissipates more power. But think about it: the smaller resistor draws more current from the supply, so it converts more energy per second.

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Fuses and Circuit Protection

A fuse is a thin wire that melts and breaks the circuit when the current exceeds a certain value, protecting the circuit and wiring from overheating.

To select the correct fuse rating:

1. Calculate the normal operating current using I = P/V.
2. Choose a fuse with a rating just above the normal operating current.

Worked Example 10

A 920 W microwave operates on a 230 V supply. What fuse should be used? Available ratings: 3 A, 5 A, 13 A.

Solution:

• I = P/V = 920/230 = 4.0 A
• The fuse must be rated above 4.0 A, so a 5 A fuse is correct.
• A 3 A fuse would blow during normal operation. A 13 A fuse would allow dangerous overcurrents before blowing.

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Energy Dissipation and Heating

When current passes through a resistor, electrical energy is converted to thermal energy. The rate of heating is P = I²R. This has practical applications:

• Electric heaters use high-resistance elements (like nichrome wire) to maximise heat output.
• Transmission lines use low-resistance cables to minimise energy wasted as heat during power distribution.
• Computer processors generate significant heat (P = I²R in the transistors) and require cooling systems.

The total thermal energy dissipated over time t is E = I²Rt, which is sometimes called Joule heating or ohmic heating.

Worked Example 11: Power Transmission

A power station generates 100 MW of power at 25 kV. The transmission cables have a total resistance of 4.0 Ω. Calculate the power lost in the cables. Then repeat the calculation if the voltage is stepped up to 400 kV.

Solution at 25 kV:

• I = P/V = 100 × 10⁶ / (25 × 10³) = 4000 A
• P_lost = I²R = (4000)² × 4.0 = 64 MW — a staggering 64% loss

Solution at 400 kV:

• I = P/V = 100 × 10⁶ / (400 × 10³) = 250 A
• P_lost = I²R = (250)² × 4.0 = 0.25 MW — only 0.25% loss

Stepping up the voltage by a factor of 16 reduces the power loss by a factor of 16² = 256. This is why the National Grid transmits at very high voltages.

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A-Level Deep Dive: Electrical Energy and Power

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic 3.6 (electrical energy and power) establishes the three equivalent forms of electrical power dissipation $P = IV$P=IV, $P = I^2 R$P=I2R and $P = V^2/R$P=V2/R, the energy delivered over a time interval $E = Pt = IVt$E=Pt=IVt, and the practical use of the kilowatt-hour as a domestic energy unit (refer to the official Pearson Edexcel specification document for exact wording). It is the natural quantitative pay-off for the earlier sub-topics: the definitions of current $I = \Delta Q/\Delta t$I=ΔQ/Δt and potential difference $V = W/Q$V=W/Q from 3.1 combine immediately to give $P = W/t = (W/Q)(Q/t) = VI$P=W/t=(W/Q)(Q/t)=VI, while substitution of Ohm's law (3.2) yields the two derived forms. The topic is examined throughout 9PH0-01 (Paper 1) and 9PH0-03 (Paper 3, General and Practical Principles), and supplies the energy-conservation backbone that links 3.5 (emf and internal resistance) to Topic 9 (thermodynamics) and Topic 13 (oscillations and gravitation, where power is again rate of energy transfer). The Edexcel formula booklet lists $P = VI$P=VI, $P = I^2 R$P=I2R and $W = VIt$W=VIt, but candidates must recognise which form is most efficient for a given data set — that is the AO2 skill the topic rewards.

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Worked example with full mark scheme

Question (8 marks):

A small electric heater is rated $2.0\ \text{kW}$2.0 kW at $230\ \text{V}$230 V (root-mean-square values for the UK mains supply).

(a) Calculate the operating current of the heater. (2)

(b) Calculate the resistance of the heating element, assuming it behaves as an ohmic conductor at its operating temperature. (2)

(c) The heater is run for $45\ \text{minutes}$45 minutes. Calculate the electrical energy transferred, giving your answer in joules and in kilowatt-hours. (3)

(d) The domestic tariff is $28\ \text{p}\ \text{kWh}^{-1}$28 p kWh−1. Calculate the cost of running the heater for $45\ \text{minutes}$45 minutes. (1)

Solution with mark scheme:

(a) Step 1 — choose the correct form of the power equation.

Power, voltage and current are linked by $P = IV$P=IV, so $I = P/V$I=P/V.

M1 — rearrangement of $P = IV$P=IV for current.

Step 2 — substitute.

$I = \dfrac{2.0 \times 10^{3}}{230} \approx 8.7\ \text{A}$I=2302.0×103​≈8.7 A

A1 — $I \approx 8.7\ \text{A}$I≈8.7 A (or $8.70\ \text{A}$8.70 A to 3 s.f.). Common error: students confuse the rated power with the energy and divide by 60 first; this loses both marks.

(b) Step 1 — choose between $R = V/I$R=V/I and $R = V^2/P$R=V2/P.

Either approach earns the M1, but $R = V^2/P$R=V2/P avoids carrying the rounded value of $I$I from part (a) and so reduces propagation error.

$R = \dfrac{V^2}{P} = \dfrac{230^2}{2.0 \times 10^{3}} = \dfrac{52900}{2000} \approx 26\ \text{Ω}$R=PV2​=2.0×1032302​=200052900​≈26 Ω

M1 — correct form of the equation, with substitution. A1 — $R \approx 26\ \text{Ω}$R≈26 Ω (accept $26.45\ \text{Ω}$26.45 Ω to extra s.f.).

(c) Step 1 — convert time. $t = 45 \times 60 = 2700\ \text{s}$t=45×60=2700 s.

Step 2 — apply $E = Pt$E=Pt.

$E = Pt = (2.0 \times 10^{3}) \times 2700 = 5.4 \times 10^{6}\ \text{J}$E=Pt=(2.0×103)×2700=5.4×106 J

M1 — substituting consistent SI units into $E = Pt$E=Pt. A1 — $E = 5.4\ \text{MJ}$E=5.4 MJ.

Step 3 — convert to kilowatt-hours.

$E = P \times t$E=P×t where $P$P is in \text{kW} and $t$t in \text{hours}$, so$,soE = 2.0 \times 0.75 = 1.5\ \text{kWh}$.

A1 — $E = 1.5\ \text{kWh}$E=1.5 kWh. Equivalently $E = (5.4 \times 10^{6})/(3.6 \times 10^{6}) = 1.5\ \text{kWh}$E=(5.4×106)/(3.6×106)=1.5 kWh, using $1\ \text{kWh} = 3.6\ \text{MJ}$1 kWh=3.6 MJ.

(d) Cost $= 1.5 \times 28 = 42\ \text{p}$=1.5×28=42 p.

B1 — correct cost in pence (or £0.42).

Total: 8 marks (M1 A1 + M1 A1 + M1 A1 A1 + B1).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A $12\ \text{V}$12 V lamp is labelled "$12\ \text{V}, 24\ \text{W}$12 V,24 W".

(a) Calculate the operating current and resistance of the lamp at its rated working point. (2)

(b) The lamp is connected to a $12\ \text{V}$12 V supply through a $2.0\ \text{Ω}$2.0 Ω series resistor used to limit the current during warm-up. Calculate the initial current through the lamp if its cold resistance is $0.50\ \text{Ω}$0.50 Ω, and compare with the rated current. (2)

(c) Explain, in terms of energy transfer per unit time, why the lamp's filament reaches a steady operating temperature rather than continuing to heat up indefinitely. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — $I = P/V = 24/12 = 2.0\ \text{A}$I=P/V=24/12=2.0 A and $R = V/I = 12/2.0 = 6.0\ \text{Ω}$R=V/I=12/2.0=6.0 Ω (or $R = V^2/P = 144/24 = 6.0\ \text{Ω}$R=V2/P=144/24=6.0 Ω).
• A1 (AO1.1b) — both numerical values quoted with units.

(b)

• M1 (AO2.1) — total cold-circuit resistance $= 0.50 + 2.0 = 2.5\ \text{Ω}$=0.50+2.0=2.5 Ω, giving initial current $I_0 = 12/2.5 = 4.8\ \text{A}$I0​=12/2.5=4.8 A.
• A1 (AO2.5) — comparison: initial current is $4.8/2.0 = 2.4\times$4.8/2.0=2.4× the rated current, hence the protective need for a current-limiting resistor during warm-up.

(c)

• M1 (AO3.2a) — recognising that the filament dissipates electrical power $P_\text{in} = I^2 R$Pin​=I2R and simultaneously loses thermal power to the surroundings (radiation and convection) at a rate that increases with temperature.
• A1 (AO3.2b) — equilibrium reached when the rate of electrical energy supply equals the rate of thermal energy loss ($P_\text{in} = P_\text{out}$Pin​=Pout​), at which point the temperature is steady.

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. Edexcel deliberately mixes the three AOs on power questions: AO1 for substitution into $P = IV$P=IV, AO2 for proportional reasoning in mixed-temperature regimes, AO3 for the energy-balance argument that explains the physics behind a steady-state operating point.

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Synoptic links

Connects to:

1. Topic 3 sub-topic 3.5 — EMF and internal resistance: the maximum power transfer condition $R_\text{ext} = r$Rext​=r, in which the load resistance equals the internal resistance, is derived by maximising $P = I^2 R = \varepsilon^2 R/(R+r)^2$P=I2R=ε2R/(R+r)2 with respect to $R$R. This is a direct application of $P = I^2 R$P=I2R to a non-trivial circuit and connects power dissipation to the upstream emf treatment.

2. Topic 9 — Thermodynamics: Joule heating $P = I^2 R$P=I2R feeds the first law of thermodynamics: the electrical energy delivered to a resistor appears as internal energy of the lattice. The rate of temperature rise $dT/dt = P/(mc)$dT/dt=P/(mc) for a thermally isolated conductor is exactly an energy-conservation statement linking $P = I^2 R$P=I2R to specific heat capacity.

3. Topic 5 — Waves (electromagnetic radiation): mains electricity in the UK is alternating at $50\ \text{Hz}$50 Hz, and the rated voltage $230\ \text{V}$230 V is the root-mean-square value chosen so that AC power dissipated in a resistor equals $V_\text{rms}^2/R$Vrms2​/R — the same form as the DC equation. This synoptic link prepares candidates for the deeper treatment of RMS quantities in further-study electromagnetism.

4. Topic 13 (further mechanics) and Topic 11 (Capacitance, exam-board variant): when a capacitor charges through a resistor, instantaneous power dissipated in the resistor is $P_R(t) = I(t)^2 R$PR​(t)=I(t)2R, and the total energy dissipated as heat over the full charging cycle equals the energy stored on the capacitor ($\tfrac{1}{2}CV^2$21​CV2). This appears on Paper 2 in the capacitors topic and rests entirely on $P = I^2 R$P=I2R together with calculus.