Electrical Power and Energy

Spec mapping: OCR H556 Module 4.2 — Energy, power and resistance (electrical power $P = IV$ and its derived forms $P = I^2R$ and $P = V^2/R$ ; energy transferred $W = Pt = VIt$ ; the kilowatt-hour as a domestic energy unit; application to fuses, heating elements and transmission losses). Refer to the official OCR H556 specification document for exact wording.

Whenever charge moves through a potential difference, energy is transferred. A resistor turns that energy into heat; a lamp turns it into heat plus light; a motor turns it into mechanical work; a charger turns it into chemical potential energy in a battery. The rate of that energy transfer is the electrical power, and one equation — $P = IV$ — connects every scenario.

This lesson derives the three equivalent forms $P = IV = I^2R = V^2/R$ , computes total energy via $W = Pt$ , and brings the formula to bear on practical questions every OCR candidate should be able to solve cold: fuse selection, transmission-line losses, kWh-to-joule conversion, and the heating element design that closes the loop with resistivity (lesson 7). By the end you should treat any "power" question as a one-line substitution, not a guessing game.

Defining Electrical Power

Power is energy per unit time:

$P = \frac{W}{t}$

For electrical energy delivered to a component, energy per unit charge is the pd ( $V = W/Q$ ) and charge per unit time is the current ( $I = Q/t$ ). Multiplying:

$P = \frac{W}{t} = \frac{V Q}{t} = V \times \frac{Q}{t} = V \, I$

This gives the foundational electrical-power equation:

$\boxed{\; P = I\,V \;}$

Quantity	Symbol	SI unit
Power	$P$	watt (W) = J s $^{-1}$
Current	$I$	ampere (A) = C s $^{-1}$
Potential difference	$V$	volt (V) = J C $^{-1}$

The watt is a derived unit: $1$ W $= 1$ V A $= 1$ J s $^{-1}$ . The product of "joules per coulomb" and "coulombs per second" is "joules per second" — pure unit cancellation makes the equation memorable.

Quick Worked Calculation

A $12$ V car headlamp draws $5.0$ A. Its power dissipation is $P = IV = 12 \times 5.0 = 60$ W — enough that touching the bulb after switch-on burns your fingers. A $230$ V mains kettle drawing $9.0$ A dissipates $P = 230 \times 9.0 = 2070$ W $\approx 2.0$ kW — enough to boil 0.5 L of water in about a minute.

Three Equivalent Forms

For an ohmic resistor, $V = IR$ (lesson 5). Substituting into $P = IV$ gives two derived forms:

$P = IV \quad\Rightarrow\quad P = I(IR) = I^2 R \quad\Rightarrow\quad P = V \cdot \frac{V}{R} = \frac{V^2}{R}$

So:

$\boxed{\; P = IV = I^2 R = \frac{V^2}{R} \;}$

Mathematically equivalent for an ohmic resistor; each form is useful in different circumstances:

Form	Use when...	Reason
$P = IV$	$I$ and $V$ both known	Most general; works for any device, ohmic or not
$P = I^2 R$	the current is fixed (e.g. transmission cable)	Quadratic in $I$ — transmission-loss lever
$P = V^2/R$	the voltage is fixed (mains-rated appliance)	Quadratic in $V$ — domestic-rating lever

Important. The forms $P = I^2R$ and $P = V^2/R$ only follow when $V = IR$ holds — for an ohmic component, or for a non-ohmic component at a single operating point. For a non-ohmic component like a filament lamp, $P = IV$ is exact at every operating point but $I^2R$ and $V^2/R$ depend on the operating-point $R = V/I$ .

Worked Example 1 — Three Forms of the Same Question

A $60$ W lamp is rated for $230$ V mains. Calculate (a) the operating current, (b) its operating resistance.

(a) From $P = IV$ : $I = P/V = 60/230 = \mathbf{0.261}$ A $\approx \mathbf{260}$ mA.

(b) From $P = V^2/R$ : $R = V^2/P = 230^2/60 = \mathbf{882}$ $\Omega$ .

Cross-check: $R = V/I = 230/0.261 = 881$ $\Omega$ ✓; $P = I^2 R = 0.261^2 \times 882 = 60.1$ W ✓.

A single rating " $60$ W at $230$ V" determines both $I$ and $R$ at that operating point — three forms of the same law.

Energy Transferred Over Time

For constant power $P$ over time $t$ :

$W = P \, t = IVt = I^2 R t = \frac{V^2}{R} t$

with $W$ in joules and $t$ in seconds.

Worked Example 2 — Energy in a Kettle

A $2.0$ kW kettle runs on $230$ V mains for $3.0$ minutes.

(a) $I = P/V = 2000/230 = \mathbf{8.7}$ A.

(b) $W = Pt = 2000 \times 180 = \mathbf{3.6 \times 10^5}$ J $= \mathbf{360}$ kJ.

Cross-check: $360$ kJ is roughly the energy to raise $1$ L of water by $80$ K ( $c_w = 4200$ J kg $^{-1}$ K $^{-1}$ , $\Delta E \approx 336$ kJ). The kettle is close to 100% efficient — all resistive heating goes into the water.

The Kilowatt-Hour

Electricity bills use the non-SI unit kWh:

$1 \text{ kWh} = (1000 \text{ J s}^{-1}) \times (3600 \text{ s}) = 3.6 \times 10^6 \text{ J} = 3.6 \text{ MJ}$

Memorise this. OCR examiners use both units in the same question.

Worked Example 3 — Kettle Running Cost

$2.0$ kW for $3.0$ minutes at $30$ p/kWh.

$E = 2.0 \text{ kW} \times (3.0/60) \text{ h} = 0.10$ kWh.
Cost $= 0.10 \times 30 = \mathbf{3}$ p per cuppa.

UK average household: $\sim 3500$ kWh/year ( $\sim 13$ GJ; $\sim 400$ W continuous average).

Worked Example 4 — Joules to kWh

$0.50$ kWh $= 0.50 \times 3.6 \times 10^6 = \mathbf{1.8 \times 10^6}$ J.

Worked Example 5 — Fuse Selection

A $230$ V hair dryer rated $1200$ W. UK plugs allow $3$ A, $5$ A, $13$ A fuses.

Operating current: $I = P/V = 1200/230 = \mathbf{5.2}$ A.
$3$ A fuse blows immediately.
$5$ A fuse is marginal — slightest extra draw trips it.
The $\mathbf{13}$ A fuse is the appropriate choice.

UK rule: $3$ A fuse below $\sim 700$ W; $13$ A fuse above. The fuse exists to break the circuit on a fault, not to limit normal operating current.

Joule Heating

Joule's law $P = I^2 R$ is the rate of heating in a resistor — electrons hand directed kinetic energy to the lattice via scattering, the lattice randomises it as heat. Joule's 19th-century experiments established the equivalence of mechanical, chemical and electrical heating — the foundation of the first law of thermodynamics.

Practical consequences:

Heating elements (kettle, toaster, hair dryer) use high- $R$ nichrome to dissipate red-hot $P = I^2R$ . Resistivity (lesson 7) sets $R$ for a given length.
Connection wires use low- $R$ copper so wire dissipation is negligible compared to the appliance.
Transmission lines are long, so $R = \rho L / A$ is non-negligible. The lever to cut loss is not $R$ (economic limit) but $I$ — high-voltage transmission gives low $I$ at the same delivered power.

Worked Example 6 — Transmission Losses (The Grid)

A power station delivers $P_{\text{deliv}} = 10$ MW through a cable of resistance $R_{\text{cable}} = 5.0$ $\Omega$ .

Case A — Naive low-voltage at $V_A = 1000$ V:

$I_A = 10 \times 10^6 / 1000 = 10\,000$ A.
$P_{\text{loss,A}} = I_A^2 R = 10\,000^2 \times 5.0 = \mathbf{5.0 \times 10^8}$ W $= 500$ MW.

Losses are 50 $\times$ the delivered power. Impossible — the cable would melt.

Case B — Grid at $V_B = 400$ kV:

$I_B = 10 \times 10^6 / 400\,000 = 25$ A.
$P_{\text{loss,B}} = 25^2 \times 5.0 = \mathbf{3125}$ W $\approx 3.1$ kW.

Losses are $\mathbf{0.031\%}$ of delivered. Stepping $V$ up $400\times$ cuts loss by $400^2 = 160\,000\times$ , because $P_{\text{loss}} \propto I^2 \propto 1/V^2$ at constant delivered power.

This is the reason the UK grid runs at 132 kV, 275 kV, 400 kV — transformers step up at the station, down at the substation.

Worked Example 7 — Toaster Element (Synoptic with Lesson 7)

A toaster dissipates $900$ W on $230$ V. Element is nichrome ( $\rho = 1.1 \times 10^{-6}$ $\Omega$ m), diameter $0.50$ mm.

$R = V^2/P = 230^2/900 = \mathbf{58.8}$ $\Omega$ .
$A = \pi (0.25 \times 10^{-3})^2 = \mathbf{1.96 \times 10^{-7}}$ m $^2$ .
$L = RA/\rho = 58.8 \times 1.96 \times 10^{-7} / 1.1 \times 10^{-6} \approx \mathbf{10.5}$ m.

Coiled into a compact ribbon. Demonstrates the interlock between lesson 7 (resistivity) and this lesson (power): given a target power and a material, you compute the length.

Worked Example 8 — Battery Run-Time

A car battery: $12$ V, $60$ Ah. Powers a $55$ W headlamp.

$I = P/V = 55/12 = \mathbf{4.58}$ A.
$Q = 60 \text{ Ah} = 60 \times 3600 = 2.16 \times 10^5$ C.
$t = Q/I = 4.72 \times 10^4$ s $\approx \mathbf{13}$ hours.

Real lead-acid batteries deliver less than rated capacity because terminal voltage falls as chemistry depletes (lesson 9).

Worked Example 9 — Non-Ohmic Filament Lamp

A filament lamp at three operating points:

$V$ (V)	$I$ (A)	$P = IV$ (W)	$R = V/I$ ( $\Omega$ )
1.0	0.45	0.45	2.2
4.0	1.05	4.20	3.8
12.0	2.00	24.0	6.0

$R$ rises as $V$ rises because tungsten's positive temperature coefficient makes the filament more resistive at higher operating temperatures ( $\sim 2500$ K at full brightness). At full power, $P = IV = 12 \times 2.00 = 24$ W matches the rating.

Key lesson: for non-ohmic components, $P = I^2 R$ gives different answers at different operating points because $R = V/I$ depends on the operating point. Use $P = IV$ as the primary form.

Worked Example 10 — Specific Heat Capacity by Electrical Heating (PAG 4 + 5)

A standard OCR PAG involves measuring the specific heat capacity of a metal block by electrical heating. A heater of resistance $R = 4.0$ $\Omega$ is embedded in a $0.50$ kg aluminium block. The heater is connected to a $12$ V supply for $5.0$ minutes; the block's temperature rises from $20$ $^\circ$ C to $42$ $^\circ$ C.

Energy supplied electrically:

$W_{\text{elec}} = \frac{V^2}{R} t = \frac{12^2}{4.0} \times (5.0 \times 60) = 36 \times 300 = \mathbf{10\,800 \text{ J} = 10.8 \text{ kJ}}$

Heat absorbed by the block (assuming no losses):

$Q = m c \Delta T = 0.50 \times c \times (42 - 20) = 0.50 \times c \times 22 = 11 c$

Setting $W_{\text{elec}} = Q$ gives $c = 10\,800/11 = \mathbf{980 \text{ J kg}^{-1} \text{ K}^{-1}}$ . Accepted value for aluminium: $897$ J kg $^{-1}$ K $^{-1}$ . The 9% discrepancy is consistent with heat losses to the surroundings during the 5-minute heating — a sensible result for an A-Level practical.

Real practicals use lagging (foam insulation) and shorter heating times to minimise losses, or apply a correction by extrapolating the cooling curve back to the heating-stop time. The basic energy bookkeeping is $W_{\text{elec}} = mc\Delta T + Q_{\text{losses}}$ , with the electrical input from this lesson, the thermal output from Module 5.1.

Worked Example 11 — Domestic Energy Audit

A typical UK household has the following appliances running for the stated daily durations.

Appliance	Power	Hours/day	Daily energy (kWh)
Lighting (LED, total)	80 W	5	0.40
Refrigerator	100 W avg	24	2.40
Kettle	2200 W	0.25	0.55
TV / electronics	150 W	4	0.60
Washing machine	800 W avg	1	0.80
Standby loads	30 W	24	0.72
Total			5.47

Electrical Power and Energy

Electrical Power and Energy

Defining Electrical Power

Quick Worked Calculation

Three Equivalent Forms

Worked Example 1 — Three Forms of the Same Question

Energy Transferred Over Time

Worked Example 2 — Energy in a Kettle

The Kilowatt-Hour

Worked Example 3 — Kettle Running Cost

Worked Example 4 — Joules to kWh

Worked Example 5 — Fuse Selection

Joule Heating

Worked Example 6 — Transmission Losses (The Grid)

Worked Example 7 — Toaster Element (Synoptic with Lesson 7)

Worked Example 8 — Battery Run-Time

Worked Example 9 — Non-Ohmic Filament Lamp

Worked Example 10 — Specific Heat Capacity by Electrical Heating (PAG 4 + 5)

Worked Example 11 — Domestic Energy Audit

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