Electrical quantities

The idea

Everything in this module is built from a handful of quantities. Think of a circuit like water in pipes: charge is the water, current is how fast it flows, voltage is the pressure pushing it, and resistance is how narrow the pipe is.

Quantity	Symbol	Unit	Meaning
Charge	$Q$	coulomb (C)	Amount of electricity
Current	$I$	ampere (A)	Charge per second: $I=Q/t$
Voltage	$V$	volt (V)	Energy per unit charge, the “push”
Resistance	$R$	ohm (Ω)	Opposition to current
Power	$P$	watt (W)	Energy per second
Energy	$E$	joule (J) or kWh	Power × time

Ohm’s law and resistance

$V=IR$

The resistance of a wire depends on what it’s made of and its shape:

$R=\rho \frac{L}{A}$

where $\rho$ is the material’s resistivity (Ω m), $L$ the length and $A$ the cross-sectional area. Longer or thinner means more resistance. Resistors are specified by resistance, power rating (how much heat they can take) and tolerance (e.g. ±1%).

Power and energy

Combine $P=VI$ with Ohm’s law and you get three forms. Use whichever matches what you know:

$P=VI=I^{2}R=\frac{V^2}{R}E=Pt$

• 1 J = 1 W for 1 s. Electricity bills use kWh: 1 kW for 1 hour = 3.6 MJ.
• Cost = energy in kWh × price per kWh. Watch the units: W × h gives Wh, so divide by 1000.

Efficiency

No real machine converts all its input power to useful output. The rest becomes heat.

$\eta =\frac{P_{out}}{P_{in}}\times 100\%P_{loss}=P_{in}-P_{out}$

Stages in series multiply: a supply (95%) feeding a drive (97%) feeding a motor (92%) gives $0.95\times 0.97\times 0.92=84.8\%$. The overall efficiency is always lower than the worst stage.

Small efficiency gains are worth a lot over time. A machine running 16 h a day, 300 days a year, at 3 kW output, wastes about £640 a year at 85% efficiency but only about £310 at 92% (at £0.25/kWh).

Fuses

A fuse is a wire designed to melt above a set current, opening the circuit before anything overheats. Choose the smallest standard rating above the normal operating current. A 1 kW charger on 240 V draws 4.2 A, so it gets a 5 A fuse.

Worked example

A robot’s 24 V battery supplies a drive motor drawing 4 A for 30 minutes. The motor’s mechanical output is 80 W. Find the power drawn, the charge and energy used, the motor’s resistance as seen from the supply, its efficiency, and a suitable fuse.

Answer

$P=VI=24\times 4=96$ W $Q=It=4\times 1800=7200$ C $E=Pt=96\times 0.5\text{h}=48$ Wh $=0.048$ kWh $=172.8$ kJ $R=V/I=24/4=6\Omega$ $\eta =80/96=83.3\%$, so 16 W is lost as heat. Normal current 4 A → 5 A fuse.

Common mistakes

• Mixing up W and Wh. Power is a rate; energy is power × time.
• Forgetting to convert minutes to hours (for kWh) or to seconds (for J).
• Writing $R=I/V$ instead of $V/I$. Check the units: Ω = V/A.
• Giving a voltage in Ω or a current in C. Units are marked.

Check yourself

1. What's the resistance of 10 m of copper wire with a 0.5 mm² cross-section? ( ${\rho }_{Cu}=1.72\times 10^{-8}$ Ω m)

$R=1.72\times 10^{-8}\times 10/(0.5\times 10^{-6})=0.344\Omega$. Remember mm² = $10^{-6}$ m².

2. A 2 Ω winding carries 5 A. How much power does it waste?

$P=I^{2}R=25\times 2=50$ W.

3. Three stages at 90%, 95% and 85%. Overall efficiency?

$0.9\times 0.95\times 0.85=72.7\%$.