Electricity in five ideas

Every machine in this course, a pump, an elevator, a crane, turns electrical energy into mechanical work. Before any motor or drive makes sense, you need five quantities and two formulas.

FoundationStart here if this is new to you

Think of a circuit as water in pipes.

• Voltage is the pressure that pushes. It is measured in volts (V).
• Current is how much flows per second. It is measured in amperes (A).
• Resistance is how narrow the pipe is. It is measured in ohms (Ω).

More pressure pushes more water through. A narrower pipe lets less through. That is all Ohm's law says.

Voltage, current and resistance

For a resistor, current is proportional to voltage:

$$V = R \cdot I$$

If you double the voltage across the same resistor, the current doubles. If you double the resistance at the same voltage, the current halves.

Power

Power is the rate at which energy is transferred, in watts (W). In an electric circuit it is voltage times current. For a resistor, Ohm's law gives two more forms:

$$P = V \cdot I = R \cdot I^2 = \frac{V^2}{R}$$

Energy is power multiplied by time. Electricity is billed in kilowatt-hours: 1 kWh is 1 kW for one hour, or 3.6 MJ.

Predict first

A drive's DC bus sits at 760 V. You replace a 40 Ω braking resistor with a 20 Ω one. What happens to the power it can absorb?

It halvesIt stays the sameIt doublesIt quadruples

Voltage V

760 V

Resistance R

35.0 Ω

Current I

21.71 A

Power P (heat in R)

16.50 kW

I = V / R P = V·I = R·I² = V² / R

Set the bench to 760 V and find the resistance that absorbs 16.4 kW. That is the braking resistor of Lab III.

Efficiency

No machine is perfect. Efficiency η (eta) is useful output power divided by input power:

$$\eta = \frac{P_\text{out}}{P_\text{in}} < 1$$

The difference becomes heat. In a motor it comes from three places:

• copper losses: $I^2 R$ in the windings
• iron losses: the magnetic core heating as it is magnetised back and forth
• mechanical losses: bearing friction and air resistance

A motor with η = 0.9 that delivers 18 kW to its shaft draws 20 kW from the supply, and 2 kW becomes heat. That heat is what limits how hard you can work a motor: its insulation has a maximum temperature.

ExplorerGo deeper: derivations and open questions

Why $I^2R$ matters at scale. Transmission lines carry power at high voltage because, for the same power $P = VI$, a higher voltage means a lower current, and losses in the line scale with $I^2$. Doubling the voltage cuts line losses to a quarter.

Open question. A 15 kW motor runs 8 000 hours a year. How much money does raising its efficiency from 90 % to 94 % save at 0.15 € per kWh? (Answer: input power drops from 16.67 kW to 15.96 kW, saving about 0.71 kW × 8 000 h ≈ 5 700 kWh, or about 850 € a year.)