Physics for ME · Chapter 14 of 16 · Advanced · Mechatronics preparation
Electricity, Circuits, and Magnetism for Mechanical Engineers
Sensors, motors, actuators, and instrumentation all speak volts and amps. This chapter teaches enough E&M to be dangerous in the right way.
Readiness check
From Chapters 1 and 6. Tick only what you can do closed-notes.
- Keep W, V, A, and Ω straight as units.
- Run a power audit: input, output, losses (Chapter 6).
- Rearrange three-symbol formulas instantly.
- Use the right-hand rule from Math Chapter 3's cross product.
- Read a simple schematic symbol set (battery, resistor, switch).
The core idea
Voltage pushes, current flows, resistance resists, and power is their product. Magnetism converts it all to force.
V = IRP = VI = I²RF = BILOhm's law plus the power relations run every circuit estimate. A current in a magnetic field feels force (motors); a moving conductor in a field generates voltage (generators, induction sensors). Same physics, both directions.
What this chapter covers
- 14.1 Charge, current, voltage: the flow picture.
- 14.2 Ohm's law and resistance: V = IR, series and parallel.
- 14.3 Circuit power and losses: P = VI and the I²R heat tax.
- 14.4 DC circuits for instrumentation: dividers, bridges, sensor basics.
- 14.5 Magnetic force on currents: F = BIL, the motor principle.
- 14.6 Induction: moving conductors and changing fields make voltage.
- 14.7 Motors, generators, and actuators: the duality in hardware.
Engineering connection: sensors, motors, actuators, mechatronics, instrumentation; benchmark MIT 8.02.
Worked example: the motor's power budget
A 24 V DC gearmotor draws 5 A at its working load and delivers 90 W of mechanical power at the shaft. Its winding resistance is 0.8 Ω. Build its power budget: input, copper loss, other losses, and efficiency.
- ProblemComplete the power budget for the motor in Figure 1.
- Given / findV = 24 V, I = 5 A, R = 0.8 Ω, Pshaft = 90 W. Find input, losses, efficiency.
- AssumptionsSteady operation; all unaccounted power becomes friction, brush, and iron losses.
- ModelElectrical input P = VI; copper loss I²R; the remainder splits between shaft output and other losses.
- EquationsPin = VI PCu = I²R η = Pout/Pin
- SolvePin = 24 × 5 = 120 W. Copper: 25 × 0.8 = 20 W. Other losses: 120 − 20 − 90 = 10 W. Efficiency: η = 90/120 = 75%.
- CheckThe budget closes: 90 + 20 + 10 = 120. Sanity: 75% is typical for a small geared DC motor; the 30 W of heat must leave through the case, which is a Chapter 11 cooling question.
- ConclusionVI, I²R, and a closing audit characterized the machine without opening it: this is how mechatronics engineers vet every actuator datasheet, and why stalled motors (all 120 W into 20-becomes-576 W of copper heat at stall current) burn out.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Voltage flowing through things | "24 volts flow into the motor" | "Which quantity moves, and which pushes?" | Current flows through; voltage is the push across. Pressure and flow, not two flows. |
| Power confused with current | Fuses sized in watts, motors in amps alone | "Is this a rate of energy or of charge?" | P = VI joins them. Fuses guard current; energy bills count watts times time. |
| Series and parallel rules swapped | Resistances added in parallel | "Same current through, or same voltage across?" | Series: same current, resistances add. Parallel: same voltage, conductances add. |
| Stall treated like heavy load | Burnt windings after a jam | "What limits current when the shaft cannot turn?" | Only R: stall current = V/R = 30 A here, with I²R = 720 W of heat. Jams kill motors in seconds. |
Practice ladder
A 12 V heater element has resistance 6 Ω. Find the current and power.
Show answer
I = 2 A; P = VI = 24 W (or I²R = 24 W: same number, good cross-check).
A strain-gauge bridge is fed 10 V through two equal 350 Ω arms in series. What is the midpoint voltage, and why do sensor circuits love this divider?
Show answer
Equal resistances halve the supply: 5 V. A tiny resistance change in one arm shifts the midpoint proportionally: the divider converts resistance change (strain) into measurable voltage.
A linear actuator's conductor carries 8 A across a 0.5 T field over an active length of 0.12 m. Find the force, and the power converted at 1.5 m/s.
Show answer
F = BIL = 0.5 × 8 × 0.12 = 0.48 N per conductor; with 50 conductors: 24 N. Power at 1.5 m/s: P = Fv = 36 W. F = BIL scaled by turns is exactly how voice coils and linear motors are sized.
Take a real motor datasheet (drone, printer, e-bike). Extract voltage, rated and stall currents, and resistance if given; build the full power budget at rated load and the heat figure at stall.
What good work looks like
The budget closing within datasheet rounding, the stall I²R computed with its survival-time implication, and one sentence on the cooling provision the design carries.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Measure a real device's electrical appetite: with a plug-in power meter or USB tester, log one device's voltage, current, and power in two operating states. Build its budget and estimate where the losses go.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Define current, voltage, and resistance with their water analogues.
Current (A): charge flow, like volume flow. Voltage (V): the push across, like pressure difference. Resistance (Ω): opposition, like a narrow pipe.
2. Write Ohm's law and the three power forms.
V = IR; P = VI = I²R = V²/R: pick the form matching what you know.
3. State the motor and generator principles in one line each.
Motor: current in a field feels F = BIL. Generator: a conductor moved through a field develops voltage (induction).
4. Why is stall the worst case for a motor?
No back-conversion limits current; I = V/R is maximal and I²R heating is at its peak with zero cooling airflow.
5. What does a Wheatstone-style divider do for sensing?
It converts small resistance changes (strain, temperature) into a voltage an instrument can read.
Textbook mapping
| Item | Mapping |
|---|---|
| Main source | OpenStax University Physics Vol. 2, Electric Charge through Electromagnetic Induction (selected sections) |
| Benchmark / reference | MIT 8.02 Electricity and Magnetism · Young and Freedman |
| Core topics | 14.1 Charge, current, voltage · 14.2 Ohm's law · 14.3 Power and losses · 14.4 Instrumentation circuits · 14.5 F = BIL · 14.6 Induction · 14.7 Motors and generators |
| Engineering connection | Sensors, actuators, instrumentation, electrical machines: the runway for Mechatronics. |
| Read next | Chapter 15: Measurement, Uncertainty, and Experimental Physics. |