Physics for ME · Module 14 of 16 · Mechatronics preparation
Electricity, Circuits, and Magnetism for Mechanical Engineers
Sensors, motors, actuators, and instrumentation all speak volts and amps. This module teaches enough E&M to be dangerous in the right way.
Readiness check
From Modules 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 (Module 6).
- Rearrange three-symbol formulas instantly.
- Use the right-hand rule from Math Module 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.
The skills, taught in order
Electricity yields to the same habits as the rest of this course: a few defining relations, an energy audit, and a flow analogy. Seven skills give a mechanical engineer enough to read a circuit and a motor or sensor datasheet.
14.1 Charge, current, and voltage
Current I is the rate of charge flow, measured in amperes; voltage V is the energy given to each unit of charge, the push, measured in volts. The water analogy holds throughout: current is the flow rate, voltage is the pressure driving it.
14.2 Ohm's law and resistance
For ordinary conductors, current is proportional to voltage: V = IR, with resistance R in ohms. In series the same current flows and resistances add; in parallel the same voltage applies and the combined resistance drops below the smallest.
| Connection | Shared quantity | Combine resistances |
|---|---|---|
| Series | same current | R = R₁ + R₂ + … |
| Parallel | same voltage | 1/R = 1/R₁ + 1/R₂ + … |
Micro-example. Two 100 Ω resistors make 200 Ω in series but only 50 Ω in parallel.
14.3 Circuit power and losses
Electrical power is P = VI, and using Ohm's law also P = I²R = V²/R. The I²R form is the heat tax every wire, winding, and resistor pays, and it is why thick cables and cooling matter.
Micro-example. A 5 A current through a 0.8 Ω winding wastes I²R = 20 W as heat, whatever useful work the device does.
14.4 DC circuits for instrumentation
A voltage divider and its balanced cousin, the Wheatstone bridge, turn a small resistance change into a readable voltage. This is how strain gauges, thermistors, and many sensors report a physical quantity as a signal.
14.5 Magnetic force on currents
A current of length L in a field B feels a force F = BIL, perpendicular to both, its direction given by the right-hand rule. Repeated over many turns, this is the force behind every motor, loudspeaker, and solenoid actuator.
14.6 Induction
The reverse also holds: move a conductor through a field, or change the flux through a loop, and a voltage appears. A rod of length L moving at speed v across a field B makes a motional EMF = BLv, and the induced current always opposes the change that made it (Lenz's law). Worked Example 2 follows this through.
14.7 Motors, generators, and actuators
Force-from-current and voltage-from-motion are the same physics read in opposite directions. A motor driven backward is a generator; an actuator pushed by hand becomes a sensor. Every electric machine lives on this duality.
| Law | Relation | Unit |
|---|---|---|
| Ohm's law | V = IR | V, A, Ω |
| Electrical power | P = VI = I²R = V²/R | W |
| Force on a current | F = BIL | N |
| Motional EMF | EMF = BLv | V |
Engineering connection: sensors, actuators, instrumentation, and electrical machines, the runway for Mechatronics.
Worked example 1: 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 Module 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.
Worked example 2: a rod generator
A conducting rod of length L = 0.25 m slides at v = 6 m/s along frictionless rails through a magnetic field B = 0.4 T pointing into the page. The rails are joined by a resistor R = 2 Ω. Find the generated voltage, the current, the force needed to keep the rod moving, and the power converted.
- ProblemFor the rod generator in Figure 2, find the EMF, the current, the force needed to keep the rod moving, and the power converted.
- Given / findL = 0.25 m, v = 6 m/s, B = 0.4 T (perpendicular), R = 2 Ω. Find EMF, I, F, and P.
- AssumptionsFrictionless rails of negligible resistance, uniform field perpendicular to the loop, the rod held at constant speed.
- ModelA conductor cutting field lines develops a motional EMF = BLv; that drives a current through R; the current in the field then feels a retarding force BIL that the applied force must balance.
- EquationsEMF = BLv I = EMF/R F = BIL P = Fv = EMF·I
- SolveEMF = 0.4 × 0.25 × 6 = 0.6 V. I = 0.6/2 = 0.3 A. Drag (and applied) force F = BIL = 0.4 × 0.3 × 0.25 = 0.03 N. Power: Fv = 0.03 × 6 = 0.18 W.
- CheckElectrical power EMF·I = 0.6 × 0.3 = 0.18 W, and I²R = 0.3² × 2 = 0.18 W: all three agree, so the mechanical power supplied equals the electrical power dissipated. Energy is converted, not created.
- ConclusionPush a conductor through a field and you pay mechanical power to harvest electrical power, the exact mirror of the motor in Example 1. This is every generator, the braking force in regenerative brakes, and the drag an eddy-current damper turns into smooth motion.
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.