A motor converts electrical energy to mechanical; a generator runs the same physics backward. Both rely on the interaction of magnetic fields from a stationary and a rotating part.
Source (supplies electrical energy), stator (stationary electromagnets), rotor (spinning part that produces torque), and shaft (delivers mechanical output).
Electrical (\(I^2R\) in the windings), magnetic (core losses from the alternating field), mechanical (friction and windage), and stray. All of it becomes heat.
\(\eta = P_\text{out}/P_\text{in} \times 100\%\). It is always less than 100%, and the missing power is left as heat.
In three steps. Their product is the apparent power, \(S = V_\text{rms}I_\text{rms}\) (VA) — not \(P\). The power factor converts it to real power, \(P = S \times \text{pf}\) (W). Efficiency then uses the real power at both ends, \(\eta = P_\text{out}/P_\text{in}\) — it never divides by \(S\). Because \(S \geq P\), dividing by \(S\) always makes the machine look worse than it is.
\(P_\text{in} = P_\text{out}/\eta\). Dividing by \(\eta\) moves upstream; multiplying moves downstream.
Series, so there is one current: \(I_\text{rms} = 120/(1+9) = 12\ \text{A}\). Purely resistive, so \(\text{pf} = 1\) and \(S = P\). Then \(P_\text{load} = 12^2(9) = 1296\ \text{W}\) and \(P_\text{loss} = 12^2(1) = 144\ \text{W}\), so \(\eta = 1296/1440 = 90.0\%\). Modeling the waste as a series resistor gives the same number straight from the resistances, \(\eta = R_\text{load}/(R_\text{load} + R_\text{loss}) = 9/10\).
They multiply: \(\eta_\text{total} = \eta_1\eta_2\eta_3\cdots\). Three stages at 90% each give \(0.9^3 = 72.9\%\), not 90%.
\(1\ \text{hp} = 745.7\ \text{W}\). Convert before doing anything else.
Because the load's demand is usually the known quantity. Starting at the source means guessing and iterating.
Mechanical output: \(P_\text{out} = 5 \times 745.7 = 3728.5\ \text{W}\). From the line, \(S = (240)(20) = 4800\ \text{VA}\), so \(P_\text{in} = S \times \text{pf} = 4800(0.85) = 4080\ \text{W}\). Then \(\eta = 3728.5/4080 = 91.4\%\) and \(P_\text{loss} = 4080 - 3728.5 = 351.5\ \text{W}\) of heat the cooling has to remove. The trap: dividing by \(S\) gives \(3728.5/4800 = 77.7\%\), which is wrong — \(S\) is what the line must carry, \(P_\text{in}\) is what the motor is actually given.
The least efficient one — and where it sits in the chain makes no difference. Efficiencies multiply, so lifting one stage from \(\eta\) to \(\eta'\) scales the source power by \(\eta/\eta'\) wherever that stage is; only its own \(\eta\) enters. (Position matters only for the different question of how far a watt of that stage's own loss propagates back upstream.)
It is \(P_\text{out}/S_\text{in}\) — overall conversion effectiveness, the useful work a source gets per volt-ampere it must supply. It is not a new efficiency: \(\eta\) itself still never divides by \(S\). It answers a different question — what burden per volt-ampere a load places on its source — and it is what sizes the wiring, \(I = P_\text{out}/(V\,\eta\,\text{pf})\), so a 92%-efficient motor at pf 0.70 (product 0.64) stresses the supply more than an 88% one at pf 0.90 (0.79). It does not chain, because real power is conserved from stage to stage (which is why the \(\eta\) values multiply) but apparent power is not. Treat \(\eta \times \text{pf}\) as a measurement at one interface — the wall plug, say — never as a per-stage property.