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An actuator turns supplied energy and a control input into a mechanical action, such as a displacement, force or torque. An electric motor produces rotation; a hydraulic cylinder converts fluid pressure into a linear push. The sensor reports a system's state, while the actuator changes it.

In an ideal cylinder with piston area $A$ and pressure difference $\Delta p$, $F=A\Delta p$. Moving at speed $v$ requires volume flow $Q=Av$, so mechanical power $Fv$ equals hydraulic power $\Delta pQ$ before losses. This relation shows why force capacity, speed and supply power must be considered together. In a controlled mechanism, actuator saturation and response time constrain achievable feedback control.

An electric machine uses electromagnetism to transfer energy between electrical and mechanical forms, or between electrical circuits. An electric motor converts electrical input into mechanical output; an electrical generator converts mechanical input into electrical output. The wider machine category also includes the stationary electrical transformer. A rotating electric machine combines force on current-carrying conductors with electromagnetic induction; the same rotating conductors that develop torque can develop an opposing induced electric potential difference.^[ref]

For an original idealized direct current motor example, let $e=k_e\omega$, $\tau =k_{t}I$, and $V=e+IR$, where $e$ is the induced opposing electric potential difference, $\omega$ the angular velocity, $\tau$ the torque, and $R$ the winding electrical resistance. In coherent SI units, assume $k_e=k_t=0.1$, $\omega =100\mathrm{rad}{\mathrm{s}}^{-1}$, $I=2\mathrm{A}$, and $R=1\Omega$. Thenand the electromechanical conversion power balance isThe $20/24$ conversion fraction is about $83.3\%$ before friction and other losses. At rest the induced term vanishes, so applying the same $12\mathrm{V}$ gives $12\mathrm{A}$ in this simplified model. This makes the opposing induced electric potential difference experimentally visible: at fixed applied electric potential difference, a shaft-speed measurement and an electric current measurement distinguish a turning motor from its stalled state. Actual starting control and additional losses change the idealized numbers, not the need to balance electrical input, mechanical output, and heating.

An electric motor converts electrical energy into mechanical work, usually rotation, through electromagnetic forces. For a coil with $N$ turns, area $A$ and electric current $I$ in a uniform magnetic field $B$, the torque magnitude iswhere $\theta$ is the angle between the coil's normal and the magnetic field. Repeated rotation requires the current or effective field arrangement to keep supplying torque in the intended direction. Electrical input power must cover useful output plus losses.

A coil with $N=100$, $I=0.5\mathrm{A}$, $A=0.01{\mathrm{m}}^2$ and $B=0.2\mathrm{T}$ has maximum torque $0.1\mathrm{N}\mathrm{m}$. At angular speed $\omega$, instantaneous mechanical power is $P=\tau \omega$; the maximum-torque value alone does not specify a motor's sustained output.

Michael Faraday's 1821 rotation experiment made the conversion visible: a current-carrying wire moved around a fixed magnet while a pool of Mercury maintained electrical contact. The Royal Institution preserves his 1822 apparatus, consisting of a wire, glass vessel and fixed bar magnet connected through the conducting liquid. Continuous rotation, rather than a single deflection, established a route from steady electric current to repeated mechanical motion.^[ref]

Electromechanical conversion power balance applies conservation of energy to an electric machine's electrical and mechanical ports, internal storage and losses. In an electric motor sign convention, with electrical input positive and mechanical output positive,The internal term can include a changing magnetic field and rotational kinetic energy. Omitting it is justified only at the relevant steady operating point or after averaging over a cycle with no net storage change.

An original constant-field direct current motor model has winding electrical resistance $R$, inductance $L$, opposing induced electromotive force $e$, and current $i$:Multiplication by $i$ givesFor an ideal conversion coupling, $e=k_e\omega$ and electromagnetic torque ${\tau }_e=k_{t}i$ give $ei={\tau }_e\omega$ when the numerical constants agree in coherent SI units. The induced electrical term is converted power, not a resistive loss. DOE's motor analysis derives current from applied minus opposing induced electric potential difference and relates torque to field and armature current.^[ref]

Suppose $v=24\mathrm{V}$, $i=2\mathrm{A}$, $R=1\Omega$, $L=0.5\mathrm{H}$ and $di/dt=4\mathrm{A}{\mathrm{s}}^{-1}$. Then $e=20\mathrm{V}$ andThe terms are winding heating, increasing magnetic energy, and mechanical conversion. If $10\mathrm{W}$ is additionally going into rotational acceleration and mechanical losses are neglected, only $30\mathrm{W}$ reaches the shaft load at that instant. Calling all $40\mathrm{W}$ useful shaft output would omit a second storage term.

A proposed experiment records terminal electric potential difference, electric current, shaft speed and load torque during acceleration and after settling. The discrepancy between electrical input and load output can include storage as well as heating. For an electrical generator, reversing the port directions gives the corresponding mechanical-input balance; the physical energy accounting is unchanged.

An H-bridge is a four-switch arrangement that can apply either polarity of a DC supply across a two-terminal electrical load. Two switches form each side, with the load joining their midpoint nodes. Power inverters and bidirectional electric motor drives use this switching geometry. The primary TI lesson illustrates its components, forward/reverse states and inductive-current paths.^[ref]

In an idealized switching model, let $q_A,q_B\in \{0,1\}$ select the lower or upper rail at midpoint A and B, with supply magnitude $V_d$. ThenOpposite choices give $+V_d$ or $-V_d$ across the load. Equal choices give zero terminal difference even though both midpoints can be at the upper rail. This zero-output state does not mean every switch is open.

Each leg's upper and lower devices must not simultaneously provide a low-resistance path between supply rails: that is H-bridge shoot-through. Bridge switching dead time introduces a nonoverlap interval during a transition. Inductive load electric current still needs a path during that interval, often involving diodes or controlled conducting devices. The actual current path depends on the bridge state and current direction.

An original inductor example shows why reversing the commanded terminal polarity need not instantly reverse electric current. Neglect winding electrical resistance and opposing motor electromotive force, and take $L=10\mathrm{mH}$ with initial current $i=2\mathrm{A}$. Applying $v_{AB}=-12\mathrm{V}$ givesCurrent reaches zero only after about $1.667\mathrm{ms}$ if this voltage is maintained. Initially $v_{AB}i=-24\mathrm{W}$: the inductive load is returning stored energy through the bridge rather than absorbing it. Its initial magnetic energy is $L{i}^2/2=0.020\mathrm{J}$.

During rapid ideal bipolar switching, applying $+V_d$ for fraction $d$ and $-V_d$ for the rest gives interval-average output $(2d-1)V_d$. With $V_d=12\mathrm{V}$ and $d=0.75$, the average is $6\mathrm{V}$. This average does not specify instantaneous current or heating without a load model.

A proposed low-voltage observation records both midpoint potentials, differential load electric potential difference and load electric current. Comparing them through reversal distinguishes a changed switch command from changed current direction. Driver delays, device drops, recirculation and the supply's ability to accept returned energy determine how closely a real bridge follows the ideal states.

James L. Kirtley Jr. is an electrical engineer whose work concerns electric machines and electric power systems. MIT's departmental profile identifies him as a professor emeritus of electrical engineering, with energy and robotics among his research areas.^[ref]

The engineering problem is to match a motor's behaviour to its application. MIT reported in 2016 that Kirtley worked on customized electric motors for the MIT Cheetah robot, where brief peaks of torque serve running and jumping. The same interview describes his interest in electric motors interacting with microgrids, particularly their dynamic behaviour during supply disturbances.^[ref] A short burst and sustained operation place different demands on the system; the largest instantaneous output alone does not describe its operating role.

His AC Power Flow in Linear Networks by James L. Kirtley Jr. connects the phasor representation of electric potential difference and electric current to physical energy transfer. In its peak-amplitude convention, complex electrical power is $S=V_{\mathrm{pk}}{I}_{\mathrm{pk}}^{\ast }/2$. With RMS phasors $V=V_{\mathrm{pk}}/\sqrt{2}$ and $I=I_{\mathrm{pk}}/\sqrt{2}$, this becomes $S=V{I}^{\ast }$. The complex conjugate ensures that the relative phase enters the active power.^[ref]

For an original load example in sinusoidal steady state, take $V=20\mathrm{V}$ and electrical impedance $Z=(3+4i)\Omega$, using RMS phasors and the passive sign convention: electric current enters the positive-potential terminal. A series resistor and inductor realize this model at the specified angular frequency when $R=3\Omega$ and $\omega L=4\Omega$. ThenThus active power is 48 watts, reactive power is 64 vars, apparent power is 80 volt-amperes, and power factor is $48/80=0.6$. Write $\varphi =\arctan (4/3)$ and $x=\omega t$. The corresponding electric potential difference and electric current areMultiplying and applying a trigonometric identity gives instantaneous powerIts cycle average is 48 watts, while its range is $[-32,128]$ watts. The negative portions represent temporary return of stored energy from this ideal load; the electrical resistance still dissipates a positive average. A synchronized product of the two waveforms reveals this behaviour, whereas multiplying their root mean square magnitudes gives only the apparent power.

Jeffrey Lang is an electrical engineer whose work concerns electromechanical conversion power balance, control and small-scale devices. MIT's Research Laboratory of Electronics identifies the same professor as Jeffrey H. Lang and describes research on electric motors, electrical generators and microelectromechanical systems, together with teaching in circuits and electromagnetism.^[ref]

The identified MIT profile provides an externally hosted portrait of Jeffrey H. Lang.

For an original motor-scale illustration, a shaft delivering torque $\tau =2\mathrm{mN}\mathrm{m}$ at angular velocity $\omega =1000\mathrm{rad}{\mathrm{s}}^{-1}$ delivers $P_m=\tau \omega =2\mathrm{W}$. If electrical input is $3\mathrm{W}$ and the operating point is steady, the remaining $1\mathrm{W}$ must be accounted for by losses or another specified output. During acceleration some input can instead increase rotational kinetic energy. Reporting only shaft speed would not establish either output power or conversion efficiency.

With Anant Agarwal, Lang is credited on the MIT course material State and Memory by Anant Agarwal and Jeffrey Lang. It explains how a capacitor's state summarizes the part of past input needed to predict its future response.^[ref] For a series RC circuit with a constant later input $V_f$,Two different earlier input histories that leave the same initial capacitor electric potential difference give the same subsequent response under the same future input. For stipulated $RC=1\mathrm{ms}$, $v_C(0)=1\mathrm{V}$ and $V_f=3\mathrm{V}$, the value after one time constant is $3-2/e\approx 2.2642\mathrm{V}$. This original numerical example shows what it means for the state to retain enough of the past without retaining the entire waveform.

A machine is a physical system arranged to transmit or transform forces, motion and energy to perform a task. A lever changes the relation between applied and delivered force; an electric motor converts electrical energy into rotation; a pump transfers mechanical work to a fluid.

For an ideal slowly moving mechanism with no stored-energy change or dissipation, input and output work agree. If constant forces act through distances $d_{\mathrm{in}}$ and $d_{\mathrm{out}}$, then $F_{\mathrm{in}}{d}_{\mathrm{in}}=F_{\mathrm{out}}{d}_{\mathrm{out}}$. A larger output force therefore accompanies a smaller output displacement. Real machines also dissipate energy through friction, vibration and heating, so motion geometry and energy dissipation must be assessed separately.