= James L. Kirtley Jr.
{c}
\b[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.<James Kirtley Jr. profile by MIT EECS>[Faculty title and Research Areas]{ref}
\Image[https://energy.mit.edu/sites/default/files/styles/2560w/public/news/james-kirtley.jpg.webp?itok=XAldEvn-]
{title=James L. Kirtley Jr., portrait in MIT Energy Initiative's 2016 interview}
{description=Identified James Kirtley photograph linked from the institutional interview; served directly by MIT.}
{height=280}
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.<Electric motors find new roles in robots, ships, cars, and microgrids by MIT Energy Initiative>[Sections Electrifying the Cheetah robot and Improving microgrids with smart motors]{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 $S=V_{\mathrm{pk}}I_{\mathrm{pk}}^*/2$. With <RMS phasors> $V=V_{\mathrm{pk}}/\sqrt2$ and $I=I_{\mathrm{pk}}/\sqrt2$, this becomes $S=VI^*$. The <complex conjugate> ensures that the relative phase enters the <active power>.<AC Power Flow in Linear Networks by James L. Kirtley Jr.>[Energy and Power, equations45–63; Section7.1, RMS amplitudes]{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 $\omega L=4\,\Omega$. Then
$$
I=\frac VZ=(2.4-3.2i)\,\mathrm A,
\qquad
S=VI^*=(48+64i)\,\mathrm{VA}.
$$
Thus <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 $\phi=\arctan(4/3)$ and $x=\omega t$. The corresponding <electric potential difference> and <electric current> are
$$
v(t)=20\sqrt2\cos x,
\qquad
i(t)=4\sqrt2\cos(x-\phi).
$$
Multiplying and applying a <trigonometric identity> gives <instantaneous power>
$$
\boxed{p(t)=48+80\cos(2x-\phi)\ \mathrm W.}
$$
Its 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>.
\Image[/kirtley-ac-power-waveform.png]
{title=A phase-shifted load returns energy during part of each cycle while absorbing positive average power}
{description=Original 20 volt RMS example with impedance $3+4i$ ohms. The voltage and current have power factor 0.6; their instantaneous product ranges from -32 to 128 watts and averages 48 watts. Values are illustrative rather than measured motor data.}
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