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Keep the illumination and cell temperature approximately constant. Connect an ammeter and an adjustable resistor as a series circuit across the solar cell; connect a high-resistance voltmeter across its terminals. Vary the electrical resistance and record pairs of electric potential difference $V$ and delivered electric current $I$. This measures the illuminated current-voltage characteristic.

A load close to a short circuit estimates $I_{\mathrm{sc}}$ at $V\simeq 0$. With the load disconnected, an open circuit measurement using a high-resistance voltmeter estimates $V_{\mathrm{oc}}$ at $I\simeq 0$. Neither endpoint delivers useful power: $P=VI$ vanishes there. The intermediate readings locate the maximum of $VI$ and determine the solar-cell fill factor. Holding illumination fixed matters because the photogenerated electric current changes with light intensity; holding temperature fixed matters because the diode saturation current and junction response change with it.

This experiment distinguishes producing an electric potential difference from delivering energy to a load. To determine conversion efficiency as well, measure the incident power on the illuminated area, rather than equating it to a lamp's electrical rating.

An ideal open circuit permits no conduction electric current through a terminal pair, corresponding to infinite load electrical impedance. At the end of a lossless transmission line, its voltage-wave reflection coefficient is $+1$ and the terminal current is zero. Real open terminals may retain parasitic capacitance and radiation loss.

Take the longitudinal scalar piezoelectric coefficient appropriate to the load and electrode direction. For a uniformly stressed slab of face area $A$ and thickness $t$, the source charge is $\Delta Q=d\Delta F=dA\Delta \sigma$. Its capacitance is $C={ϵ}_0\kappa A/t$, so the ideal initially uncharged, electrically isolated sensor givesThe sign depends on the poling direction and electrode convention. Equivalently, the incremental constitutive law is $\Delta D=d\Delta \sigma +{ϵ}_0\kappa \Delta E$; an ideal open circuit keeps free electrode charge, hence $D$, fixed and gives $\Delta E=-d\Delta \sigma /({ϵ}_0\kappa )$. area cancels because both generated charge and capacitance scale with area. Thickness does not cancel: the material comparison is through $d/\kappa$ only when geometry is fixed. Here $\kappa$ must be the relative permittivity under the relevant mechanical and frequency conditions.

The original diagram has an ideal $5\mathrm{V}$ electric battery feeding series resistors of $8\Omega$ above the tapping point and $2\Omega$ below it. The long battery bar is positive, so terminal $A$ is above terminal $B$ in electric potential. With an open circuit at the output, the internal electric current is $5/(8+2)=0.5\mathrm{A}$. The voltage divider therefore givesFor the Thévenin resistance, suppress the ideal voltage source by replacing it with a short circuit. The two resistors are then in parallel as seen from $A,B$:By Thévenin's theorem, the terminal voltage under an outward load electric current $I$ is $V_A-V_B={\mathcal{E}}_0-I{R}_{\mathrm{Th}}$. Consequentlyflowing through an external short circuit from $A$ to $B$. Directly shorting the original terminals bypasses the $2\Omega$ resistor and leaves $5\mathrm{V}$ across $8\Omega$, independently confirming the result.

At direct current an ideal capacitor is an open circuit. A finite measured electrical resistance therefore requires the resistor and capacitor to form a parallel circuit, with $R=100\Omega$. Use phasors proportional to $e^{i\omega t}$. The electrical admittance and complex impedance areInterpreting the stated AC value as the measured impedance magnitude, $\omega =2\pi (50){\mathrm{s}}^{-1}$ and $|Z|=20\Omega$ give $\omega RC=\sqrt{24}$. ThusThe electric current leads the applied voltage by ${\tan }^{-1}\sqrt{24}=78.46^{\circ }$. Strictly, AC “resistance” is ambiguous: if it means $\mathrm{Re}Z=20\Omega$, the same parallel circuit instead gives $\omega RC=2$, $C=63.7\mu \mathrm{F}$, and $Z=20-40i\Omega$. A measurement of $|V|/|I|$ selects the boxed interpretation.

A suitable electrical circuit has $R$ in series with the input and a shunt branch consisting of $C$ in parallel with the lossy inductor $R_L+i\omega L$; take the output voltage across that shunt branch. This is the middle arrangement in the circuit figure. In terms of its complex impedance,At direct current the capacitor is an open circuit and the inductor contributes only $R_L$, so $H(0)=R_L/(R+R_L)=0.01$. ThereforeWhen $R_L$ is neglected, $Z_p=i\omega L/(1-{\omega }^{2}LC)$. At the antiresonance ${\omega }_0^{2}LC=1$ its electrical admittance vanishes, so $|H|$ reaches its maximum value of one. ConsequentlyThis value uses precisely the requested lossless approximation; keeping $R_L$ changes both the peak position and its height. A transfer function peak and one DC measurement do not uniquely identify every possible two-port network, but the displayed arrangement meets both observations to the stipulated approximation. Also, the angle brackets in the source must denote AC wave amplitudes or root mean square values: the literal time averages of sinusoidal voltages are zero, whose ratio would be undefined.

At zero angular frequency, the lower right capacitor is an open circuit, while the upper right electrical impedance tends to $R_x$. With the detector removed, no electric current flows down that branch, so $V_R=V_0$ and $V_L=V_0/2$. ThereforeFor the connected ideal ammeter, the resulting electric current from right to left is $V_0/(R+2R_x)$, despite its zero terminal voltage. This follows from $I_A=V_{\mathrm{oc}}/Z_{\mathrm{th}}$ with $Z_{\mathrm{th}}=R/2+R_x$.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.