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In December 1947, John Bardeen and Walter Houser Brattain obtained transistor action at Bell Telephone Laboratories. William Shockley proposed the distinct bipolar junction transistor within the following month.^[ref]

Their 1948 paper describes ground, etched germanium, initially also anodized and rinsed. Surface contacts separated by $50$–$250\mu \mathrm{m}$ share a broad base contact. Positive emitter biasing injects electron holes; negative collector biasing draws them toward the collector. An input signal changes emitter electric current, and a collector electrical load converts the resulting change into voltage.^[ref]

Early electric field control experiments gave unexpectedly weak responses. John Bardeen proposed that surface states screened the interior. Surface charge could create a barrier and a hole-rich semiconductor inversion layer. Replacing an electrolyte control with a gold contact reversed the response sign, suggesting electron hole carrier injection. Closely spaced contacts made from a gold-coated wedge split with a razor then enabled power gain. The relative importance of surface and bulk transport depended on preparation and collector forming.^[ref]

The 1948 authors found that collector-current increments could exceed emitter-current increments; their current gain depended on biasing.^[ref] A terminal-current ratio alone cannot identify the microscopic path. Surface states, the local barrier, carrier injection and contact preparation prevent a single ideal modern bipolar junction transistor model from explaining every point-contact transistor.

For the conventional electric current directions drawn in the figure, $I_E$ enters the emitter, $I_C$ leaves the collector and signed $I_B$ enters the base. Kirchhoff's current law requiresThus a local current gain $\alpha >1$ requires an accompanying base-current change. The collector increment is not simply a count of injected electron holes, and it does not violate electric charge conservation.

For an experiment following the functional arrangement, first establish each contact's separate current-voltage characteristic relative to the base. Then hold collector biasing fixed, vary a small emitter signal, and compare the collector electrical load voltage with the input. A decisive power gain check measures both input electric current and voltage and the output power in a known electrical load; voltage enlargement alone is insufficient. These are proposed measurement controls, not additional claims about the 1947 laboratory record.

Repeating a setting tests drift; reversing or removing emitter biasing tests whether the response requires carrier injection. Contact motion, surface contamination, temperature, direct electrical coupling and noise can all affect the inference. A small-signal model describes measured local terminal response without deciding every microscopic mechanism.

A useful short-interval approximation replaces a battery by an ideal electromotive force $\mathcal{E}$ in series with a positive internal electrical resistance $r$. This is the Thévenin terminal response with open-circuit voltage $\mathcal{E}$ and Thévenin resistance $r$. The model describes the terminals without asserting that there is a separate physical resistor hidden inside the cell. Both $\mathcal{E}$ and $r$ can change with composition, temperature, electric current and history.^[ref]

For a resistive electrical load $R>0$, take $I>0$ to mean discharge: conventional current leaves the positive terminal. Kirchhoff's voltage law and Ohm's law giveFor the stipulated $\mathcal{E}=3\mathrm{V}$, $r=1\Omega$, $R=4\Omega$,An open-circuit voltage of $3\mathrm{V}$ therefore does not promise $3\mathrm{V}$ at this electrical load. Increasing the load electrical resistance reduces electric current and brings $V_L$ closer to $\mathcal{E}$, as in Thévenin load response.

The signed relation also explains charging within this model. If an external source drives electric current into the positive terminal, $I<0$ and $V_L=\mathcal{E}+|I|r$. The terminal electric potential difference then exceeds $\mathcal{E}$. This algebra is not evidence that every battery is rechargeable.^[ref]

In the constant-parameter battery internal resistance model, the source's available electrical power splits between the electrical load and internal dissipation:The stipulated $3\mathrm{V}$, $1\Omega$, $4\Omega$ calculation gives $P_L=1.44\mathrm{W}$, $P_{\mathrm{int}}=0.36\mathrm{W}$, and $\mathcal{E}I=1.80\mathrm{W}$. A resistive electrical load converts its share to Joule heating; another electrical load might convert some of it to mechanical work or light. The source term is available electrical work per time, not a complete cell heat budget: reaction entropy can also matter.

Define the delivery efficiency relative to the source's available electrical work, rather than to all chemical energy or to a complete charging cycle:Here $\eta =0.80$. The largest load power occurs at $R=r$: since $(R+r{)}^2\ge 4Rr$,Thus maximum load power and maximum delivery efficiency ask different questions. As $R\to \infty$, $V_L\to \mathcal{E}$ and $\eta \to 1$, but $I\to 0$ and $P_L\to 0$: a high nearly unloaded voltage supplies almost no power. As $R\to 0$, the model gives $V_L\to 0$, $P_L\to 0$ and internal power tending to ${\mathcal{E}}^2/r$. Real high-current operation can invalidate constant $r$ and $\mathcal{E}$; the mathematical maximum does not establish a permissible operating point for an actual battery.

For a linear filter with chosen nonzero reference voltage gain $H_0$, define its effective noise bandwidth byThis is the width of an ideal rectangular linear filter with gain magnitude $|H_0|$ passing the same mean square of input white noise. ConsequentlyUse the full voltage transfer from the open-circuit voltage source to the readout, including any electrical load, when applying this formula. A nominal signal bandwidth stated between two cutoff frequencies does not specify the integral: gradual tails and a nonuniform passband matter. For an ideal rectangular unit-gain passband $[f_1,f_2]$, $B_{\mathrm{n}}=f_2-f_1$. For an unloaded RC low-pass filter with $H_0=1$ and cutoff frequency $f_c=1/(2\pi RC)$,The last equality follows by substituting $x=f/f_c$ and integrating $1/(1+x^2)$ as $\arctan x$. Using the nominal $f_c$ as the noise bandwidth underestimates the variance by a factor $\pi /2$, or the root mean square voltage by $\sqrt{\pi /2}$.

An electric battery converts the available chemical Gibbs free energy of one or more galvanic cells into electrical work delivered to an electrical load. The reacting materials are a finite reservoir: a battery can sustain a terminal electric potential difference while discharging, but cannot supply energy indefinitely. Unlike a capacitor, whose capacitor energy resides in an electric field, a battery chiefly stores its available energy in the chemical state of its materials. The detailed half-reactions and apparatus of a Daniell cell explain this conversion at the electrode level.

During discharge, electrons move through metallic wires from the negative anode, through the electrical load, to the positive cathode; external conventional current points the other way. Inside the battery, ions carry electric current through the electrolyte, and redox chemical reactions connect ionic transport to electronic transport at the electrodes. The same electron does not make a lap through both wire and electrolyte. Electric charge conservation closes the circuit; changing chemical composition supplies the energy. See galvanic cell for the two transport paths and electrolyte for how oppositely moving ions contribute to one net electric current.

The source's electromotive force $\mathcal{E}$ measures its driving work per unit electric charge, in volts; the loaded terminal electric potential difference $V_L$ measures the electrical work per unit electric charge delivered outside the battery. They agree in an ideal lossless source, or approximately near open circuit, but generally differ during discharge.^[ref]

In the ideal fixed-voltage model used in fixed-voltage capacitor energy balance, a battery maintains a prescribed terminal electric potential difference while electric charge flows to or from the capacitor. This is a constraint for that calculation, not an inexhaustible reservoir. The fixed-voltage capacitor energy balance also accounts for electrical energy returned to the source; an actual cell can accept sustained charging only if its chemistry and operating conditions permit it.

Connect a passive electrical load of electrical resistance $R_L>0$ from the junction to the reference. In the steady direct current model, $R_L$ and $R_2$ share the same electric potential difference and are connected in parallel. Their equivalent electrical resistance isThis is the replacement required when a voltmeter is connected across $R_2$. The inspected teaching guide explicitly models the meter this way.^[ref]

The Kirchhoff's current law proof makes the changed assumption visible:The upper resistor carries the sum of the lower and load electric currents, so the two original resistors no longer have equal electric currents. In the stipulated $R_L=10\mathrm{k\Omega }$ example, $R_p=5\mathrm{k\Omega }$, the upper current is $2/3\mathrm{mA}$, and each lower branch carries $1/3\mathrm{mA}$.

For a digital multimeter used as a voltmeter, let its input electrical resistance on the selected voltage range be $R_m$. With no other load, set $R_L=R_m$; if an intended load is also present, use $R_L\parallel R_m$ in place of $R_L$. Two instruments connected simultaneously add two branches. Their ideal indicated value equals the voltage of this altered circuit; voltmeter loading can exist even with perfect measurement calibration. Real readout additionally has measurement uncertainty, offsets, and finite resolution. Consult the actual range and input specifications rather than assigning every digital multimeter the same $R_m$.

For an oscilloscope, the resistance-only model applies to settled direct current with a compatible input coupling setting. The oscilloscope input and cable also have capacitance; changing signals can therefore have loading that depends on frequency. A probe changes the impedance seen at its tip and the relation between tip voltage and displayed voltage; use the probe-and-input combination's specification and attenuation setting. NJIT's guide models the parallel electrical resistance and capacitance and adjusts probe compensation with a square-wave readout.^[ref]

Photovoltaic mode uses no external reverse bias: a transimpedance amplifier can hold zero terminal electric potential difference, while an open electrical circuit develops a photovoltage. Photoconductive mode biases the diode cathode positive relative to the diode anode, generally reducing junction capacitance and increasing speed and dark current.^[ref]

A simple calculation separates the readout from the conversion. If a linear transimpedance amplifier has feedback electrical resistance $R_f=1.00\mathrm{M\Omega }$, the $650\mathrm{nm}$ model above produces an output change of magnitude $R_f{I}_{\mathrm{ph}}=0.393195\mathrm{V}$. The output sign depends on the connection. Increasing $R_f$ increases that electric potential difference, but does not create more collected charge carriers.

For a reverse-biased photodiode feeding an electrical load $R_L$, Ohm's law gives a readout magnitude $V_L\simeq I_{\mathrm{ph}}{R}_L$ while the response remains linear. With negligible dark current and other series drops, the remaining reverse bias is $V_R\simeq V_b-I_{\mathrm{ph}}{R}_L$, where $V_b$ is the supply magnitude. Thus the chosen electrical load can consume the available bias: a larger readout does not automatically reduce measurement uncertainty. For the same constructed example, $V_b=5\mathrm{V}$ and $R_L=1\mathrm{M\Omega }$ leave about $4.60680\mathrm{V}$; this algebra alone does not establish a real device's linear range.

Thorlabs' silicon FDS100 experiment sends a collimated, fibre-coupled laser beam through a beam splitter to the photodiode and a reference optical power meter. Varying reverse bias and electrical load tests linearity. The measured readout approaches saturation when its electric potential difference consumes the bias.^[ref] A separate low-signal setup uses a $637\mathrm{nm}$ laser, polarizers, an optical chopper, optical attenuators and a lock-in amplifier inside a light-tight enclosure. These are actual apparatus choices, not the parameters of the constructed example above.^[ref]

For an independent dark current check, Thorlabs enclosed a photodiode in nested metal boxes to exclude light and suppress interference, used a Keithley 6487 ammeter supplying reverse bias, and monitored temperature. Only one specimen per type was tested; the nearby temperature sensor did not directly contact the semiconductor. Such a dark control measures the baseline but cannot certify spectral responsivity.^[ref]

To infer spectral responsivity at a selected wavelength, measure the radiant flux actually reaching the test photodiode, rather than just the reference arm. A reference-arm measurement calibration must include the beam splitter ratio and path losses. With signed readouts in one fixed orientation, subtract a beam-blocked baseline and usefor direct electric current readout or a calibrated linear transimpedance amplifier, respectively. Repeating at several incident powers tests whether this ratio is constant. Blocking only the source estimates ambient light plus electronic baseline; covering the detector excludes ambient light too. Their difference distinguishes optical background from dark current and readout offset. For a broad optical spectrum, the ratio is a spectrum-weighted response, not generally $R$ at one wavelength.

For independent Poisson distribution collection, the root mean square shot noise in a one-sided rectangular frequency band of signal bandwidth $B>0$ isHere $I_d$ is reverse dark current; thermal electrical noise and amplifier noise also contribute.^[ref]

Subtracting a mean dark current removes its offset, not its fluctuations: subtracting a constant leaves the variance unchanged. Consequently $I_{\mathrm{ph}}\gg I_d$ makes dark current shot noise small relative to signal shot noise; it does not by itself establish high signal-to-noise ratio. With an independently measured baseline, the variance of that estimate must also be propagated. Increasing the measurement signal bandwidth raises the admitted noise. Calibration of wavelength, reference radiant flux, readout gain and collection geometry, together with drift and linearity, determines what measurement uncertainty can be claimed.

A resistor with a movable contact that selects a point along its resistive path. In the Thévenin measurement experiment, a contact and one end provide an adjustable resistive electrical load.

An RC low-pass filter places a resistor $R>0$ between input and output, with a capacitor $C>0$ from output to their common return. The ideal model prescribes the input electric potential difference and leaves the output unloaded. It is a first-order low-pass filter with unit eventual gain.^[ref]

For electrical resistance $R$ and capacitance $C$, Ohm's law gives the electric current through the resistor as $(v_{\mathrm{in}}-v_{\mathrm{out}})/R$. The capacitor relation gives $Cd{v}_{\mathrm{out}}/dt$. Equating them by Kirchhoff's current law yields the ordinary differential equationThe time constant $\tau$ has units of time. A rapid change in electric potential difference requires a large electric current into the capacitor; the finite electrical resistance limits that rate. A slowly maintained voltage instead lets the capacitor approach the input voltage. The same equation therefore predicts both smoothing of sustained oscillations and a rounded response to a sudden change.

For the original constructed example throughout this article, takeThese stipulated ideal values specify an analytical model, rather than measurements of particular components.

A real electrical load can draw electric current from the output. For a load resistor $R_L>0$ across the capacitor, Kirchhoff's current law instead givesTherefore the loaded continuous-time frequency response isLoading reduces the eventual gain and shortens the time constant. With the constructed values and $R_L=10\mathrm{k}\Omega$, $K_L\approx 0.81967$, ${\tau }_L\approx 84.754\mu \mathrm{s}$, and the corner relative to the loaded low-frequency gain moves to $1.8778\mathrm{kHz}$. A $3\mathrm{V}$ step now approaches $2.4590\mathrm{V}$. A quicker rise alone would conceal the error in the final voltage.

With the ideal input voltage source set to zero, the output impedance seen by the electrical load isThis follows from Thévenin's theorem: the resistor and capacitor are then in parallel. A sufficient small-loading condition is $|Z_L(\omega )|\gg |Z_{\mathrm{out}}(\omega )|$ over the frequencies of interest. An oscilloscope is itself an electrical load: its input electrical resistance and cable or probe capacitance can both change the response. Pure extra parallel capacitance increases $C$ and lowers the corner, whereas a finite parallel electrical resistance gives the different gain-and-corner changes above.

The definition of input also matters. If voltage is prescribed inside a source with series electrical resistance $R_s$, the unloaded response to that internal source uses $R+R_s$. If the actual voltage at the filter input terminal is measured, the ratio of output to that measured input still uses $R$. Measuring both terminals avoids mistaking a source setting for the voltage delivered to the filter.

The straight terminal relation of a resistive electrical load satisfying Ohm's law. Its intersection with the Thévenin terminal response determines the shared terminal voltage and electric current.

Resistor tolerance specifies the permissible manufacturing deviation of a resistor's electrical resistance from its nominal value under stated conditions. For nominal electrical resistance $R_{\mathrm{nom}}>0$ and fractional resistor tolerance $0\le \tau <1$, the specified range is $R_{\mathrm{nom}}(1-\tau )\le R\le R_{\mathrm{nom}}(1+\tau )$. This differs from measurement uncertainty: resistor tolerance describes a component specification, while measurement uncertainty describes knowledge of a measured value; the tolerance alone is not a probability distribution for that value.

Two nominally equal resistors can have different actual electrical resistances, changing their voltage divider ratio even without an external electrical load. In a loaded voltage divider, use the actual electrical resistances to calculate both the unloaded output and the Thévenin resistance governing voltage divider loading error. A mismatch from resistor tolerance and a change from voltmeter loading are distinct effects that can occur together.

An external electrical load changes the terminal noise. Connecting a purely resistive load $R_L>0$ to the noisy resistor gives the voltage divider ratio $R_L/(R+R_L)$. The source resistor's contribution at the load isIn a rectangular signal bandwidth $B$, the power delivered by that source contribution to the load isbecause $(R-R_L{)}^2\ge 0$. Equality occurs at $R_L=R$: this is matched available noise power. Matching halves the voltage, quarters its power spectral density, and gives $P_{\mathrm{av}}=k_{\mathrm{B}}TB$. Dividing the squared open-circuit voltage by $R$ would instead give four times this available power.

These expressions isolate the source resistor's contribution. A real load resistor also generates thermal electrical noise. At the same temperature its independent contribution produces the equal reverse power flow, so the net thermal power transfer is zero at thermal equilibrium. For example, two equal resistors at the same temperature have total terminal power spectral density $2k_{\mathrm{B}}TR$, equivalent to $4k_{\mathrm{B}}T(R/2)$. One contribution alone is $k_{\mathrm{B}}TR$. The fictitious noiseless series $R$ in the Thévenin equivalent is not an additional physical source: adding its noise again would double count the original resistor.

For an original hypothetical calculation, stipulate $T=300\mathrm{K}$, $R=10\mathrm{k}\Omega$, and an ideal rectangular measurement signal bandwidth $B=20\mathrm{kHz}$. Using $k_{\mathrm{B}}=1.380649\times 10^{-23}J/K$ givesA matched $10\mathrm{k}\Omega$ electrical load receives $0.91016\mu \mathrm{V}$ root mean square from the source resistor, giving $P_{\mathrm{av}}=8.283894\times 10^{-17}\mathrm{W}$. These are stipulated model predictions, not laboratory measurements. Increasing $R$ raises the open-circuit voltage noise but does not increase the matched available power. Restricting the measurement signal bandwidth reduces root mean square noise as its square root; it improves signal-to-noise ratio only if the desired signal is preserved.

In the pictured original linear resistive circuit, the ideal voltage source keeps its upper node at $12\mathrm{V}$ relative to $b$. Let $R_1=2\mathrm{k\Omega }$ connect that node to $a$, and $R_2=4\mathrm{k\Omega }$ connect $a$ to $b$. With the external electrical load removed, the voltage divider givesTo find the Thévenin resistance, Thévenin source suppression replaces the ideal $12\mathrm{V}$ voltage source by a wire. Both resistors now join $a$ to $b$ in parallel, givingThis is the electrical resistance looking into the original network with its independent electrical source suppressed; the external electrical load is excluded.

An independent check uses Kirchhoff's current law at $a$, without suppressing anything:The Thévenin terminal response thus follows directly from the original topology. Shorting $a$ to $b$ in this ideal calculation gives $I_{\mathrm{sc}}=6\mathrm{mA}$: the lower resistor has zero voltage and the upper resistor limits the electric current. All numerical values here and in the figures are stipulated model values, not laboratory data.

A positive resistive electrical load obeys Ohm's law, $V_L=I{R}_L$. Its resistive load line intersects the Thévenin terminal response atprovided $R_{\mathrm{th}}+R_L\ne 0$. For the Thévenin divider example, $R_L=1\mathrm{k\Omega }$ gives $I=24/7\mathrm{mA}$ and $V_L=24/7\mathrm{V}$; $R_L=4\mathrm{k\Omega }$ gives $I=1.5\mathrm{mA}$ and $V_L=6\mathrm{V}$. At $R_L=R_{\mathrm{th}}=\frac{4}{3}\mathrm{k\Omega }$, the voltage is $4\mathrm{V}$, half the open-circuit voltage.

For this positive $R_{\mathrm{th}}$, increasing $R_L$ reduces the drawn electric current and raises $V_L$ toward $V_{\mathrm{oc}}$. This explains voltmeter loading: a voltmeter itself is an electrical load. The equivalence also applies to a nonlinear electrical load through intersection with its terminal relation, but an arbitrary load need not give a unique operating point. If negative $R_{\mathrm{th}}$ cancels positive $R_L$, the displayed formula is singular; a real active circuit's limits and dynamics must then be considered.

An authored teaching experiment uses an adjustable steady-voltage supply, a digital multimeter, fixed resistors, and a $10\mathrm{k\Omega }$ potentiometer or resistance decade box. Its original circuit uses a $10\mathrm{V}$ supply and $3.3$, $6.8$, and $4.7\mathrm{k\Omega }$ resistors; the student compares $8.2$ and $2.2\mathrm{k\Omega }$ electrical loads. The readout is load voltage, recorded against prediction and deviation. The student then removes the electrical load to measure the open-circuit voltage, disconnects the supply and substitutes a wire to measure the Thévenin resistance with the multimeter's resistance mode, and restores the supply for an independent method: adjust the variable electrical load until the voltmeter reads half the unloaded voltage. Finally the student builds the equivalent and repeats both load measurements. Its tables are blank recording forms, not reported observations. ^[ref]

The half-voltage inference assumes negligible voltmeter loading and positive $R_{\mathrm{th}}$. A finite voltmeter input electrical resistance $R_m$ actually gives an apparent unloaded readingWhen the same voltmeter remains connected across a variable electrical load $R_L$, the network sees $R_p=R_L{R}_m/(R_L+R_m)$. Setting its reading to $V_m/2$, rather than to the true $V_{\mathrm{oc}}/2$, givesThus the uncorrected half-reading method underestimates positive $R_{\mathrm{th}}$. In the original $12\mathrm{V}$ example, a stipulated $R_m=10\mathrm{k\Omega }$ gives $V_m=120/17\mathrm{V}\approx 7.059\mathrm{V}$ and half-reading load $R_L=20/17\mathrm{k\Omega }\approx 1.176\mathrm{k\Omega }$, instead of $\frac{4}{3}\mathrm{k\Omega }$. These are model predictions for a deliberately loading meter, not the specification or observed readout of the source experiment's instrument.

Finite readings can also estimate the terminal line without a physical short. For two distinct load settings with corrected delivered electric currents $I_1,I_2$ and terminal voltages $V_1,V_2$,If the voltmeter remains across each electrical load, the delivered electric current is $I_k=V_k(1/R_{L,k}+1/R_m)$, including the meter branch. Omitting it biases the inferred slope.

Useful additional controls are to measure each resistor separately before assembly, verify terminal polarity, repeat an earlier load setting to detect supply drift, and compare a third setting with the inferred Thévenin terminal response. These are proposed controls, not reported outcomes of the source experiment. Closely spaced electric currents make the slope estimate sensitive to voltage resolution and measurement error; widely separated settings may trigger source current limiting or Joule heating that changes electrical resistance. Lead and contact electrical resistance, meter input uncertainty and rounding limit the inference. An ohmmeter injects its own test electric current; its reading on a powered network is not the suppressed-network ratio. Measurements with independent sources physically removed do not characterize an active dependent electrical source whose control or supply has also been disconnected.

To determine the Thévenin resistance, remove the external electrical load and set each independent electrical source to zero in the circuit model. A zero ideal voltage source enforces zero electric potential difference, so replace it by a short connection. A zero ideal current source enforces zero electric current, so replace it by an open connection. Retain every resistor, including any explicitly modeled supply electrical resistance. Turning an actual power supply's front-panel control to zero is not evidence that it physically becomes an ideal wire.

Retain every dependent electrical source and its controlling variable: its response to an electrical test source is part of the network's electrical resistance. Impose terminal voltage $v_t$ with an electrical test source and calculate $j_t$ entering the suppressed network at $a$. For a nonzero test yielding a finite ratio,Alternatively inject a known entering electric current and solve for $v_t$. The test electric current is opposite to the delivered-current convention of the original Thévenin terminal response. ^[ref]

For an original illustration, put a positive resistor $R$ across $a,b$ and a dependent electrical source in parallel that injects electric current $g{v}_t$ from $b$ toward $a$, where $g$ has units of electrical conductance. Its control is the same terminal voltage $v_t$. Kirchhoff's current law givesWith $R=1\mathrm{k\Omega }$ and $g=0.5\mathrm{mS}$, a $1\mathrm{V}$ test draws $0.5\mathrm{mA}$, giving $R_{\mathrm{th}}=2\mathrm{k\Omega }$. Removing the dependent electrical source would falsely give $1\mathrm{k\Omega }$. With $g=2\mathrm{mS}$, the model instead has $R_{\mathrm{th}}=-1\mathrm{k\Omega }$; with $g=1\mathrm{mS}$ it draws zero test electric current for any imposed voltage and has no finite ratio. An active physical realization requires a supply and has a limited operating range.

For a finite nonzero $R_{\mathrm{th}}$ and a well-defined short-circuit response, $I_{\mathrm{sc}}=V_{\mathrm{oc}}/R_{\mathrm{th}}$. Thus $R_{\mathrm{th}}=V_{\mathrm{oc}}/I_{\mathrm{sc}}$ is usable when $I_{\mathrm{sc}}\ne 0$. A source-free positive resistor has both $V_{\mathrm{oc}}=0$ and $I_{\mathrm{sc}}=0$, so this quotient cannot reveal its electrical resistance; an electrical test source can.

Thévenin's theorem replaces the terminal behavior of a linear resistive circuit by an ideal voltage source in series with a Thévenin resistance, whenever its terminal electric potential difference is a well-defined affine function of the delivered electric current. Fix terminals $a,b$, write $V=\varphi (a)-\varphi (b)$, and take $I$ positive out of the network at $a$ into the electrical load, returning through $b$. ThenThe open-circuit voltage $V_{\mathrm{oc}}$ is the intercept at $I=0$; $-R_{\mathrm{th}}$ is the slope. Thus two parameters predict what happens when many different electrical loads are connected. The Wheatstone bridge detector loading calculation uses precisely this replacement to find the detector reading even when the bridge's two voltage dividers interact through its detector.

For example, a $12\mathrm{V}$ ideal voltage source with $2\mathrm{k\Omega }$ above and $4\mathrm{k\Omega }$ below a voltage divider junction supplies $V_{\mathrm{oc}}=8\mathrm{V}$ and $R_{\mathrm{th}}=\frac{4}{3}\mathrm{k\Omega }$ at that junction relative to the lower terminal. Attaching a $4\mathrm{k\Omega }$ electrical load reduces its voltage to $6\mathrm{V}$ and draws $1.5\mathrm{mA}$. The equivalent explains why an unloaded divider ratio alone cannot predict a loaded output.

The equivalence concerns the chosen terminals, including the power $VI$ delivered there. It does not identify internal branch electric currents or internal Joule heating. At open circuit the original divider still draws $2\mathrm{mA}$ from its $12\mathrm{V}$ supply, while the equivalent's series branch carries no electric current. ^[ref]