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A direct current is an electric current that flows in one direction; its magnitude may vary with time. A constant or steady electric current additionally has a value independent of time. Thus a direct current need not be constant. The steady direct current model of a voltage divider assumes a constant input electric potential difference and fixed electrical resistances.

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.

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]

Southern Illinois University's ET 304A guide, Lab 4: Voltage Measurements and Meter Loading Effects, documents an instructive test: build equal-arm voltage dividers on a $20\mathrm{V}$ supply, then measure across the lower arm with a digital multimeter, analog multimeter, and oscilloscope. Starting with two $10\mathrm{k\Omega }$ resistors, repeat with equal pairs from $220\mathrm{k\Omega }$ to $4.7\mathrm{M\Omega }$. It requires separate recorded readings, unloaded predictions, and predictions including each instrument's input electrical resistance. The guide obtains meter resistances from specification sheets and scope resistance from the input-panel marking. Equal ratios control the unloaded target while changing susceptibility to voltmeter loading. The provided tables are blank: the guide specifies a procedure, not measured outcomes.^[ref]

Preparation and accuracy checks are made concrete by Berkeley's Lab 1: Introductory Experiments and Linear Circuits I. Its different divider uses nominal $470\mathrm{k\Omega }$ and $10\mathrm{k\Omega }$ resistors and a $24\mathrm{V}$ supply on a breadboard. Measure isolated resistors first, record actual supply voltage, recalculate with these measured values, then record the junction voltage with the digital multimeter. It explicitly compares actual values with resistor tolerance and the final voltage with a range-dependent accuracy bound from the instrument manual. This is an uncertainty check, not an independent measurement calibration.^[ref]

For a controlled comparison, additionally record the selected instrument range, probe factor, and input coupling; connect each instrument separately to the same electrical reference potential, and monitor the input rather than treating the supply label as an exact voltage. These are proposed controls for interpreting the documented procedures. A voltage indication and a resistance indication from the same digital multimeter do not independently validate its measurement calibration; a traceable reference comparison would be needed for that. Resistor tolerance describes component variation, while measurement uncertainty describes what is known about measured values. Using nominal arms alone can confuse mismatch with loading: equal nominal arms each within one-percent tolerance can give unloaded ratios from $0.495$ to $0.505$.

The inference to test is specific: once actual arm values, input voltage, and meter input electrical resistance are accounted for, the readout should approach the loaded voltage divider prediction within the remaining measurement uncertainty. A discrepancy can instead reflect contacts, changing temperature, source droop, or an input that is not a fixed passive resistor. A source with positive series electrical resistance $R_s$ can be modeled by replacing $R_1$ with $R_1+R_s$ only when $V_s$ denotes its ideal internal voltage; when the actual voltage across the divider is measured, that terminal value is already $V_{\mathrm{in}}$. During settling, or with appreciable input capacitance and changing signals, the steady direct current calculation is insufficient. Neither institutional procedure supplies measured values for the original $10\mathrm{V}$ example above.

First disconnect the detector. Each side is a voltage divider. Ohm's law and Kirchhoff's voltage law give the open-circuit intermediate electric potentials and their difference:Setting this difference to zero and cross-multiplying givesThe balance compares ratios, not necessarily equal arm values or equal branch electric currents. At balance the detector carries no electric current, so Kirchhoff's current law again makes the electric current constant along each two-arm path.^[ref]

For a worked null measurement, suppose $R_1=1.000\mathrm{k\Omega }$ and $R_2=2.000\mathrm{k\Omega }$, and the detector null occurs at adjustable $R_3=1.500\mathrm{k\Omega }$. ThenFor $V_s=3.000\mathrm{V}$, both intermediate electric potentials are $2.000\mathrm{V}$. The left branch carries $1.000\mathrm{mA}$ and the right branch $0.6667\mathrm{mA}$: the zero detector reading does not mean that the unknown carries no electric current. These are stipulated ideal values illustrating the calculation, not experimental observations.

A real voltmeter has finite input electrical resistance $R_d$; a galvanometer also has a finite winding electrical resistance. Away from balance, its electric current changes both intermediate electric potentials. This is voltmeter loading when the detector is a voltmeter. Model the detector as a passive resistor, with no offset or injected electric current. Thévenin's theorem supplies the full response without pretending the connected bridge remains two independent voltage dividers.

With the detector removed, the equivalent source is $V_{\mathrm{oc}}$. Replace the ideal supply by a wire to calculate the source electrical resistance seen between $B,C$: nodes $A,D$ merge, making $R_1,R_2$ parallel and $R_3,R_x$ parallel. ConsequentlySince the attenuation factor is strictly positive for $R_d>0$, the null remains $R_2{R}_3=R_1{R}_x$. A smaller detector electrical resistance reduces the measured electric potential difference off balance but does not shift the ideal null. Its electric current sensitivity instead depends on $1/(R_{\mathrm{th}}+R_d)$; detector gain and resolution also matter in deciding how closely a null can be located.

In the plotted example, $R_x=1.100\mathrm{k\Omega }$ gives $V_{\mathrm{oc}}=-119.048\mathrm{mV}$ and $R_{\mathrm{th}}=1.02381\mathrm{k\Omega }$. A detector with $R_d=1.000\mathrm{k\Omega }$ instead reads $V_d=-58.824\mathrm{mV}$. Using that loaded reading in the unloaded voltage divider formula would underestimate the deviation of $R_x$ from $1\mathrm{k\Omega }$. Balancing the bridge avoids this particular systematic measurement error.

The local derivative makes the sensitivity explicit. For $R_1=R_2=R_3=R$ and $R_x=R+\delta R$, the first-order Taylor expansion isAt the plotted null, the open-circuit slope is $-1.25mV/\Omega$, compared with $-0.625mV/\Omega$ for $R_d=1\mathrm{k\Omega }$. Increasing the supply improves this ideal sensitivity but also increases Joule heating. A finite source electrical resistance requires recomputing $R_{\mathrm{th}}$; the displayed parallel-pair expression assumes an ideal supply.