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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.

Two long round wires of radius $a_w\ll d$ carrying opposite electric currents form a loop with the displayed leading external self-inductance per unit length. Integrating their summed magnetic field across a connecting ribbon gives the logarithm; magnetic-energy integration gives the same leading value. The zero-radius limit diverges. Uniform direct current inside two nonmagnetic round wires contributes another ${\mu }_0/(4\pi )$, while an ideal surface current has no interior contribution.

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]

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.

An electrical transformer transfers alternating electrical power between windings through a shared changing magnetic flux. A core with large magnetic permeability confines most of the flux, so both windings link approximately the same flux $\Phi (t)$. Faraday's law gives induced electromotive forcesFor an ideal electrical transformer, negligible winding resistance and leakage flux imply the terminal-voltage magnitude ratioThe relative phase depends on the winding sense and terminal convention. At a loaded secondary, Lenz's law makes the secondary current oppose the flux change. The primary draws an additional current which approximately cancels the secondary magnetomotive force: $N_p{I}_p\simeq N_s{I}_s$ in magnitude when the magnetizing current is negligible. Consequentlyfor a resistive load, using root mean square values. More generally the ideal device conserves instantaneous power, with the appropriate current signs. A load impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_p/N_s{)}^2{Z}_L$, enabling impedance matching and electrical isolation.

For sinusoidal flux, $\Phi ={\Phi }_{\max }\sin \omega t$ gives $V_{\mathrm{rms}}=\omega N{\Phi }_{\max }/\sqrt{2}$. At a given supply voltage, reducing frequency increases the required peak flux and can drive the core into magnetic saturation. Real losses include winding Joule heating, magnetic hysteresis, eddy currents and leakage flux; a laminated core suppresses eddy-current loops. A steady direct current produces no continuing secondary electromotive force, while its small winding resistance can allow a damaging primary current after switching transients.

Write an ideal point-transfer model using the Dirac delta function:For this arrival stream with independent increments, distinct arrivals contribute the product of the means, which cancels when centering. The same-arrival term leaves the autocovarianceIts Fourier transform gives the two-sided power spectral density, while folding negative frequencies onto positive ones gives the one-sided convention:These spectra describe $\delta i$, excluding the spectral line from the mean direct current. The amplitude spectral density is $\sqrt{2qI}$ in $\mathrm{A}/\sqrt{\mathrm{Hz}}$; it is not a root mean square electric current until a measurement signal bandwidth is specified. Approximate white noise requires frequencies low enough that the single-transfer pulse shape and any correlations can be neglected. Neither the ideal impulses nor a flat spectrum extending to infinity describe a real detector at every timescale.

All spectra above are per hertz, with $f$ in cycles per second. One possible angular frequency convention defines a density ${\mathcal{S}}_I^{(2)}(\omega )$ integrated using $d\omega$ alone. Since $\omega =2\pi f$, it hasAnother common convention keeps the numerical density $qI$ but integrates it with $d\omega /(2\pi )$. State the integration measure as well as whether a spectrum is one-sided: a change of variable and a folding of signs are separate operations.

An original teaching model exposes the roles of input loading, output loading and the supply. Choose a small-signal model with positive input electrical resistance $r_i$, transconductance $g_m>0$, and output electrical resistance $r_o>0$. Its controlled output current source draws $g_m{v}_i$ from the output node toward the reference node. This chosen sign gives an inverting voltage gain; it is not the historical contact circuit's sign convention.

Assume a fixed operating point, negligible reverse coupling, a frequency range where coupling capacitors pass the signal, and negligible other capacitances and transit delays. A collector resistor $R_D$ connects the output to a steady ideal supply $V_{DD}$; an external electrical load $R_L$ is coupled through a capacitor, so it draws no direct current. The input source has series electrical resistance $R_s$. All voltages below are small departures from the operating point. An ideal fixed supply has zero incremental voltage and therefore becomes the reference node in the small-signal model.

Define $R_O=R_D\parallel r_o$ and $R_P=R_O\parallel R_L$, where $a\parallel b=ab/(a+b)$. Input voltage division and output Kirchhoff's current law giveConsequently the port voltage gain, overall source-to-output voltage gain, and load current gain magnitudes areThis current gain uses load electric current, not all the controlled-source electric current. It is also a different port definition from the historical emitter-to-collector current gain. Larger $R_s$ attenuates the input; smaller $R_L$ attenuates the output.

For a sinusoid input signal, use root mean square voltages. The power gain from accepted input signal power $P_i=v_i^2/r_i$ to useful load power $P_L=v_o^2/R_L$ isUnlike $|A_i|$, this tends to zero as either $R_L\to 0$ or $R_L\to \infty$. Differentiation givesThus matching the output electrical load maximizes delivered power in this unilateral resistive model. For fixed source voltage, input loading multiplies $P_L$ by $[r_i/(R_s+r_i){]}^2$; the accepted-input power gain above excludes the loss in $R_s$.

Choose $r_i=R_s=1\mathrm{k\Omega }$, $g_m=2\mathrm{mS}$, $r_o=20\mathrm{k\Omega }$, and $R_D=2\mathrm{k\Omega }$. Then $R_O=20/11\mathrm{k\Omega }$. With $R_L=2\mathrm{k\Omega }$ and $v_s=20\mathrm{mV}$ root mean square, the input is $10\mathrm{mV}$ and the inverted output has magnitude $19.0476\mathrm{mV}$ root mean square. Hence $P_i=0.100\mu \mathrm{W}$ and $P_L=0.181406\mu \mathrm{W}$: $G_P=1.81406$. These are calculated values of the chosen model, not historical measurements.

With the same parameters but $R_L=100\Omega$, $|A_i|=1.89573$ while $G_P=0.359381<1$. At $R_L=10\mathrm{k\Omega }$, $|A_v|=3.07692$ while $G_P=0.946746<1$. Neither current gain nor voltage gain alone establishes power gain.

For the direct current load line in the figure, choose $V_{DD}=6\mathrm{V}$, $I_Q=1\mathrm{mA}$ and $V_Q=4\mathrm{V}$, consistent with $V_Q=V_{DD}-R_D{I}_Q$. Locally extend the model as $I=I_Q+g_m{v}_i+(V-V_Q)/r_o$. With the external electrical load isolated at direct current, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_m{R}_O{v}_i$. Keeping $|v_o|$ small compared with the available voltage headroom is essential; extrapolation to arbitrarily large signals would be unphysical.

At the chosen operating point the collector supply delivers $V_{DD}{I}_Q=6\mathrm{mW}$: $2\mathrm{mW}$ is dissipated in $R_D$ and $4\mathrm{mW}$ in the output device. Under a small signal, that energy flow is redistributed to provide load power. The incremental controlled current source hides the supply's steady energy flow; it does not create energy. A real complete amplifier also needs its input biasing network and has further losses. This calculation establishes neither its total efficiency nor a universal law for point-contact transistors.

A voltage divider uses two resistors connected in series to obtain a fraction of an applied electric potential difference. Let the upper electrical resistance be $R_1>0$, the lower one $R_2>0$, and the input be $V_{\mathrm{in}}=V_s$ relative to a common electrical reference potential. The output is the junction's electric potential difference relative to that same reference, across $R_2$. The familiar ratio assumes zero output electric current: no external electrical load draws from the junction. In a steady direct current model with fixed resistors obeying Ohm's law and an ideal voltage source,Here $V_0$ is the open-circuit voltage; $R_1,R_2$ correspond to $R_a,R_b$ in the header formula. With no junction branch, Kirchhoff's current law makes the same electric current pass through both resistors, and Ohm's law gives the two voltage drops. For $V_s\ne 0$, division gives $V_0/V_s=R_2/(R_1+R_2)$; if $V_s=0$, the predicted output is zero but this ratio is undefined.

An original stipulated example has $R_1=R_2=10\mathrm{k\Omega }$ and $V_{\mathrm{in}}=10\mathrm{V}$. With zero output electric current, $I=0.5\mathrm{mA}$ and $V_0=5\mathrm{V}$. An attached $10\mathrm{k\Omega }$ electrical load instead gives $V_{\mathrm{out}}=10/3\mathrm{V}$; a $1\mathrm{M\Omega }$ load gives $V_{\mathrm{out}}=1000/201\mathrm{V}=4.975124378\dots \mathrm{V}$. These are exact model predictions, not measured readings or specifications of a particular multimeter or oscilloscope.

The divider is useful for scaling a voltage or converting a change in a sensor's electrical resistance into an electric potential difference. Its beauty is also its trap: multiplying both electrical resistances by the same positive factor preserves the unloaded ratio, but changes the effect of any fixed electrical load. A divider therefore supplies a predictable ratio under specified loading conditions; it is not a source that maintains that voltage for arbitrary output electric currents. A Wheatstone bridge compares two divider junctions, while Thévenin's theorem describes the divider's response to many different loads.

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.

Voltmeter loading is the change in an electric potential difference caused by the electric current drawn by the measuring voltmeter.

For a loaded voltage divider in steady direct current, a voltmeter connected across the lower resistor places its finite input electrical resistance $R_m$ in parallel with the lower arm's electrical resistance $R_2$ and any intended electrical load of electrical resistance $R_L$. With positive, fixed electrical resistances, an upper resistor of electrical resistance $R_1$, and an ideal voltage source supplying electric potential difference $V_{\mathrm{in}}$, Kirchhoff's current law and Ohm's law giveHere $V_m$ is the connected circuit's electric potential difference across the voltmeter; omit $1/R_L$ if there is no intended electrical load. The parallel connection and need for large input electrical resistance are standard voltmeter operating principles.^[ref]

Voltmeter loading is a bias relative to the target electrical circuit with the voltmeter absent and any intended electrical load still present. Voltage divider loading error instead compares the attached-load output with the wholly unloaded output. An accurate readout of $V_m$, supported by measurement calibration, can still differ from either target: calibration accuracy concerns the indication of the connected electric potential difference, whereas voltmeter loading changes that electric potential difference itself.