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Consider two voltage dividers across a common source, with impedances $Z_0,Z_1$ in one branch and $R_1,Z_p$ in the other. Equal midpoint voltage phasors imply $Z_0{Z}_p=R_1{Z}_1$. If $Z_0=R_0+1/(i\omega C_0)$, $Z_1=1/(i\omega C_1)$ and $Z_p=(1/R+i\omega C{)}^{-1}$, then separating real and imaginary parts proves $R_0=R_{1}C/C_1$ and $C_0=R{C}_1/R_1$. The topology, especially the parallel standard arm, is essential.

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.

Keep the source diagram's topology: the arm from the left supply node to $X$ contains $R_0$ and $C_0$ in series; the arm from $X$ to the right supply node is $C_1$; the arm from the left node to $Y$ is $R_1$; and the arm from $Y$ to the right node contains $R$ and $C$ in parallel. Their complex impedances areFor this capacitor bridge with series loss and a parallel standard arm, the detector carries no electric current at balance. Taking the right supply node as zero electric potential and the left one as the source voltage $V_s$, the two voltage dividers give $V_X=V_s{Z}_1/(Z_0+Z_1)$ and $V_Y=V_s{Z}_p/(R_1+Z_p)$. Equality of these complex voltages means equality at every time, rather than merely equality of their magnitudes. Cross multiplication yields $R_1{Z}_1=Z_0{Z}_p$. ThusEquating real and imaginary parts givesThe two adjustable standards determine the two components of the unknown complex impedance. Treating the lower-right arm as a series combination would give a different and incorrect balance equation for the source circuit. The result also gives the dimensionless dielectric loss tangent $\omega C_0{R}_0=\omega RC$ for the series-loss model.

The original diagram's upper arm is a series resistor and capacitor; its lower arm is a parallel resistor and capacitor. The output is the upper-minus-lower voltage across that parallel arm. Assume an ideal input voltage source and an unloaded output. Their electrical impedances areFor this series-parallel RC band-pass filter, the voltage divider gives, with $w=\omega RC$,Taking the real part of $V_{0}H{e}^{i\omega t}$ yields the steady outputInitial charging contributes decaying transients if the source is switched on at a finite time; the displayed expression is the sinusoidal steady state. The output polarity matters: reversing the indicated output voltage would multiply $H$ by $-1$.

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.

The output is measured across the capacitor. The voltage divider gives the RC low-pass filter transfer functionThereforeThe amplitude ratio tends to $1$ and the phase shift to zero as $\omega \to 0$. For $\omega \to \infty$, the amplitude ratio is asymptotic to $1/(\omega RC)$ and the phase shift tends to $-\pi /2$.

Use complex phasors with time dependence $e^{i\omega t}$ and assume an unloaded output. The capacitor has electrical impedance $Z_C=1/(i\omega C)$. The voltage divider formed with the resistor gives the transfer functionFor input $V_i(t)=V_{i0}\cos \omega t$, the output amplitude and phase shift areso $V_o(t)=|{\tilde{V}}_o|\cos (\omega t-{\tan }^{-1}(\omega RC))$.

Here $\tau =RC=16.0\mu \mathrm{s}$, and the cutoff frequency is $f_c=1/(2\pi RC)=9.947\mathrm{kHz}$. This is an RC low-pass filter: at small frequency the capacitor has large electrical impedance and the output nearly equals the input; at large frequency it increasingly shunts the signal to ground. The amplitude then falls as $1/f$, and the phase shift approaches $-90^{\circ }$.

No numerical input amplitude is supplied, so the output voltages are specified as fractions of $V_{i0}$. An application is suppression of high-frequency electrical noise on a slowly varying signal. A finite load changes the transfer function and must be included if its electrical impedance is comparable with the circuit's output electrical impedance.

The voltage divider relation for the capacitor output gives the transfer functionConsequently the relative wave amplitude isFor input $V_0\cos \omega t$, the output is $V_0|H|\cos (\omega t+\varphi )$, where the continuous phase shift isFor positive frequency and $R>0$, this phase shift goes from $0$ towards $-\pi$, passing through $-\pi /2$ at ${\omega }_0$. A one-argument arctangent without a quadrant correction would wrongly predict a positive phase shift above ${\omega }_0$. The capacitor follows a slowly varying input voltage, while its output falls as $1/({\omega }^{2}LC)$ at high frequency.

At the balance angular frequency, $Z=W$ and the two voltage dividers have the same ratio. ThusThis zero detector electric current is the operational condition used to find $R_v$ and ${\omega }_0$.

For sinusoidal steady state with the $e^{i\omega t}$ convention, complex impedance is the phasor ratio $Z=\tilde{V}/\tilde{I}$, encoding both amplitude ratio and phase. From $V=LdI/dt$, an ideal inductor has $Z_L=i\omega L$; from $I=CdV/dt$, a capacitor has $Z_C=(i\omega C{)}^{-1}$.

The original diagram has $R_1,L_1$ in series, and $R_4,C_4$ in parallel. ThusWith the right supply node as the reference, the two unloaded voltage divider outputs are ${\tilde{V}}_A=V_0{R}_2/(Z_1+R_2)$ and ${\tilde{V}}_B=V_0{Z}_4/(R_3+Z_4)$. Subtracting givesThe series-RL parallel-RC bridge balance is $Z_1{Z}_4=R_2{R}_3$. Equating its real and imaginary coefficients givesThese ideal components balance the bridge at any frequency, not just the test frequency.

At $\omega =200\mathrm{rad}{\mathrm{s}}^{-1}$, the current amplitude through $R_2$ is $5/|50+50i|=0.0707\mathrm{A}$. Using peak rather than RMS supply amplitude,In the DC steady state the inductor is a short and the capacitor is open. Both dividers have equal resistors, so each midpoint is at2.5V and the bridge stays balanced. The upper-branch current is0.100A, giving $\boxed{P_{R_2,\mathrm{DC}}=0.250\mathrm{W}}$.

For the high-input-resistance voltmeter, hold the other components fixed. At the DC balance,The DC bridge resistance resolution is a local readout resolution, or0.08% of $R_1$, not a total uncertainty allowing for supply, component calibration or voltmeter loading. ^[ref]

Use an RC low-pass filter: put the resistor in series with the input and the capacitor between the output node and the return conductor. Measure the output voltage across the capacitor, with negligible output loading. The voltage divider then has transfer functionAt the high frequency, the attenuation requirement isAt the low frequency, approximately unchanged amplitude means $2\pi f_{l}RC\ll 1$, soThe upper relation is an asymptotic design condition, because “approximately” specifies no numerical tolerance. For example, retaining at least $99\%$ of the low-frequency amplitude would sharpen the upper bound to approximately $68.7\mu \mathrm{F}$.

A practical illustrative choice is $C=22\mu \mathrm{F}$. Its voltage gain is approximately $0.99896$ at $1\mathrm{Hz}$ and $0.001096$ at $20\mathrm{kHz}$: the latter amplitude is reduced by a factor of about $912$. Its cutoff frequency is $1/(2\pi RC)\simeq 21.9\mathrm{Hz}$.

Complex impedance. In sinusoidal steady state write physical voltage and electric current as real parts of $\tilde{V}{e}^{i\omega t}$ and $\tilde{I}{e}^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is a ratio of phasors; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in ohms.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=Cdv/dt$. Consequently $\tilde{I}=i\omega C\tilde{V}$, givingWith this time convention the voltage lags the electric current by $\pi /2$.

Bridge output. The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible electric current, the two voltage dividers giveTheir difference is thereforeThe reactive-arm electrical impedances are

Balance. Setting $V_{AB}=0$ and cross-multiplying gives $R_3{Z}_2=R_4{Z}_1$. For this Wien bridge,The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts givesPositive capacitances require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to

Band-pass output and phase. Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the Wien bridge band-pass response isAt finite positive frequency, its imaginary part vanishes exactly when $u-u^{-1}=0$. ThusThe magnitude $|H|=[9+(u-u^{-1}{)}^2{]}^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a band-pass filter.

A convenient continuous, unwrapped wave phase isIt decreases from $-\pi /2$ as $\omega \to 0^{+}$, passes through $-\pi$ at ${\omega }_r$, and tends to $-3\pi /2$ as $\omega \to \infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same phasor. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.

Use the complex amplitude convention $v(t)=\mathrm{Re}(\hat{V}{e}^{\mathrm{i}\omega t})$, where $\omega =2\pi f$ is the angular frequency and $f$ is the frequency in cycles per second. Substitution into the RC low-pass filter equation replaces the time derivative with multiplication by $\mathrm{i}\omega$, giving its continuous-time frequency responseHere $\varphi$ is the output-minus-input phase difference; a negative value means the output lags. Equivalently, the voltage divider formed from the electrical impedances $R$ and $1/(\mathrm{i}\omega C)$ gives the same ratio. For an input $A\cos (\omega t)$, the sustained output is $A|H|\cos (\omega t+\varphi )$ after the initial exponential decay has died away.

Define the cutoff frequency by a voltage amplitude ratio $1/\sqrt{2}$ relative to the low-frequency value:On a Bode plot, the voltage gain in decibels isAt the cutoff frequency it is $-3.0103\mathrm{dB}$, and far above that frequency it approaches a decrease of $20\mathrm{dB}$ per factor of ten in $f$. There is a smooth transition rather than an abrupt wall. The squared voltage ratio is one half at the corner; converting that statement to a power ratio requires the same real electrical resistance for the compared voltages. It does not mean that an unloaded capacitor dissipates half the source power.

For $R=2.2\mathrm{k}\Omega$ and $C=47\mathrm{nF}$, the cutoff frequency is $f_c\approx 1.5392\mathrm{kHz}$. A $1\mathrm{V}$ peak input at $0.1f_c$, $f_c$, and $10f_c$ produces peak output amplitudes of approximately $0.9950\mathrm{V}$, $0.7071\mathrm{V}$, and $0.09950\mathrm{V}$, with phase differences $-5.711^{\circ }$, $-45^{\circ }$, and $-84.289^{\circ }$. Thus a wanted oscillation at $0.1f_c$ is almost preserved while an unwanted one at $10f_c$ is reduced by about a factor of ten in voltage. The superposition principle lets these two responses be added when both oscillations occur in the same signal.

The University of Colorado Boulder's Lab 3: Filters, §§7.1–7.4, specifies measuring the resistor and capacitor, then driving the assembled RC low-pass filter with a function generator. A two-channel oscilloscope reads actual input and output using high electrical impedance. Sweep the frequency, record the voltage amplitude ratio, and locate $1/\sqrt{2}$. A control compares a cable with a $10\times$ oscilloscope probe on a resistive voltage divider, isolating measurement capacitance before using the probe for the filter. A slow square wave then lets discharge timing estimate the time constant independently. Agreement supports the RC low-pass frequency response; source output impedance, probe loading, and instrument signal bandwidth limit it. These are prescribed measurements, with no measured dataset supplied here.^[ref]

The two-channel setup can also test the predicted phase difference directly: for positive frequency $f$, measure the output's time lag $\Delta t$ between corresponding features of the two sinusoids. For this single-stage model the phase lies between $-\pi /2$ and $0$, so choose the lag within a quarter cycle and calculate $\varphi =-2\pi f\Delta t$. At the cutoff frequency the predicted lag is one eighth of a cycle. This tests the timing part of the frequency response as well as the voltage ratio.

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.

A series resistor $R$ and capacitor $C$ feed a parallel pair with the same values. Measuring the parallel pair's voltage relative to the return gives this unloaded transfer function for phasors proportional to $e^{i\omega t}$. The voltage divider applied to $Z_s=R+1/(i\omega C)$ and $Z_p=R/(1+i\omega RC)$ proves the formula. The amplitude peaks at $\omega =1/(RC)$ with ratio $1/3$ and zero relative phase. At low and high positive frequencies, the transfer function tends respectively to $i\omega RC$ and $-i/(\omega RC)$. Reversing output polarity changes its sign, while loading changes the response itself.

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.

Use the A-positive terminal order of Seebeck effect for each thermocouple. Its thermocouple measuring junction has temperature $T_{h,i}$ and its reference connections have temperature $T_{c,i}$. Define the difference of Seebeck coefficients as ${\alpha }_i(T)=S_{A,i}(T)-S_{B,i}(T)$. Connect the B reference end of one thermocouple to the A reference end of the next at a common local temperature, so the terminal electric potential differences add:The first A reference terminal is the positive meter input; the final B reference terminal is the negative input. Following this connected path and adding the individual Seebeck effect integrals proves the result. Matched meter leads with common endpoint temperatures cancel as in an isothermal terminal block. The manufacturer's sum formula also allows unequal individual contributions.^[ref]

If all pairs share $T_h,T_c$ and the same nearly constant difference $\alpha$ of Seebeck coefficients throughout that temperature interval, thenThis $\alpha$ is the pair's relative Seebeck coefficient, not either material's individual coefficient or a universal device constant. As an original constructed example, choose $N=20$, $\alpha =40\mu \mathrm{V}{\mathrm{K}}^{-1}$ and $\Delta T=5\mathrm{K}$. Each thermocouple contributes $0.20\mathrm{mV}$ andReversing the meter connections changes the sign. Exchanging $T_h$ and $T_c$ reverses the integrals and hence the electromotive force; this reversible sign response does not make a detector with heat conduction a reversible thermodynamic process. Reversing the wiring of one pair instead gives $(N-2)\alpha \Delta T$, because one contribution is subtracted rather than added. For this example the result is $3.6\mathrm{mV}$, and at equal junction temperatures every contribution vanishes.

The open-circuit condition matters. If the array's total electrical resistance is $R_{\mathrm{pile}}$ and the meter input has electrical resistance $R_m$, the elementary voltage divider model gives $V_m=V_{\mathrm{oc}}{R}_m/(R_m+R_{\mathrm{pile}})$. Thus $R_m\gg R_{\mathrm{pile}}$ is needed to suppress voltmeter loading. Appreciable electric current also permits Joule heating and the thermoelectric heat transfer to disturb the assumed junction temperatures.

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.