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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.

An RL circuit with the output measured across its resistor is a first-order low-pass filter. For an ideal inductor, an ideal voltage source and a high-resistance output instrument, the electrical impedances $i\omega L$ and $R$ giveThe gain approaches one at low angular frequency and $R/(\omega L)$ at high angular frequency; at ${\omega }_c$ it is $1/\sqrt{2}$. The time-domain equation is $L\dot{I}+RI=V_{\mathrm{in}}$ and $V_{\mathrm{out}}=RI$, so the time constant is $L/R$. Instrument loading, source resistance and nonideal inductance can change the response.

Consider a stipulated series circuit containing an ideal resistor of electrical resistance $R=100\Omega$ and an ideal capacitor of capacitance $C=1\mu \mathrm{F}$. The same electric current passes through both. Their terminal electric potential differences add by Kirchhoff's voltage law, soA quick derivation uses $i=Cd{v}_C/dt$: taking the derivative of the complex exponential multiplies the capacitor voltage phasor by $\mathrm{i}\omega$, hence ${\hat{V}}_C=\hat{I}/(\mathrm{i}\omega C)$. Dimensionally, $1/(\omega C)$ has units $s/F=V/A=\Omega$, matching $R$.

At frequency $f=1000\mathrm{Hz}$, the angular frequency is $\omega =2\pi f$. The signed capacitive electrical reactance, modulus and complex argument of the total electrical impedance areApply $V_{\mathrm{rms}}=1.00\mathrm{V}$ across the complete series circuit. Its root mean square electric current isWith voltage as the zero-phase reference, the current phasor has complex argument $+57.858092^{\circ }$: electric current leads. The phasor drops have moduli $V_{R,\mathrm{rms}}=0.532018\mathrm{V}$ and $V_{C,\mathrm{rms}}=0.846733\mathrm{V}$; these add as complex numbers, not as positive magnitudes. Their squared magnitudes sum to $(1.00\mathrm{V}{)}^2$.

The mean absorbed power uses the root mean square values:The ideal capacitor has zero cycle-averaged absorbed power; all the mean absorbed power becomes Joule heating in the resistor.^[ref] To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt{2}{V}_{\mathrm{rms}}\cos \omega t$ and $i(t)=\sqrt{2}{I}_{\mathrm{rms}}\cos (\omega t-\varphi )$. The cosine addition formula reduces the average to $V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(C{v}_C^2/2)/dt$, and the capacitor energy returns to its starting value after each cycle.

At fixed applied $V_{\mathrm{rms}}$, the limit $\omega \to 0^{+}$ gives $|Z|\to \infty$, $\varphi \to -\pi /2$ and $I_{\mathrm{rms}}\to 0$: the ideal capacitor blocks steady electric current. The limit $\omega \to \infty$ gives $Z\to R$, $\varphi \to 0$ and $I_{\mathrm{rms}}\to V_{\mathrm{rms}}/R$. These are limits of the ideal lumped electrical circuit; unwanted lead inductance and component losses can invalidate that model at high frequency. At $\omega RC=1$, the equal legs of the impedance right triangle give $|Z|=\sqrt{2}R$ and $\varphi =-\pi /4$ without numerical evaluation. Measuring both the magnitude and the sign of the phase difference distinguishes this capacitor response from an equally large positive electrical reactance.

An ideal short circuit fixes zero electric potential difference across a terminal pair, corresponding to zero load electrical impedance. At the end of a lossless transmission line, its voltage-wave reflection coefficient is $-1$ and the terminal voltage is zero. The available current is still limited by source and circuit impedances; real short circuits can dissipate substantial Joule heating.

MIT's guide uses an illuminated vacuum photodiode, with collecting anode positive relative to the emitting cathode. A transimpedance amplifier gives $V_1=R_{F}I$; a multimeter reads its mean. The fluctuation path rejects below $100\mathrm{Hz}$, adds roughly tenfold voltage gain, then nominal hundredfold preamplification and a $1$–$50\mathrm{kHz}$ band-pass filter; the readout measures root mean square voltage. With diode bias off, a sinusoid at the test port calibrates $g(f)$. Restore bias, retain gain settings, vary lamp brightness, and repeat both readings. The guide warns about overload; it supplies a procedure, not measured results.^[ref]

The inference can be expressed without relying on nominal passband edges. Let $g(f)$ denote the calibrated voltage transfer after the $R_F$ conversion, so $K(f)=R_{F}g(f)$ and define $\mathcal{K}={\int }_0^{\infty }|K(f){|}^{2}df$. Under independent arrivals and an unchanged additive background,The slope of voltage variance against mean electric current can therefore estimate $q$. A free intercept accounts for a constant background; forcing the line through zero generally biases the slope. The root mean square readings must have their mean removed and then be squared. Subtracting two root mean square amplitudes does not subtract independent variances. Estimate the variance from successive records; averaging the waveforms first would suppress the fluctuation being measured.

Additional controls sharpen this interpretation. Block the light with detector bias, temperature, and gain unchanged; this preserves dark current and electronics while removing photocurrent. Repeat this dark comparison between illumination settings to check drift. If independent dark arrivals have mean magnitude $I_d$ and illuminated arrivals add $I_{\mathrm{ph}}$, their spectra add: $2q(I_d+I_{\mathrm{ph}})$. Subtracting the unchanged dark variance isolates $2q{I}_{\mathrm{ph}}\mathcal{K}$. This subtraction model assumes unchanged transfers and independence; switching off bias for gain calibration is a different control and need not preserve the dark background.

Vary measured $I_{\mathrm{ph}}$ rather than assuming lamp knob position is proportional to photocurrent. Check collection versus bias to identify a range where incomplete collection is negligible; a space charge limited tube can correlate transfers. Verify that the transimpedance amplifier remains linear and that its frequency response is unchanged over the illumination range. Changing signal bandwidth should change the excess variance through $\mathcal{K}$; compare inferred input spectra in several bands. Correlated fluctuations of the light source can add a contribution scaling approximately as $I_{\mathrm{ph}}^2$, while an electronics floor can hide the linear term at small electric currents. A fitted line over a short range alone does not establish independent transfers.

The gain calibration must include the same detector-to-readout path used for noise, or explicitly supply any omitted transfer. Detector capacitance, finite collection time, amplifier voltage and current noise, thermal electrical noise, pickup, clipping, and finite recording precision limit the inference. Current-dependent electrical impedance or heating can make a background current-dependent too. Since $\mathcal{K}$ contains squared gain, a small fractional error in gain produces approximately twice that fractional error in the predicted variance scale. Knowing a nominal gain and two cutoff settings is insufficient.

MIT's laboratory guide mounts resistors on an aluminum box beneath a metal beaker shield, with short twisted leads. Its SR560 amplifier uses nominal gain $1000$ before a Krohn-Hite $1$–$50\mathrm{kHz}$ band-pass filter; an oscilloscope reads root mean square voltage. Calibration sweeps an attenuated sinusoid and measures the actual gain $g(f)$. Students alternate the resistor and a short circuit, subtracting squared readings $V_R^2-V_S^2$, and disconnect the ohmmeter during noise acquisition. Lead and input capacitance enter the gain-squared integral. With settings unchanged, varying electrical resistance or temperature tests the predicted mean square. The guide prescribes repeated readings and warns against averaging noise waveforms to zero. This is a documented teaching protocol, not a supplied dataset of measured outcomes.^[ref]

The subtraction requires care: shorting the input changes the source electrical impedance. An amplifier's input current noise can therefore produce different voltage noise in the two configurations; a shorted baseline alone does not remove that contribution. Gain drift, interference and insufficient observation time also limit the inference. These are limitations of interpreting the measurement, not corrections that can be inferred from the nominal signal bandwidth alone. In an independent analysis one would compare the measured difference withwhere $C$ represents the measured shunt capacitance and $g$ excludes that separately modeled shunt. If calibration already includes the same shunt response, applying the denominator again counts the attenuation twice. The thermal noise worked example and both original figures illustrate ideal models; they claim no measured agreement or measured value of the Boltzmann constant.

An electrical transformer transfers energy between windings through electromagnetic induction. Its operation depends on changing magnetic flux, rather than a static magnetic field alone.

An alternating-current electrical circuit comparing a parallel resistor-capacitor arm with a series resistor-capacitor arm. Let the upper-left arm have electrical impedance $Z_1=R_1/(1+i\omega R_1{C}_1)$, the lower-left arm $Z_2=R_2+1/(i\omega C_2)$, and the upper-right and lower-right arms have electrical resistances $R_3,R_4$. For input between the upper and lower nodes and an unloaded difference measurement between the arm junctions, voltage divider balance requires $Z_2/Z_1=R_4/R_3$. Separating real and imaginary parts givesThe balance angular frequency is positive for positive components. Equal $R_1=R_2=R$ and $C_1=C_2=C$ require $\omega =1/(RC)$ and $R_4=2R_3$. Reversing which reactive arm is upper changes the divider ratios, so the connections must always be specified.