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The capacitor stores electrostatic energy, the inductor stores magnetic energy, and the resistor converts energy into heat. Their instantaneous values areThe sign convention for electric current does not change these expressions. With the initial voltage, $E_C(0)=C{V}_0^2/2=90\mu \mathrm{J}$ and $E_I(0)=P_R(0)=0$.

After switching, Kirchhoff's voltage law gives the free series RLC circuit equationIts characteristic roots are $-\gamma \pm i{\omega }_d$, whereA real, nonzero oscillation frequency requires $R^2<4L/C$. For a passive resistor,The endpoint $R=0$ is the undamped limit; strictly decaying oscillations have $0<R<894\Omega$. The critical endpoint itself does not have a nonzero ${\omega }_d$.

Let $Q=C{V}_0$ denote the initial electric charge, so $Q=30\mu \mathrm{C}$ and $\dot{q}(0)=0$. Distinguish $Q$ from the amplitude coefficient $q_0$ in an exponential-times-cosine representation: imposing zero initial electric current generally requires a nonzero phase.

The original diagram places the output across the $10\Omega$ resistor, in series with an ideal $1\mathrm{mH}$ inductor. Treat the voltages as phasors with time dependence $e^{i\omega t}$. The electrical impedances are $Z_R=R$ and $Z_L=i\omega L$, so the same electric current flows through both:This is an RL low-pass filter, with corner angular frequencyThe frequency response tends to one for $\omega \ll {\omega }_c$, has magnitude $1/\sqrt{2}$ at the corner, and tends to ${\omega }_c/\omega$ for $\omega \gg {\omega }_c$. The requested sketch uses a logarithmic horizontal axis and a linear voltage-gain axis:

At the two specified angular frequencies,The relative attenuation by a linear filter is the ratio of these gains:Thus the noise amplitude is attenuated by a factor of 10 relative to the signal. A squared-voltage ratio would decrease by a factor of $100$; the question's voltage transfer function gives the amplitude comparison.

Keep $+x$ along the upper-sheet current and $+z$ upward, putting the lower sheet at $z=0$ and upper sheet at $z=d$. Between them, the upper sheet's lower-side magnetic field is $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The lower sheet carries current in $-x$, so its upper-side magnetic field is also $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The superposition principle therefore givesEach perfect conductor is an equipotential. The circuit's resistor carries current from upper to lower, so Ohm's law fixes their electric potential difference:The broad parallel faces and their constant potentials make the electric field perpendicular to them and independent of the coordinates along them, away from edges. There is no volume electric charge in the gap. Apply Gauss's law to a small pillbox wholly within the gap, with equal face areas $A$ at heights $z_1,z_2$. Its side flux is zero, soThus $E_z$ is independent of height as well. Since $V_{\mathrm{upper}}-V_{\mathrm{lower}}=-{\int }_0^d{E}_{z}dz$,It points from the higher-potential upper sheet to the lower sheet. Gauss's law requires the planar symmetry argument: zero enclosed charge alone would not prove a uniform electric field in a general geometry. These combined fields characterize parallel current sheets with a resistive termination.

Equipartition gives mean energy $k_{B}T/2$ for each independent quadratic term in a classical canonical Hamiltonian. For example, $p^2/(2m)$ and $k{x}^2/2$ each contribute this amount in a harmonic oscillator. It is a thermal average, not an instantaneous equality; constraints, nonquadratic terms or quantum level spacings large compared with $k_{B}T$ require a different calculation.

The fluctuation-dissipation theorem connects equilibrium fluctuations with the dissipative part of linear response. If an applied force couples as $-f(t)x$, use response to $e^{-i\omega t}$ and define the two-sided angular-frequency spectrum $S_{xx}(\omega )=\int e^{i\omega t}⟨\delta x(t)\delta x(0)⟩dt$. The classical relation isFor a viscously damped particle, $m\dot{v}=-\zeta v+\xi (t)$, the corresponding white-noise force satisfies $⟨\xi (t)\xi (t^{\prime })⟩=2\zeta k_{B}T\delta (t-t^{\prime })$. Noise replenishes energy lost by friction; their strengths cannot be varied independently at fixed equilibrium temperature.

Electrically, the one-sided voltage-noise spectrum per hertz of an equilibrium resistor is $4k_{B}TR$. A capacitor in parallel filters it toThe same equilibrium voltage variance follows from equipartition. Increasing R makes the spectrum taller and narrower without changing its integrated area. This thermodynamic connection between dissipative conductance and thermal electrical fluctuations is the content of the classical noise calculation.^[ref]

A photodiode is a semiconductor diode producing photocurrent from photon absorption. Exciting an electron across the band gap leaves an electron hole. The depletion region's electric field separates the charge carriers; others can arrive by diffusion before electron-hole recombination. Contacts complete the electrical circuit.^[ref]

The useful bridge is from light to a measurable electric current: spectral responsivity converts incident radiant flux into photocurrent, while quantum efficiency counts collected signal charge carriers per incident photon. These answer different questions: how much electric current one watt produces, and how many incident photons contribute to the signal. The worked calculation below makes their relationship explicit.

A resistance thermometer using platinum as its sensing material. It provides the temperature readout in the saturation-pressure equilibrium measurement.

Transduction turns electrical resistance into a temperature indication. A uniform wire has $R={\rho }_{\mathrm{el}}\ell /A$, with electrical resistivity ${\rho }_{\mathrm{el}}$, length $\ell$, and cross-sectional area $A$. Stronger atomic vibrations at higher temperature scatter electrons more, increasing platinum's electrical resistivity; impurities and crystallographic defects also contribute.^[ref]

A small known electric current $I$ and measured electric potential difference $V$ give $R=V/I$ through Ohm's law; a resistance bridge can instead compare the element with a reference resistor. Inverting the calibrated response then gives the sensing element's temperature. This makes electrical readout convenient, but the platinum resistance thermometer lead resistance error, resistance thermometer self-heating, and resistance thermometer immersion test explain why a stable display alone is insufficient. The gas thermometer supplies a contrasting route through an equation of state; the saturation-pressure equilibrium measurement shows why a trustworthy independent temperature readout matters.

A 1998 NIST account describes stirred comparison baths, a reference SPRT with a 30 Hz resistance bridge, and a temperature-controlled $100\Omega$ reference resistor. The test IPRTs used a $1\mathrm{mA}$ current source, a scanner, and an 8½-digit voltmeter. Separate current and voltage leads made each test probe a four-terminal device; readings in both current directions supplied voltage ratios to the reference resistor.^[ref]

Its program bracketed test readings with reference SPRT readings and repeated a block if bath drift was excessive. A check SPRT monitored the system; reference water-triple point measurements checked stability. Sensor depths were matched, and unfamiliar probes underwent immersion tests. Ice and water-triple point readings before and after calibration checked changes. Additional points could test the fitted curve. Reported calibration uncertainties excluded transport, subsequent drift, hysteresis, and the user's measurement system. These are controls and limits of that documented apparatus, not an accuracy promise for all platinum resistance thermometers.^[ref]

The inference is a pair: reference temperature from the SPRT, test electrical resistance from the electrical readout. Record measuring electric current, wiring, immersion, and the stated reference scale with each pair. A bath controller's setpoint is not itself the calibrated reference temperature. Returning to a reference and checking intermediate points distinguishes a fitted relation from a reproducible calibration over the intended interval.

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 ideal constant-voltage source $V$ charging an initially uncharged capacitor through a series resistor gives $Q(t)=CV(1-e^{-t/(RC)})$ and $I(t)=(V/R)e^{-t/(RC)}$. These follow from Kirchhoff's voltage law and $I=dQ/dt$. A smaller time constant means faster fractional settling; it does not fix the initial current or final charge unless the separate values of $R$ and $C$ are specified.

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.

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.

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.

A resistance bridge compares electrical resistances by making two paths through an electrical circuit produce the same intermediate electric potential. The Wheatstone bridge turns this into a null measurement: adjust one arm until a detector joining the two intermediate points reads zero, then infer the unknown from a ratio. This is useful because the balance relation does not require an accurately calibrated detector gain or supply electric potential difference, provided the supply is nonzero and the resistors obey Ohm's law.

Connect one physical resistor $R$ across an ideal lossless capacitor $C$, with no applied source and an ideal readout that draws no electric current. Represent the resistor by its noisy Thévenin equivalent. The resulting RC low-pass frequency response gives the stationary capacitor voltage power spectral densityIts mean square follows by the effective noise bandwidth calculation:This is also the equipartition theorem: the mean capacitor energy is $\frac{1}{2}C⟨v_C^2⟩=\frac{1}{2}{k}_{\mathrm{B}}T$. Raising $R$ makes the low-frequency spectrum taller and its width narrower in exactly compensating proportions. It changes the equilibration time $RC$, not the final variance.

The integral to infinity belongs to the ideal classical model. For a physical electrical circuit, assume thermal equilibrium, linear components with negligible extra dissipation, and a corner far below the frequencies where parasitic effects, frequency-dependent electrical resistance, or $hf\sim k_{\mathrm{B}}T$ invalidate the model; the omitted high-frequency tail must be negligible. A finite-bandwidth instrument measures only part of this variance. Its input capacitance contributes to the total $C$, and its finite input electrical resistance introduces both loading and its own thermal electrical noise.

For $T=300\mathrm{K}$ and $C=1\mathrm{nF}$, the predicted equilibrium variance is $4.141947\times 10^{-12}{\mathrm{V}}^2$ and the root mean square voltage is $2.0352\mu \mathrm{V}$. At $R=10\mathrm{k}\Omega$, $RC=10\mu \mathrm{s}$ and $B_{\mathrm{n}}=25\mathrm{kHz}$. At $R=100\mathrm{k}\Omega$, $RC=100\mu \mathrm{s}$ and $B_{\mathrm{n}}=2.5\mathrm{kHz}$; the equilibrium root mean square voltage is identical.

To obtain the growth curve, prepare $v_C(0)=0$ exactly and connect the resistor at $t=0$ while maintaining its temperature. The RC low-pass filter's causal impulse response is $h(t)=e^{-t/(RC)}/(RC)$ for $t\ge 0$. Classical ideal white noise with two-sided power spectral density $2k_{\mathrm{B}}TR$ givesThe factor $2k_{\mathrm{B}}TR$ here is the two-sided time-domain noise intensity; the equilibrium spectral integral above used the one-sided $4k_{\mathrm{B}}TR$. They give the same long-time limit. The variance approaches equilibrium with time constant $RC/2$, although a deterministic initial voltage decays with time constant $RC$. If the capacitor already starts in thermal equilibrium, its initial variance supplies the missing exponentially decaying term and the total variance stays constant.

Whether increasing electrical resistance increases Joule heating depends on what the supply holds fixed. Consider separate ideal resistor experiments with $R=5\Omega$ and $R=10\Omega$, each maintained at its specified temperature. At fixed electric current $I_0=1\mathrm{A}$, doubling $R$ changes the required electric potential difference from $5$ to $10\mathrm{V}$ and doubles power from $5$ to $10\mathrm{W}$. At fixed electric potential difference $V_0=5\mathrm{V}$, doubling $R$ halves electric current from $1$ to $0.5\mathrm{A}$ and halves power from $5$ to $2.5\mathrm{W}$. These are stipulated model values, not measured data.

The comparison also predicts feedback when a particular resistor's electrical resistance rises with temperature: a supply maintaining electric current increases its heating power, whereas one maintaining terminal electric potential difference reduces it. Predicting the actual temperature still needs the material response $R(T)$ and its heat transfer to the surroundings; a nominal supply setting does not establish which drive is actually constant at the resistor.

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 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.

For equal frequency-setting resistors $R$ and capacitors $C$, write $\tau =RC$. The denominator is $1+(3-K)s\tau +s^2{\tau }^2$. With positive $\tau$, asymptotic stability requires $K<3$; at $K=3$ the ideal poles are imaginary, and larger gain gives right-half-plane poles. In particular $K=5$ gives $(1-s\tau {)}^2$, a repeated unstable pole. A finite formal gain curve must not be confused with stable filter operation.

An ideal voltage source drives a series resistor feeding a parallel inductor and capacitor. In the $e^{i\omega t}$ convention, the branch voltage ratio is the title expression. It is a band-pass response with unit peak at $1/\sqrt{LC}$, phase from $+\pi /2$ through zero to $-\pi /2$, and half-power bandwidth $1/(RC)$. At resonance the input current vanishes while equal and opposite inductor/capacitor currents remain finite. Reactive branch-current magnitudes must be combined as phasors, not added as positive scalars.