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Interchange the input and feedback components: put the capacitor $C$ between the input and the inverting terminal, and the resistor $R$ between the output and the inverting terminal. Keep the non-inverting terminal grounded, as in the lower circuit above. For an ideal operational amplifier under unsaturated negative feedback, the inverting node remains at virtual ground.

The input capacitor current is $I_C=Cd{V}_{\mathrm{in}}/dt$. It flows through the feedback resistor, givingThis is an inverting op amp differentiator. Its ideal gain magnitude increases with frequency, so it strongly amplifies high-frequency noise. Practical differentiators include components that restrict the useful band and stabilize feedback; that qualification does not change the displayed ideal low-amplitude relation within the operating band.

The original circuit has $R_1$ and $C$ in series between the input and the inverting node, with $R_2$ providing feedback and the noninverting node grounded. An ideal operational amplifier in stable linear negative feedback draws no input electric current and holds the inverting node at virtual ground. For phasors proportional to $e^{i\omega t}$, the input complex impedance is $Z_1=R_1+1/(i\omega C)$. Kirchhoff's current law gives ${\tilde{V}}_{\mathrm{in}}/Z_1=-{\tilde{V}}_{\mathrm{out}}/R_2$, henceThis is an inverting high-pass filter. Its phase can be chosen continuously asThus $V_{\mathrm{out}}(t)=V_0|H|\cos (\omega t+\arg H)$. The ratio above is a phasor ratio, not an instantaneous ratio at input zero crossings. Setting $R_1=R_2$ does not remove the capacitor's phase: the output lags by $90^{\circ }+{\tan }^{-1}(\omega R_{1}C)$, between $90^{\circ }$ and $180^{\circ }$. At $\omega R_{1}C=1$ the lag is $135^{\circ }$; only for $\omega R_{1}C\gg 1$ does it approach $180^{\circ }$. Being within the amplifier's own bandwidth does not specify the external RC corner frequency.

The electromagnetic energy density in vacuum is $u_E={ϵ}_0{E}^2/2$. For a length ${\ell }_z$ of the coaxial capacitor, integrate over annular shells:Since $C=C^{\prime }{\ell }_z=2\pi {ϵ}_0{\ell }_z/\eta$, this becomesThe integral is over the inter-cylinder electric field, and gives the stored energy, independently of which electrode is assigned zero electric potential.

Use phasors with time dependence $e^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is the complex number relating the sinusoidal voltage and electric current: $|Z|$ is their amplitude ratio and $\arg Z$ is the voltage phase relative to the electric current. The component laws giveThus the resistor has zero relative phase, the capacitor voltage lags its electric current by $\pi /2$, and the inductor voltage leads its electric current by $\pi /2$. The ideal capacitor and inductor store and return energy; the real electrical impedance of the resistor describes dissipation.

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 $-1$.

For positive angular frequency, the squared amplitude response of the series-parallel RC band-pass filter isIts denominator is minimized exactly when $w=w^{-1}$, soThis is a band-pass filter maximum produced entirely by the two RC arms; the drawn electrical circuit contains no inductor. The low- and high-frequency limits areThus the output amplitude tends to zero linearly with $\omega$ at low frequency and as $1/\omega$ at high frequency. In real time,The upper series capacitor suppresses low frequencies, while the lower shunt capacitor suppresses high frequencies.

Use an ideal inverting summing amplifier, grounding the non-inverting input and connecting the two sources through resistances $R$ and $R/2$. A feedback resistance $R_f=2R$ gives the desired weights. Operational-amplifier negative feedback makes the inverting node a virtual ground, and negligible input current lets Kirchhoff's current law give

Replacing the feedback resistor by a capacitor $C$ makes a weighted op amp integrator:The initial capacitor charge determines the integration constant. A nonzero DC weighted input produces a ramp; a real operational amplifier eventually reaches its output-voltage limit. A parallel feedback resistance can limit drift while preserving integration over an appropriate frequency range.

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 $-\pi /2$.

When $\omega RC\gg 1$, the capacitor voltage changes little during each half-cycle. Thus ${\dot{V}}_{\mathrm{out}}\simeq V_{\mathrm{in}}/(RC)$, giving an approximately triangular output. Its extrema areThe finite but small triangle is centered on zero for the chosen zero-mean input. A nonzero input mean shifts the output mean by the same amount.

For a capacitor, $I_C=CdV/dt$, soThe electric current leads the voltage by $\pi /2$.

For this question use phasors with time factor $e^{i\omega t}$, so an electrical impedance is $Z=\tilde{V}/\tilde{I}$. The same electric current flows through all components of a series circuit, and their voltages add by Kirchhoff's voltage law. Hence the total series electrical impedance isThe individual steady sinusoidal currents and their phase relations are derived in the following three parts. The diagram normalizes each current by its own amplitude, so it compares phase differences without mixing unlike physical units.

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.

For an ideal operational amplifier under negative feedback, the grounded non-inverting terminal makes the inverting terminal a virtual ground, and no electric current enters either input. The feedback resistor and capacitor have parallel electrical impedanceKirchhoff's current law at the inverting node gives ${\tilde{V}}_i/R_1+{\tilde{V}}_o/Z_f=0$, and henceWith the stated components, $R_{2}C=16.0\mu \mathrm{s}$ and $R_2/R_1=5$. Thus this inverting low-pass amplifier has exactly the same ideal cutoff frequency and normalized frequency response as the passive RC low-pass filter, with output multiplied by $-5$:Its amplitude ratios at the three stated frequencies are $4.99975$, $3.52616$, and $0.97560$. One principal choice of phase shift is $180^{\circ }-{\tan }^{-1}(\omega \tau )$: respectively $179.424^{\circ }$, $134.848^{\circ }$, and $101.252^{\circ }$. Adding or subtracting $360^{\circ }$ describes the same output; the extra inversion must be retained.

Use phasors with time dependence $e^{i\omega t}$ and electrical impedance $Z=\tilde{V}/\tilde{I}$. The element laws $V=IR$, $I=CdV/dt$ and $V=LdI/dt$ giveA complex number records both amplitude and phase, turning the linear differential equations of a sinusoidally driven electrical circuit into algebraic equations. A resistor has in-phase voltage and electric current and dissipates average power. For an ideal capacitor, the electric current leads the voltage by $90^{\circ }$; for an ideal inductor, it lags by $90^{\circ }$. Their imaginary electrical impedances express reversible energy storage and return, with zero net average power over a period. Using $e^{-i\omega t}$ would conjugate the displayed reactive signs without changing predictions.

In the illustrated Maxwell bridge, the left and right vertices are the supply terminals and the upper and lower vertices are detector terminals. Insert a sensitive AC null detector between the latter and drive the former with a sinusoidal source. The upper-left arm is $Z_x=R_x+i\omega L_x$, the upper-right arm is $R_3$, the lower-left arm is $R_2$, and the lower-right arm is the parallel combinationAt a null no detector electric current flows, so the two voltage dividers giveMultiplying by the parallel-arm electrical admittance yieldsEquating real and imaginary parts therefore givesTwo independent adjustments are needed to annul the in-phase and quadrature detector signals. For example, change $R_2$ to set $L_x$ and $R_1$ to set $R_x$, iterating until the detector reads zero. Alternatively, with fixed $R_2,R_3$, independently vary $R_1$ and $C$.

The Maxwell bridge compares an unknown lossy inductor with a stable standard capacitor, avoiding the need for an accurately known reference inductor. The ideal balance equations contain no frequency, and a null measurement need not rely on absolute detector calibration. In practice choose a suitable frequency below the coil's self-resonance, where the series $R_x,L_x$ model and the standard capacitor remain accurate.

If the initial setting is far from balance, the bridge as drawn needs adjustable components and a detector, not merely a more precise readout. Fit resistance boxes or switched ranges for $R_1$ and a ratio arm, and a selectable standard $C$. Choose ranges aroundThen approach the null with coarse adjustments followed by finer ones. A large required $R_1$ can be brought into range by increasing $C$ and decreasing $R_2{R}_3$ proportionally. Start with reduced excitation or reduced detector sensitivity to avoid overload, then increase sensitivity near the null. Changing frequency alone cannot repair an impossible component range, since the ideal balance conditions are frequency-independent. Stray capacitance and lead electrical resistance set practical accuracy limits and may require shielding and short connections.

Let $v_C$ denote the capacitor voltage with the same polarity as the applied voltage. By Kirchhoff's voltage law, Ohm's law and $I=Cd{v}_C/dt$,Solving this first-order differential equation gives the RC low-pass step responseThe capacitor voltage is continuous at switching because an instantaneous finite jump would require an impulsive electric current.

For the square-wave response of an RC low-pass filter, take equal positive and negative half-periods. Let $A=10\mathrm{V}$, $T=1/f$, $h=T/2$ and $\alpha =e^{-h/\tau }$. After the transient, symmetry makes the capacitor voltage alternate between $-v_p$ and $+v_p$. Over the positive half-period,Requiring $v_C(h)=v_p$ givesThe next half-period is the sign-reversed copy. These exact expressions explain the requested sketches:

With phasors proportional to $e^{i\omega t}$, the resistor, inductor, and capacitor have complex impedances $R$, $i\omega L$, and $1/(i\omega C)$. In a series circuit they carry the same electric current, so their voltage drops add:The real part describes Joule heating; the imaginary part changes sign between capacitive and inductive response, passing through zero at ${\omega }_0=1/\sqrt{LC}$.

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 ${\omega }_0=1/\sqrt{LC}$ the inductive and capacitive electrical reactances cancel, so the complex impedance is $R$ and the electric current amplitude is $V_0/R$. The capacitor voltage amplitude is $V_0/(R{\omega }_{0}C)$. Thus the specified quality factor isA capacitor output exceeding the input voltage is consistent with a passive electrical circuit: the inductor and capacitor voltages largely cancel in their sum. The expression assumes positive $R$; the ideal undamped limit $R=0$ has no finite steady resonant response.

Put $x=\omega /{\omega }_0$. Maximizing the output wave amplitude is equivalent to minimizingIts derivative is $D^{\prime }(x)=2x[2x^2-2+Q^{-2}]$. An interior minimum at positive frequency therefore requires $Q>1/\sqrt{2}$, and occurs atIndeed $D^{\prime \prime }(x_{\mathrm{res}})=8x_{\mathrm{res}}^2>0$, and the peak value is $|H{|}_{\max }=Q/\sqrt{1-1/(4Q^2)}$. For $0<Q\le 1/\sqrt{2}$, $D$ increases for $x>0$ and the largest output is the DC value $|H(0)|=1$. At equality the putative peak has reached zero frequency; there is no distinct nonzero-frequency resonance. This threshold concerns capacitor voltage peaking: the series RLC circuit's electric current still has its maximum at ${\omega }_0$ for every positive $R$.

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.