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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.

The passive RC low-pass filter is simple, inexpensive, requires no power supply, and has no operational amplifier saturation or finite bandwidth limitation. It offers no voltage gain; its output impedance is finite, and both source electrical resistance and loading can change the observed frequency response. Its resistor also produces thermal noise.

The active inverting low-pass amplifier provides a low-frequency voltage gain of magnitude five and, within the usable bandwidth, a small output impedance, so it can drive a load with much less change in frequency response. The input electrical resistance is approximately $R_1=1\mathrm{k}\Omega$ and may itself load the source. It needs a power supply and adds electrical noise and offset; finite operational amplifier gain, bandwidth, slew rate, available output swing, and output electric current limit the ideal formula. Its inversion must also suit the application. The best choice depends on required voltage gain, source and load electrical impedances, and the signal's frequency and amplitude.

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.

The component of magnetic force along the slope has magnitude $IBs\cos \varphi$. The rate at which the sliding bar does work against this magnetic braking isBy Ohm's law, $\mathcal{E}=IR$, soThis exactly equals the Joule heating in the resistor in the negligible-self-inductance approximation. The normal force does no work because it is perpendicular to the velocity.

The gravitational power is instead $mgv\sin \varphi$. By the work-energy theorem,It equals the Joule heating only at constant speed. For a concrete counterexample, during the initial acceleration the kinetic energy increases, so some gravitational power remains mechanical.

Use the coordinates in 8. At downhill speed $v$, the moving bar has $\mathbf{v}\times \mathbf{B}=-Bv\cos \varphi \hat{\mathbf{b}}$. The electromotive force and induced electric current are thereforeHere $I$ is positive along $-\hat{\mathbf{b}}$ across the sliding bar, then returns through the rails and resistor. This direction makes the magnetic force oppose downhill motion, as required by Lenz's law. The factor $\cos \varphi$ is essential: only the component of magnetic field normal to the rail plane threads the circuit. Rail and bar electrical resistances and circuit self-inductance are neglected relative to the stated large $R$ model.

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 ${\omega }_0=1/\sqrt{LC}$.

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.

Use the following arrangement: an inverting operational amplifier with input resistor $R_a$ and feedback resistor $R_f$, followed by an op amp integrator with input resistor $R_i$ and feedback capacitor $C_i$. Ground both non-inverting inputs. Each ideal operational amplifier has infinite input impedance, so its inputs draw no electric current; negative feedback makes the inverting input a virtual ground.

Kirchhoff's current law at the first summing junction gives $V_1=-(R_f/R_a)V_{\mathrm{in}}$. At the second summing junction it gives $V_1/R_i+C_{i}d{V}_{\mathrm{out}}/dt=0$. Consequently the two inversions cancel:The initial output voltage records the initial capacitor charge. A reset switch across $C_i$ sets it to zero. The formula holds while both stages remain within their supply rails; a persistent input voltage otherwise produces amplifier saturation.

Write $V_{\mathrm{in}}=v(t)+V_{\mathrm{off}}$, with constant unwanted offset $V_{\mathrm{off}}$. Add an adjustable reference voltage $V_b$ through another resistor $R_b$ to the first inverting summing junction, as the dashed branch in the circuit figure shows. Kirchhoff's current law now givesChoose the reference polarity and magnitude so thatA potentiometer between positive and negative reference supplies permits either sign. The first stage then has zero output voltage when $v=0$, and the op amp integrator integrates only $v(t)$:Merely placing a large resistor across the integrating capacitor limits drift but also changes the transfer function; cancellation at the summing junction preserves ideal integration of the desired signal.

Use sinusoidal phasors with physical values $v(t)=\mathrm{Re}(\hat{V}{e}^{j\omega t})$, $i(t)=\mathrm{Re}(\hat{I}{e}^{j\omega t})$, and $j^2=-1$. The complex impedance is $Z=\hat{V}/\hat{I}=X+jY$: its modulus is the voltage-to-current amplitude ratio, and its complex argument is the voltage-minus-current phase difference. Here $X$ is the electrical resistance part and $Y$ the signed electrical reactance; the source's $X,Y$ notation differs from the common $R,X$ notation.

The component laws $v_R=Ri$, $v_L=Ldi/dt$ and $i_C=Cd{v}_C/dt$ give the complex impedances of an ideal resistor, inductor and capacitor:In a series circuit the same electric current flows through both components, so Kirchhoff's voltage law adds their voltage phasors. The resulting series RC impedance isFor positive $R,C,\omega$, the negative phase difference means that electric current leads the applied electric potential difference.

Position A disconnects the battery and closes the resistor-inductor loop. Taking clockwise electric current as positive, Kirchhoff's voltage law for this RL circuit givesThe electric current through an ideal inductor remains continuous across a switch with no impulsive voltage. Thus $I(0^{+})=I_0$, and separation of variables givesIf the switch remains at A, the same result holds for every $t>0$. The time constant is $L/R$; the stored magnetic energy is dissipated by Joule heating.

Let the source's top terminal have voltage phasor $V_0$ and its bottom terminal zero. Denote the left and right junctions by $L$ and $R$. The two left resistors are equal, so at balance $V_L=V_0/2$. Zero electric current through the ammeter requires $V_R=V_L$, and therefore equal upper and lower right-arm electrical impedances:Write $x=\omega R_x{C}_x$. Separating real parts and imaginary parts givesThe second equation implies $1+x^2={\omega }^2{R}_x^{2}C{C}_x$. Subtract $x^2={\omega }^2{R}_x^2{C}_x^2$ to obtainA finite balance with positive components requires $0<C_x<C$, and then $0<R_v<R_x$. This equal-ratio RC bridge uses equality of arm electrical impedances, independent of the conjugate-matching convention in part (a).

Detector convention. The stated negligible-electrical impedance ammeter is an ideal short circuit, whose actual terminal voltage is zero at every angular frequency. The nontrivial bridge output usually intended is the open-circuit junction difference, obtained with the detector removed or replaced by a high-input impedance voltmeter. Define its polarity byFor a detector of finite electrical impedance $Z_A$, its terminal voltage is insteadHere $Z_{\mathrm{th}}$ is the Thévenin equivalent electrical impedance, found by shorting the ideal source. This exhibits why the ideal-ammeter limit $Z_A\to 0$ has zero terminal voltage even when the bridge is unbalanced.

An ideal operational amplifier has infinite input impedance and open-loop gain, zero output impedance and no offsets or response limits. Its input currents vanish. Stable operational-amplifier negative feedback, a finite output and unsaturated linear operation then imply a virtual short, $v_{+}=v_{-}$. Real amplifiers have finite frequency-dependent gain, input bias currents, offset voltage, output loading, finite bandwidth, slew rate and supply-rail limits. Equal input voltages cannot be assumed in an open-loop or saturated circuit.

Here the noninverting input is grounded. In the ideal-feedback limit the inverting node is a virtual ground, so the resistor current charges the feedback capacitor:The initial capacitor charge fixes the integration constant. This is an op amp integrator, with $H(i\omega )=-1/(i\omega RC)$.

A tiny direct-current input, input bias current or offset can drive the output towards saturation. A parallel feedback resistor $R_f$ provides a direct-current return path, giving $H(s)=-(R_f/R)/(1+s{R}_{f}C)$; choose $\omega R_{f}C\gg 1$ over the intended integration band. A reset switch sets initial charge; low-offset components and suitable signal amplitude reduce drift and saturation. The useful band must also remain below the amplifier's bandwidth and slew-rate limits.

For the stipulated finite but constant real gain $A=100$, infinite input impedance and negligible output impedance, write the inverting-node voltage as $v=-V_{\mathrm{out}}/A$. With phasors proportional to $e^{i\omega t}$,ThereforeThe low-frequency gain is $-100$. The corner is ${\omega }_c=1/(101RC)$; above it the amplitude falls as $100/(101\omega RC)$ and the phase approaches $+\pi /2$. phase $+\pi$ and $-\pi$ describe the same low-frequency inversion. This finite-gain circuit is a low-pass response with an integrating high-frequency regime, not an ideal integrator down to zero frequency. A genuinely frequency-dependent $A(i\omega )$ would introduce additional high-frequency limitations beyond this specified model.

An ideal operational amplifier draws no current at either input. With stable operational-amplifier negative feedback in its unsaturated linear range, infinite open-loop gain also gives a virtual short, $V_{+}=V_{-}$. The amplifier adjusts its output to enforce that equality; it is not a conducting connection between the inputs. Here $V_{+}=0$, so the ideal summing node is a virtual ground. Kirchhoff's current law gives $V_{\mathrm{in}}/R_1=-V_{\mathrm{out}}/R_2$, hence the ideal gain is $-R_2/R_1$.

For the finite-input-resistance inverting amplifier let the inverting-node voltage be $v$, and continue to neglect output impedance. The controlled output and node-current balance areEliminating $v$ givesDividing the denominator by $R_2$ gives the requested resistance-sum form. The ideal result follows whenever $|A|\gg 1+R_2/R_1+R_2/R_{\mathrm{in}}$; taking $R_{\mathrm{in}}\to \infty$ alone is insufficient if $A$ is small. At finite frequency $A$ and the gain are phasor ratios, subject to feedback stability.

With equal resistors and sufficiently large loop gain, $V_{\mathrm{out}}\simeq -V_{\mathrm{in}}$. This gives a controlled sign reversal or 180-degree phase inversion, useful for subtraction and opposite-phase signal paths, with amplitude set by a resistor ratio rather than uncertain open-loop gain. The source still sees approximately $R_1$, so this is not a high-input-impedance voltage follower. For the ideal-input, single-pole model, the noise gain of an operational amplifier is two and the closed-loop bandwidth is approximately half the gain-bandwidth product. Output swing, slew rate and load constraints must also permit linear operation.

For an experimental-noise interpretation of Nyquist's theorem, a resistor $R$ in thermal equilibrium at temperature $T$ has one-sided open-circuit voltage-noise spectral densityin the classical range $hf\ll k_{B}T$, over an approximately rectangular bandwidth $\Delta f$. Across a matched load, half the open-circuit voltage appears, giving available noise power $k_{B}T\Delta f$. This is thermal agitation, present without a mean current, and is distinct from shot noise. A real readout weights the spectrum with its transfer function: $⟨v_{\mathrm{out}}^2⟩={\int }_0^{\infty }{S}_V(f)|H(f){|}^{2}df$. Cooling, reduced bandwidth and suitable impedance/noise matching reduce measured noise; the resistor's noise does not simply disappear in an ideal amplifier model.^[ref]

The bare name also has a sampling interpretation. The Nyquist–Shannon sampling theorem permits exact reconstruction of a bandlimited signal from sufficiently fast uniform samples. The practical condition is $f_s>2f_{\max }$, with an anti-aliasing filter before sampling; otherwise distinct continuous frequencies produce identical samples. Ideal reconstruction uses shifted sinc functions. Critical-rate equality needs an appropriate signal class and can fail for unrestricted edge-frequency sinusoids with unknown phase. Sampling fast enough prevents aliasing, but does not remove thermal noise. This distinction covers the source's unspecified use of “Nyquist's theorem”.^[ref]

The ideal operational-amplifier rules are zero input currents and, when stable negative feedback maintains unsaturated linear operation, equal input voltages. Infinite open-loop gain motivates the latter; it must not be assumed to guarantee asymptotic stability for every external network.

For the numerical sketch and usefulness discussion, use the transfer function derived in part (c). The chosen values give $K=5$ and $\tau =RC=10^{-4}\mathrm{s}$. With $x=\omega \tau$,The formal amplitude falls from 5 to zero, with high-frequency slope $-40$ dB per decade. Its unwrapped phase rises from zero through $+90^{\circ }$ at $\omega =10^4{\mathrm{s}}^{-1}$ to $+180^{\circ }$; the corresponding characteristic frequency is 1.59 kHz. At that frequency the amplitude is 2.5.

These are formal sinusoidal particular-solution curves, not a stable settled filter response. For the formal sinusoidal response of an unstable linear system, replacing $i\omega$ by the Laplace variable $s$ gives $G(s)=5/(1-s\tau {)}^2$, with a repeated pole at $s=+1/\tau$. A homogeneous output contains $(A+Bt)e^{t/\tau }$ and grows, so real perturbations lead to supply-limited nonlinear behaviour. With equal positive resistors and capacitors in the frequency-setting network, the general denominator is $1+(3-K)s\tau +s^2{\tau }^2$: the Sallen-Key equal-component stability bound is $K<3$. For example $K=3-\sqrt{2}$ gives a Butterworth filter response. Thus the Sallen-Key low-pass filter architecture is useful for active filtering with suitable gain and damping, but the stated $K=5$ component set is unsuitable for linear filtering. The component values are retained in this diagnosis.^[ref]

An ideal operational amplifier has infinite differential open-loop gain, infinite input impedance, zero output impedance and unlimited bandwidth and output capability, and zero input offset voltage and input bias current. In a stable negative-feedback circuit operating linearly, its input currents vanish and $V_{-}=V_{+}$. This equality is a feedback consequence, not a rule for a saturated or open-loop amplifier. Here the grounded non-inverting input makes the inverting node a virtual earth.

The inverting two-pole band-pass filter has a series resistor and capacitor input, while the feedback arm is a parallel resistor and capacitor. With Laplace variable $s$,Kirchhoff's current law at virtual earth gives the minus sign. Substituting $s=i\omega$ yieldsIt vanishes at zero and infinite frequency, with low-frequency $|H{|}^2\simeq (\omega R_2{C}_1{)}^2$ and high-frequency $|H{|}^2\simeq 1/(\omega R_1{C}_2{)}^2$. Differentiation gives the single maximumFor ${\tau }_1\gg {\tau }_2$, the broad interval $1/{\tau }_1\ll \omega \ll 1/{\tau }_2$ has $|H|\simeq R_2/R_1$. For ${\tau }_2\gg {\tau }_1$, the interval $1/{\tau }_2\ll \omega \ll 1/{\tau }_1$ instead has $|H|\simeq C_1/C_2$. The same rising and falling asymptotes apply outside either plateau.

Writing $a={\tau }_1+{\tau }_2$ and $b={\tau }_1{\tau }_2$, the half-power frequencies and quality factor areEqual time constants give the narrowest fractional pass-band achievable here, with ${\omega }_{\pm }=(\sqrt{2}\pm 1){\omega }_0$ andThis topology cannot make a genuinely sharp, high-$Q$ band-pass. Its poles are real, and the selectivity bound for a two-real-pole band-pass prevents arbitrarily narrow selection. A more selective resonant filter needs additional circuit structure. The gain above answers the component-ratio question for the sharpest available setting without assuming an impossible limit.

Real amplifiers have finite frequency-dependent gain and phase lag, so virtual earth becomes inaccurate, the response changes and stability must be checked. Gain-bandwidth limits, slew rate for large signals, output loading and amplifier saturation also matter. Component tolerances move the corners; capacitor parasitics, stray capacitance, resistor noise and input noise become significant. The ideal high-frequency tail should not be extrapolated beyond the amplifier and lumped-component operating ranges.

Response and resonance. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ describe inertia, linear viscous resistance and the spring's restoring force; the right side is the applied harmonic force. Substituting the complex trial motion and taking real parts givesThe steady response excludes the decaying homogeneous transient. Its magnitude is $[(k-m{\omega }^2{)}^2+b^2{\omega }^2{]}^{-1/2}$. Differentiating the denominator gives $2\omega [-2m(k-m{\omega }^2)+b^2]=0$, soA nonzero-frequency amplitude maximum exists only for $b^2<2km$; otherwise the maximum is at zero frequency. Increasing damping lowers and broadens the peak and shifts it downward. Every curve starts at $|R(0)|=1/k$ and tends to $1/(m{\omega }^2)$ at high frequency.

Series circuit. The PDF circuit has the resistor, inductor and capacitor in one loop. With current $I=\dot{q}$, Kirchhoff's voltage law gives $V=L\dot{I}+RI+q/C$, henceThus $m,b,k$ correspond to $L,R,1/C$. The resistor's rms voltage isIt is maximized exactly at ${\omega }_{\max }={\omega }_0=1/\sqrt{LC}$ for $R>0$. The charge-amplitude resonance is lower. Expanding its square root for $R^{2}C/L\ll 1$ givesThe two maxima refer to different observables: resistor voltage follows current, whereas charge follows displacement.