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Feed a node through resistance $R$ from an ideal voltage $V$, connect that node to ground through $2R$, and connect a capacitor from it to a second node grounded through $R$. The capacitor sees a Thévenin equivalent with open-circuit voltage $2V/3$ and resistance $(R\parallel 2R)+R=5R/3$. For zero initial capacitor voltage, $I_C=2V{e}^{-t/\tau }/(5R)$ and $\tau =5RC/3$. The $2R$ branch current is $V/(3R)-I_C/3$, so the two branch currents coincide at $t=\tau \log (8/5)$. This illustrates why the source-side resistor alone does not determine a charging time constant.

The group delay introduced by a capacitive amplifier load is the negative angular frequency derivative of the extra wave phase produced by that load. For the resistive output model in reactive-load phase shift in a finite-output-impedance amplifier, divide the loaded transfer function by the unloaded one:The added group delay is thereforeThis subtraction of unloaded wave phase prevents the amplifier's intrinsic group delay from being attributed to the capacitor. The terminal model uses a Thévenin equivalent; the group delay definition differentiates an unwrapped wave phase, as in the SciPy signal-processing documentation. Its digital-filter function reports samples, whereas the analog derivative above uses radians per second and gives seconds.^[ref][ref]

For an original example, choose $R=200\Omega$ and $C=10\mathrm{nF}$, hence $\tau =2\mu \mathrm{s}$. At $\omega \tau =1$, corresponding to $f=1/(2\pi \tau )\approx 79.6\mathrm{kHz}$, the added group delay is $1\mu \mathrm{s}$. The same operating point has $-{\varphi }_{\mathrm{rel}}/\omega =\pi \tau /4\approx 1.571\mu \mathrm{s}$: a single sinusoid's phase displacement and the delay of a narrow frequency packet need not agree.

A practical test applies a weakly modulated carrier, varies its central frequency, and measures the output modulation displacement while comparing loaded and unloaded operation. The group delay describes that narrow-band envelope only when the frequency response varies slowly across its bandwidth. The corresponding causal impulse response begins at zero time, $h(t)=e^{-t/\tau }/\tau$ for $t\ge 0$, and vanishes for $t<0$; it is not a rigidly translated impulse. Thus group delay versus pulse-front arrival for an RC load matters when a measured pulse is broad enough to change shape.

A voltage gain determines a power gain only when the electrical conditions at both ports are included. The familiar equality between $20{\log }_{10}|A_v|$ and $10{\log }_{10}{G}_P$ requires a particular impedance relation; it is not a conversion between arbitrary voltage and power ratios.

Use phasors whose moduli are root mean square amplitudes, at one sinusoidal frequency. At a load with electrical impedance $Z$ and electrical admittance $Y=1/Z$, the current is $I=YV$. Its complex electrical power is $V{I}^{\ast }$, with the complex conjugate arising from the relative phase of the two sinusoids. Averaging $v(t)i(t)$ over a cycle gives the active powerTo see the averaging step, write $v(t)=\sqrt{2}|V|\cos (\omega t+{\varphi }_V)$ and similarly for $i(t)$. The product's cycle average is $|V||I|\cos ({\varphi }_V-{\varphi }_I)$, the real part of $V{I}^{\ast }$. Peak-amplitude phasors instead require an extra factor $1/2$.

Let $Y_{\mathrm{in}}$ describe the actual input impedance and $Y_L$ the actual load impedance. Take $\mathrm{Re}{Y}_{\mathrm{in}}>0$ and $\mathrm{Re}{Y}_L>0$ with nonzero terminal signals, and define currents into the amplifier at the input and into the load at the output. With the measured loaded voltage gain $A_v=V_L/V_{\mathrm{in}}$,Thus $G_P=|A_v{|}^2$ requires equality of the relevant electrical conductances, not merely equality of impedance moduli. For example, input $Z_{\mathrm{in}}=50\Omega$ and load $Z_L=(30+40i)\Omega$ both have modulus $50\Omega$, but their electrical admittances have real parts $0.020$ and $0.012\mathrm{S}$. The multiplier is $0.6$, so $G_P=0.6|A_v{|}^2$.

For real resistive ports this becomesTake an original example with measured $|A_v|=0.5$, $R_{\mathrm{in}}=10\mathrm{k\Omega }$ and $R_L=50\Omega$. The voltage gain is $-6.02\mathrm{dB}$, while the power gain is $50$, or $16.99\mathrm{dB}$. An active buffer can reduce the signal's electric potential difference while delivering more signal power than it accepts at its input; the supply provides the additional energy.

The amplifier's output impedance is distinct from the connected load impedance. To show why that distinction matters, model its output by a Thévenin equivalent with open-circuit voltage $A_0{V}_{\mathrm{in}}$ and series resistance $R_o$. The voltage divider givesWith $A_0=0.6$ and $R_o=10\Omega$, the $50\Omega$ load produces the preceding $A_v=0.5$ example. Differentiating $R_L/(R_o+R_L{)}^2$ gives $(R_o-R_L)/(R_o+R_L{)}^3$, so delivered power is largest at $R_L=R_o$, as in the maximum power transfer theorem. Here its power gain is $90$. Increasing $R_L$ much further makes $A_v$ approach $A_0$ but reduces delivered power. That high-load-impedance regime is impedance bridging, useful for sensing the source's voltage with little loading.

A bench demonstration uses a function generator, a known termination and an oscilloscope. The Keysight guide explains that an expected-load setting changes how amplitude is displayed for generators with a fixed series output impedance; a high-impedance load can receive twice the voltage expected for a matched $50\Omega$ termination.^[ref] As an independent low-frequency resistive calculation, an open-circuit source of $1\mathrm{V}$ root mean square with $50\Omega$ series resistance delivers $0.5\mathrm{V}$ and $5\mathrm{mW}$ to $50\Omega$. Into $1\mathrm{M\Omega }$ it delivers about $0.99995\mathrm{V}$ but only $1.00\mu \mathrm{W}$. The larger terminal reading accompanies much less delivered power. Specify the actual load, frequency and amplitude convention when comparing gain measurements.

An ideal source $V$ feeds a series resistance $2R$, followed by parallel branches $R+L$ and $2R$. The Thévenin equivalent seen by the inductor has voltage $V/2$ and resistance $R+(2R\parallel 2R)=2R$. From zero initial current, the inductor current is $V(1-e^{-t/\tau })/(4R)$, the other branch current is $V(1+e^{-t/\tau })/(8R)$, and the source current is $V(3-e^{-t/\tau })/(8R)$. Kirchhoff's current law holds at every time, and the final stored magnetic energy is $L{V}^2/(32R^2)$.

A Thévenin equivalent voltage source $V_s$ with positive source resistance $R_s$ supplies load power $P_L=V_s^2{R}_L/(R_s+R_L{)}^2$ to a real resistive load. Differentiating gives its maximum at $R_L=R_s$. With reactive impedances and fixed source impedance, conjugate matching maximizes the average delivered power. Maximum delivered power is distinct from maximum efficiency; at a real resistive match half of the source's internally generated power is dissipated in $R_s$.

The same current flows through both resistances, $I=V_{\mathrm{oc}}/(R+R_L)$. The load terminal voltage is $I{R}_L$, smaller than the open-circuit voltage. The electrical battery symbol represents the maintained thermoelectric source under this model; heat input is still required.

Take B as reference and define positive source current $I$ upward into the upper internal node; the source symbol does not specify an arrow, so this definition fixes its sign. With A open, no current flows through the upper series resistor. The shunt resistor therefore has voltage $IR$, and the battery's longer plate is on the right, adding $+V$ toward A. ThusDeactivate the independent sources: open the ideal current source and short the ideal voltage source. The two resistors then lie in series between A and B, givingThe Thévenin equivalent is a source $IR+V$ in series with $2R$, positive toward A. If $I$ is instead defined downward, replace $I$ by $-I$; the resistance is unchanged.

The constant particular solution is $A=V/(4R)$, and the homogeneous equation has decay rate $2R/L$. Imposing $I_1(0)=0$ givesAs a separate check, the Thévenin equivalent seen by $L$ has open-circuit voltage $V/2$ and resistance $R+(2R\parallel 2R)=2R$, again giving $\tau =L/(2R)$.

Label the capacitor's left terminal $a$ and right terminal $b$, with the battery's negative node at zero. The source diagram has $2R$ from $a$ to ground, $R$ from $b$ to ground, and a source-side resistor $R$ feeding $a$ from electric potential difference $V$. The capacitor electric current $I_2$ flows from $a$ to $b$; $I_1$ flows through $2R$.

With the capacitor removed, $V_a=2V/3$ and $V_b=0$, so the Thévenin equivalent seen by the capacitor has $V_{\mathrm{th}}=2V/3$. Suppress the ideal electric potential difference source to find its electrical resistance: the left terminal sees $R\parallel 2R=2R/3$ to ground and the right sees $R$, giving $R_{\mathrm{th}}=5R/3$ between the terminals.

The initially uncharged capacitor electric potential difference $V_C=V_a-V_b$ therefore obeysSolving gives $V_C=(2V/3)(1-e^{-t/\tau })$, with $\tau =R_{\mathrm{th}}C$. The capacitor electric current is $I_2=C{\dot{V}}_C$, henceAt $t=0^{+}$, the short-like uncharged capacitor makes $V_a=V_b=2V/5$, so $I_2=2V/(5R)$ directly from the right resistor. At long times the capacitor carries no direct electric current. These endpoint checks agree with the exponential solution.

The original diagram places $R_2$ in parallel with the inductor; $R_1$ connects their common node to the source only when the switch is in position 2. Let $v(t)$ be their shared voltage and $i_L(t)$ the downward inductor electric current. Since $i_L(0^{-})=0$ and a finite voltage cannot change it instantaneously, the inductor initially behaves as an open branch. ThusAt long times, an ideal inductor carrying constant electric current has $v=L{\dot{i}}_L=0$, shorting $R_2$. Therefore $I_2(\infty )=0$ and $i_L(\infty )=V_0/R_1=10\mathrm{A}$.

For the full transient, Kirchhoff's current law gives $(V_0-v)/R_1=i_L+v/R_2$ with $v=L{\dot{i}}_L$. Equivalently, the source seen by the inductor has the Thévenin equivalentSolving $L{\dot{i}}_L+R_{\mathrm{th}}{i}_L=V_{\mathrm{th}}$ givesThe relevant resistance is the parallel combination with the ideal voltage source suppressed, not $R_1+R_2$ and not $R_1$ alone.

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.

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.

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.

Model an ideal resistor of constant electrical resistance $R>0$ at uniform absolute temperature $T$ by its Thévenin equivalent: a zero-mean random voltage source in series with a noiseless $R$. In the classical range its one-sided, open-circuit voltage power spectral density, per hertz, iswhere $k_{\mathrm{B}}$ is the Boltzmann constant. Thus a positive-frequency interval of width $df$ contributes $4k_{\mathrm{B}}TRdf$ to the mean square voltage. This is approximately white noise only where the electrical resistance is effectively constant and $hf\ll k_{\mathrm{B}}T$, with $h$ the Planck constant. The ideal constant spectrum cannot describe arbitrarily high frequencies.^[ref]

For a real voltage, an equivalent two-sided power spectral density is $S_v^{(2)}(f)=2k_{\mathrm{B}}TR$ over signed frequencies. With a real linear time-invariant system of frequency response $H(f)$, the same output mean square isAngle brackets denote an expected value; for this zero-mean noise the mean square is its variance. Do not integrate $4k_{\mathrm{B}}TR$ over both signs of frequency: that counts the same fluctuations twice. All integrals here use frequency $f$ in hertz, not angular frequency $\omega =2\pi f$. The amplitude spectral density is $\sqrt{4k_{\mathrm{B}}TR}$ in $\mathrm{V}/\sqrt{\mathrm{Hz}}$, whereas the root mean square voltage requires a specified measurement signal bandwidth.

A Thévenin equivalent represents a network at two terminals by a voltage source in series with its Thévenin resistance. It reproduces the terminal relation under the assumptions of Thévenin's theorem.