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A negative ratio of terminal voltage to entering electric current, or a negative differential slope in a more general terminal relation. A linear dependent electrical source can produce a negative Thévenin resistance without any of the network's ordinary resistors having a negative value.

An operational-amplifier configuration applying the signal to the non-inverting input and returning a fraction of output to the inverting input. In stable unsaturated ideal operation, zero input current and equal input voltages give the displayed voltage gain. The feedback resistor $R_f$ connects output to the inverting input and $R_g$ connects that input to the reference wavefunction node.

Ohm's law states that an Ohmic conductor has electric current proportional to its applied electric potential difference, with constant electrical resistance under fixed material conditions:Here $V=\varphi (A)-\varphi (B)$ is the terminal electric potential difference, and positive $I$ enters at $A$ and exits at $B$. This sign convention makes $VI$ the electrical power absorbed. The straight-line relation through the origin lets one constant predict the response at many applied electric potential differences. Merely defining $R=V/I$ at one nonzero operating point does not establish Ohm's law: a component may have a ratio that varies with applied electric potential difference. Material and temperature must be specified and held fixed; many components have a nonlinear response. ^[ref].

For the ideal resistor model $R=2\Omega$ at electric potential difference $V=5\mathrm{V}$,This power is converted to internal energy by Joule heating; a watt is a joule per second. If the model holds for time $10\mathrm{s}$, the transferred energy is $125\mathrm{J}$. Doubling electric potential difference doubles electric current and quadruples power for the same fixed $R$. The model keeps material temperature fixed even while calculating Joule heating; appreciable changes in temperature may require a changing electrical resistance. ^[ref].

In 1827, Georg Ohm published his mathematical account of electrical transport in Die galvanische Kette, mathematisch bearbeitet. On printed page 36, his relation $S=A/L$ expresses electric current in terms of total driving tension and a reduced length incorporating electrical conductivity and geometry. This is historical notation: his $A$ is not the wire cross-sectional area used here, and his $L$ is not simply the wire's physical length. The striking idea is that material response and geometry can be combined into one quantity governing an electrical circuit. ^[ref].

For an ideal operational amplifier with grounded non-inverting input, connect an input capacitor $C$ to the inverting input and a feedback resistor $R$ from output to that input. The virtual ground and Kirchhoff's current law give $C{\dot{V}}_{\mathrm{in}}=-V_{\mathrm{out}}/R$. The ideal response amplifies high-frequency noise, so practical differentiators restrict its operating band and maintain feedback stability.

An op amp integrator places a resistor $R$ between the input source and the inverting input of an operational amplifier, and a feedback capacitor $C$ between that input and the output, with the non-inverting input grounded. For an ideal operational amplifier, Kirchhoff's current law gives $v_{\mathrm{in}}/R=-Cd{v}_o/dt$, henceThe initial capacitor charge supplies the integration constant. A parallel feedback resistor $R_f$ replaces the transfer by $H(s)=-(R_f/R)/(1+s{R}_{f}C)$: the direct-current gain becomes finite while integration is recovered when $|s{R}_{f}C|\gg 1$. This limits offset-driven saturation. ^[ref]

Ask the Applications Engineer—7: Op-amp Noise. Covers resistor Johnson-Nyquist noise, amplifier voltage noise and input-current noise.

Two broad perfect conductors of width $w$, separated by gap $d\ll w$, carry opposite steady electric currents $I$ and connect through a terminating resistor $R$. Away from edges, Ampère's circuital law and the superposition principle give gap magnetic field magnitude ${\mu }_{0}I/w$. Each conductor is an equipotential, and Ohm's law sets their electric potential difference to $IR$. Planar symmetry and charge-free Gauss's law then give uniform gap electric field magnitude $IR/d$. Injection parallel to the upper current makes the electric and magnetic Lorentz forces opposite. Their cancellation selects speed $Rw/({\mu }_{0}d)$ independent of nonzero electric charge and of operating current; nonrelativistic kinetic energy is $m[Rw/({\mu }_{0}d){]}^2/2$.

An ideal resistor, inductor and capacitor connected across the same voltage have admittance $Y=1/R+i(\omega C-1/(\omega L))$ in the $e^{i\omega t}$ convention. At ${\omega }_0=1/\sqrt{LC}$ the reactive branch currents cancel, giving minimum source current $V_0/R$ and maximum impedance $R$. This differs from a lossy series coil placed across a capacitor.

Only the ideal inductor stores energy in this circuit; ideal resistors dissipate energy as heat. With $I_1(\infty )=V/(4R)$,The question excludes the battery's internal energy, and the continuously generated resistor heat is not stored electromagnetic energy.

For a linear element in sinusoidal steady state, the complex impedance is the ratio of the voltage phasor to the electric current phasor, $Z=\tilde{V}/\tilde{I}$. Use the convention that the physical signal is the real part of its phasor times $e^{i\omega t}$. Then time differentiation becomes multiplication by $i\omega$.

For a resistor, Ohm's law gives $\tilde{V}=R\tilde{I}$. For a capacitor, $I=C\dot{V}$ gives $\tilde{I}=i\omega C\tilde{V}$. For an inductor, $V=L\dot{I}$ gives $\tilde{V}=i\omega L\tilde{I}$. HenceThe imaginary part encodes the relative phase of electric current and voltage: electric current leads voltage in an ideal capacitor and lags it in an ideal inductor.

For a fixed number $N$ of identical bulbs, first approximate each as a resistor of electrical resistance $R$ connected to a fixed ideal battery electric potential difference $V$. A series circuit has electrical resistance $NR$, while a parallel circuit has electrical resistance $R/N$. ThusParallel connection therefore maximizes dissipation in this model: every bulb receives the full source electric potential difference instead of sharing it. For $N=1$ the two arrangements coincide.

An actual incandescent light bulb changes electrical resistance as its filament heats, so a single constant $R$ need not describe both operating points. The same qualitative conclusion holds for identical stable bulbs with increasing current-voltage function $i(U)$: parallel operation gives $P_p=NVi(V)$, whereas symmetric series operation gives $P_s=Vi(V/N)$. The former is larger for $N>1$. This comparison assumes a fixed collection and an ideal source as specified; an unlimited number of parallel bulbs on a zero-internal-resistance ideal battery would have no finite total-power maximum.

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.

Choose the phasor convention $V(t)=\mathrm{Re}(\tilde{V}{e}^{i\omega t})$ and similarly for the electric current. In a linear sinusoidal steady state, differentiation becomes multiplication by $i\omega$, and complex impedance is $Z=\tilde{V}/\tilde{I}$. Kirchhoff's current law and Kirchhoff's voltage law then become algebraic equations that retain both amplitude and phase information.

For a resistor, $V=RI$, so $Z_R=R$. For a capacitor, $I=C\dot{V}$, while an inductor obeys $V=L\dot{I}$. ThereforeThe resistor's electric potential difference and current are in phase. The inductor's electric potential difference leads its current by $\pi /2$; the capacitor's current leads its electric potential difference by $\pi /2$. Replacing $e^{i\omega t}$ by $e^{-i\omega t}$ conjugates every impedance but leaves these physical lead-and-lag statements unchanged. The phasor method describes the steady state; switch-on transients must be added separately when initial conditions are relevant.

Use an ideal operational amplifier with its non-inverting terminal grounded, an input resistor $R$ and a feedback capacitor $C$, as in the upper circuit. Stable unsaturated negative feedback holds the inverting terminal at virtual ground, while the input current into the amplifier itself vanishes.

Applying Kirchhoff's current law at the inverting node givesThus the circuit is an inverting op amp integrator, with its integration constant set by the initial capacitor charge. A constant input makes the output ramp until a real amplifier reaches a supply rail. A practical low-frequency design usually adds a large feedback resistor to limit drift; the ideal integral relation applies where the feedback capacitor dominates.

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.

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 $-1$.

The original diagram has an ideal $5\mathrm{V}$ electric battery feeding series resistors of $8\Omega$ above the tapping point and $2\Omega$ below it. The long battery bar is positive, so terminal $A$ is above terminal $B$ in electric potential. With an open circuit at the output, the internal electric current is $5/(8+2)=0.5\mathrm{A}$. The voltage divider therefore givesFor the Thévenin resistance, suppress the ideal voltage source by replacing it with a short circuit. The two resistors are then in parallel as seen from $A,B$:By Thévenin's theorem, the terminal voltage under an outward load electric current $I$ is $V_A-V_B={\mathcal{E}}_0-I{R}_{\mathrm{Th}}$. Consequentlyflowing through an external short circuit from $A$ to $B$. Directly shorting the original terminals bypasses the $2\Omega$ resistor and leaves $5\mathrm{V}$ across $8\Omega$, independently confirming the result.

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.

For a resistor, Ohm's law givesThe electric current and voltage are in phase.