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A resistor is an electrical circuit component whose intended behavior is electrical resistance. An ideal ohmic resistor obeys Ohm's law, $V=IR$, and converts electrical energy into heat at power $P=I^{2}R$ through Joule heating. It provides the dissipative term in a series RLC circuit.

An RL circuit combines electrical resistance $R$ and self-inductance $L$. In the series model, the same electric current $I$ flows through both components. An ideal battery maintains terminal electric potential difference $V$; the drops in the direction of positive electric current obey Ohm's law and the inductor relation:This is Kirchhoff's voltage law with the inductor's terminal drop, rather than its oppositely signed induced electromotive force. For constant $R>0$, $L>0$, and $V$, the first-order linear differential equation has solutionThe time constant $\tau$ measures exponential decay of the difference from the final electric current. ^[ref].

For a stipulated example, take electrical resistance $R=2\Omega$, self-inductance $L=0.5\mathrm{H}$, terminal electric potential difference $V=6\mathrm{V}$, and initial electric current $I(0)=0$. ThenWith $t$ measured numerically in seconds, this is $I(t)=3(1-e^{-4t})$ in amperes. At one time constant,and the terminal electric potential difference drops areTheir sum is exactly $6\mathrm{V}$. At $t=0^{+}$, the inductor takes the entire terminal electric potential difference while the electric current is still zero. At steady electric current, the inductor's terminal drop is zero and the entire drop is across the electrical resistance.

The final magnetic energy stored by the inductor is $U_{\infty }=L{I}_{\infty }^2/2=2.25\mathrm{J}$. At one time constant, the electric current has reached about $63.21\%$ of its final value, but its magnetic energy has reached only about $39.96\%$:This difference follows from the square of electric current in the inductor's magnetic energy.

For release, take $t=0$ at the start of a separate experiment with initial electric current $3\mathrm{A}$. Remove the battery from the model while maintaining a closed path through the same electrical resistance and inductor. ThenNow $v_L=-v_R$: the induced electromotive force supports the continuing positive electric current as its magnitude decreases. This closed-loop model describes exponential decay, not opening the sole path and imposing an instantaneous jump to zero electric current. The inductor's magnetic energy and the power of Joule heating areTheir time constant is $\tau /2=0.125\mathrm{s}$. The total energy dissipated by Joule heating is exactly the initial magnetic energy:The second diagram in inductor shows this energy transfer. A finite closed path lets us predict both how long the electric current persists and how much energy becomes heat. ^[ref].

For an ideal open ion channel with constant electrical conductance $g$, membrane potential $V$ and reversal potential $E_{\mathrm{rev}}$, define outward electric current as positive:This is an Ohm's law model of the open state, not a universal channel law or a rule for opening probability. In the synthetic figure, $g=10\mathrm{pS}$ and $E_{\mathrm{rev}}=0\mathrm{mV}$. Because $1\mathrm{pS}\cdot 1\mathrm{mV}=10^{-3}\mathrm{pA}$, the open-state electric currents at $-80$ and $-40\mathrm{mV}$ are $-0.8$ and $-0.4\mathrm{pA}$. Zero electric current at the reversal potential does not imply a closed ion channel. Real ion channels can have nonlinear responses, so the straight line is an explicit idealization.

For two planar homogeneous layers in series, charge conservation makes the steady normal current density the same in both layers. Combining Ohm's law with $E_1{d}_1+E_2{d}_2=V$ determines the fields when both conductivities are positive. The free interfacial surface charge density is ${ϵ}_2{E}_2-{ϵ}_1{E}_1$, whereas total free-plus-bound charge is ${ϵ}_0(E_2-E_1)$. The two notions of charge are distinct. This DC state can differ from the immediate response to switching the voltage; interfacial free charge accumulates during the transient.

For a thin layer with essentially uniform tangential electric field, integrating electrical conductivity through its thickness gives the sheet conductivity tensor ${\mathbf{\sigma }}_s$. The two-dimensional form of Ohm's law is $\mathbf{K}={\mathbf{\sigma }}_s{\mathbf{E}}_{\parallel }$, with surface current density $\mathbf{K}$. Its SI unit is $\mathrm{S}=A/V$: amperes per metre divided by volts per metre. When response is the same in every in-plane direction, the sheet conductivity reduces to a scalar $\Sigma$; for uniform bulk electrical conductivity $\sigma$ and thickness $t$, $\Sigma =\sigma t$.

Thermal resistance measures the temperature drop required to drive a given rate of heat transfer. For the steady one-dimensional bar model in thermal conduction,Here $r$ relates the temperature drop to heat flux, while $R_{\mathrm{th}}$ relates it to the rate of heat transfer. The single bar above has $r=0.050{\mathrm{m}}^2\mathrm{K}{\mathrm{W}}^{-1}$ and $R_{\mathrm{th}}=5\mathrm{K}{\mathrm{W}}^{-1}$, so its $20\mathrm{K}$ drop drives $4\mathrm{W}$.

To see why series thermal resistances add, apply Fourier's law in each layer to obtain $\Delta T_i=\dot{Q}{R}_i$. With no accumulation or side losses the rate of heat transfer is the same through both, so adding the drops gives $T_h-T_c=\dot{Q}(R_1+R_2)$. This lets us size layered walls without solving a new differential equation for every layer. The structure matches Ohm's law: temperature difference plays the role of electric potential difference, the rate of heat transfer plays the role of electric current, and thermal resistance plays the role of electrical resistance. ^[ref]

A positive resistive electrical load obeys Ohm's law, $V_L=I{R}_L$. Its resistive load line intersects the Thévenin terminal response atprovided $R_{\mathrm{th}}+R_L\ne 0$. For the Thévenin divider example, $R_L=1\mathrm{k\Omega }$ gives $I=24/7\mathrm{mA}$ and $V_L=24/7\mathrm{V}$; $R_L=4\mathrm{k\Omega }$ gives $I=1.5\mathrm{mA}$ and $V_L=6\mathrm{V}$. At $R_L=R_{\mathrm{th}}=\frac{4}{3}\mathrm{k\Omega }$, the voltage is $4\mathrm{V}$, half the open-circuit voltage.

For this positive $R_{\mathrm{th}}$, increasing $R_L$ reduces the drawn electric current and raises $V_L$ toward $V_{\mathrm{oc}}$. This explains voltmeter loading: a voltmeter itself is an electrical load. The equivalence also applies to a nonlinear electrical load through intersection with its terminal relation, but an arbitrary load need not give a unique operating point. If negative $R_{\mathrm{th}}$ cancels positive $R_L$, the displayed formula is singular; a real active circuit's limits and dynamics must then be considered.

With the passive sign convention applied to the network, its entering electric current is $J=-I$, and the same relation becomes $V=V_{\mathrm{oc}}+J{R}_{\mathrm{th}}$. The electrical load uses the passive sign convention with entering electric current $I$, so it absorbs power $VI$. A positive delivered electric current therefore produces a voltage drop for positive $R_{\mathrm{th}}$; changing the reference direction without changing the equation would give the wrong sign.

The reason is linearity of Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, and the constitutive relations of any linear dependent electrical sources. Solve once at $I=0$ with the independent electrical sources present. Subtract this solution from a solution with prescribed delivered electric current $I$: the fixed independent electrical sources cancel, leaving a homogeneous linear network whose terminal voltage change is proportional to $I$. This is a short superposition principle argument for an affine function; the fixed offset need not vanish.

The qualification about a well-defined response matters. An ideal voltage source alone has $R_{\mathrm{th}}=0$. An ideal nonzero current source alone constrains $I$ and does not have a finite open-circuit voltage or a representation $V=V_{\mathrm{oc}}-I{R}_{\mathrm{th}}$ with finite coefficients. Linear dependent electrical sources can give negative resistance, and singular ideal networks may fail to have a unique finite response. Neither a finite nor a strictly positive passive $R_{\mathrm{th}}$ follows merely from calling an electrical circuit linear. ^[ref]

Consider two identical magnetic layers separated by a nonmagnetic layer, between reservoirs that impose the same terminal electric potential difference on both spin channels. Neglect spin-flip scattering and assume each layer obeys Ohm's law. In a magnetic layer, define positive channel electrical resistances $a,b>0$. Let the spacer contribute $r\ge 0$ to each channel. These are circuit parameters, not measured iron or chromium resistivities.

For parallel layer magnetizations, one channel meets $a,r,a$ and the other $b,r,b$. For opposite layer magnetizations, each meets one $a$, one $b$, and the spacer. The two channels share the same terminal electric potential difference and run in parallel. Writing $x\parallel y=xy/(x+y)$ givesThe inequality is algebra, given the specified network. It shows how a difference between spin-channel responses can yield a difference between magnetic configurations even though all the constituent resistors are ohmic.

For chosen $a=1\Omega$, $b=4\Omega$, $r=1\Omega$, the parallel channels have total electrical resistances $3$ and $9\Omega$, while the opposite-alignment channels each have $6\Omega$. Thus $R_P=2.25\Omega$ and $R_{AP}=3.00\Omega$. At a common terminal electric potential difference $1\mathrm{V}$, the total electric currents are $4/9\mathrm{A}$ and $1/3\mathrm{A}$, respectively.

A voltage divider uses two resistors connected in series to obtain a fraction of an applied electric potential difference. Let the upper electrical resistance be $R_1>0$, the lower one $R_2>0$, and the input be $V_{\mathrm{in}}=V_s$ relative to a common electrical reference potential. The output is the junction's electric potential difference relative to that same reference, across $R_2$. The familiar ratio assumes zero output electric current: no external electrical load draws from the junction. In a steady direct current model with fixed resistors obeying Ohm's law and an ideal voltage source,Here $V_0$ is the open-circuit voltage; $R_1,R_2$ correspond to $R_a,R_b$ in the header formula. With no junction branch, Kirchhoff's current law makes the same electric current pass through both resistors, and Ohm's law gives the two voltage drops. For $V_s\ne 0$, division gives $V_0/V_s=R_2/(R_1+R_2)$; if $V_s=0$, the predicted output is zero but this ratio is undefined.

An original stipulated example has $R_1=R_2=10\mathrm{k\Omega }$ and $V_{\mathrm{in}}=10\mathrm{V}$. With zero output electric current, $I=0.5\mathrm{mA}$ and $V_0=5\mathrm{V}$. An attached $10\mathrm{k\Omega }$ electrical load instead gives $V_{\mathrm{out}}=10/3\mathrm{V}$; a $1\mathrm{M\Omega }$ load gives $V_{\mathrm{out}}=1000/201\mathrm{V}=4.975124378\dots \mathrm{V}$. These are exact model predictions, not measured readings or specifications of a particular multimeter or oscilloscope.

The divider is useful for scaling a voltage or converting a change in a sensor's electrical resistance into an electric potential difference. Its beauty is also its trap: multiplying both electrical resistances by the same positive factor preserves the unloaded ratio, but changes the effect of any fixed electrical load. A divider therefore supplies a predictable ratio under specified loading conditions; it is not a source that maintains that voltage for arbitrary output electric currents. A Wheatstone bridge compares two divider junctions, while Thévenin's theorem describes the divider's response to many different loads.

Voltmeter loading is the change in an electric potential difference caused by the electric current drawn by the measuring voltmeter.

For a loaded voltage divider in steady direct current, a voltmeter connected across the lower resistor places its finite input electrical resistance $R_m$ in parallel with the lower arm's electrical resistance $R_2$ and any intended electrical load of electrical resistance $R_L$. With positive, fixed electrical resistances, an upper resistor of electrical resistance $R_1$, and an ideal voltage source supplying electric potential difference $V_{\mathrm{in}}$, Kirchhoff's current law and Ohm's law giveHere $V_m$ is the connected circuit's electric potential difference across the voltmeter; omit $1/R_L$ if there is no intended electrical load. The parallel connection and need for large input electrical resistance are standard voltmeter operating principles.^[ref]

Voltmeter loading is a bias relative to the target electrical circuit with the voltmeter absent and any intended electrical load still present. Voltage divider loading error instead compares the attached-load output with the wholly unloaded output. An accurate readout of $V_m$, supported by measurement calibration, can still differ from either target: calibration accuracy concerns the indication of the connected electric potential difference, whereas voltmeter loading changes that electric potential difference itself.

First disconnect the detector. Each side is a voltage divider. Ohm's law and Kirchhoff's voltage law give the open-circuit intermediate electric potentials and their difference:Setting this difference to zero and cross-multiplying givesThe balance compares ratios, not necessarily equal arm values or equal branch electric currents. At balance the detector carries no electric current, so Kirchhoff's current law again makes the electric current constant along each two-arm path.^[ref]

For a worked null measurement, suppose $R_1=1.000\mathrm{k\Omega }$ and $R_2=2.000\mathrm{k\Omega }$, and the detector null occurs at adjustable $R_3=1.500\mathrm{k\Omega }$. ThenFor $V_s=3.000\mathrm{V}$, both intermediate electric potentials are $2.000\mathrm{V}$. The left branch carries $1.000\mathrm{mA}$ and the right branch $0.6667\mathrm{mA}$: the zero detector reading does not mean that the unknown carries no electric current. These are stipulated ideal values illustrating the calculation, not experimental observations.