Physics · Electricity

Ohm’s law

Q: How do you rearrange $U = R\cdot{}I$ for I or R?

Dividing $U=RI$ by nonzero resistance gives $I=\frac UR$. At nonzero current, division by I gives $R=\frac UI$. An ideal 9 V source and constant 180 Ω resistor give $I=50\,\mathrm{mA}$. The check $180\,\Omega\cdot0.050\,\mathrm A=9\,\mathrm V$ restores the source voltage, so resistance times current is the appropriate reverse calculation. Units help too: volts divided by ohms gives amperes, and volts divided by amperes gives ohms. Rearrangement is algebraic; the component model still determines whether constant resistance is physically justified. At zero current the ratio cannot be evaluated.

Ohm's law links voltage, resistance and current: the voltage across an ohmic conductor is proportional to the current.

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Formula

Calculate

U = R \cdot I

U in volts [V] · R in ohms [Ω] = [V/A] · I in amperes [A]

Variables & units

Variables & units
Symbol	Meaning	Unit
$U$	Electric voltage	V
$R$	Electrical resistance	Ω
$I$	Electric current	A

LaTeX

U = R \cdot I

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Diagram: a line through the origin in the U-I diagram; a slope triangle shows ΔU over ΔI as the resistance R. — The voltage is proportional to the current; the slope of the line is the resistance R = U/I.

Ohm's law calculator

Find voltage, current or resistance: enter two values and the third appears straight away.

Find

Voltage U

Voltage across this one component, e.g. 12 V across a small lamp on a power supply.

Resistance R

Resistance of the component, constant for an ohmic conductor, e.g. 24 Ω.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for I, substitute, evaluate

1. Rearrange for I
$I=\frac{U}{R}$
2. Substitute values (in SI units)
$I=\frac{12}{24}=0.5\,\mathrm A$
3. Result
I = 0.5 A

Result: Current I

I = 0.5 A

With U = 12 V and R = 24 Ω, I = 0.5 A.

Quick tips

Formula triangle: U on top, R and I below. Cover the quantity you need and the rest is the calculation, e.g. I = U ÷ R.
Unit check: 1 V divided by 1 Ω gives 1 A. If the answer is not a current, a unit went wrong.

Assumptions and limits

Constant positive resistance under unchanged operating conditions. Voltage and current refer to the same component using passive port orientation; both may be negative.

Derivation & background

Georg Simon Ohm’s investigations relate voltage and current in conductors with a straight characteristic through the origin. Constant resistance requires unchanged operating conditions, especially an appropriate temperature. A heated filament lamp and a diode give typical curved characteristics. A material class alone does not determine linearity in every operating range. The ratio $R=\frac UI$ at $I\ne0$ initially describes one operating point; Ohm’s law requires the same value throughout the range considered.

Exam blueprint

Validity range

Voltage across and current through the same ohmic two-terminal element under unchanged operating conditions. Its characteristic is a line through the origin over the range considered. Nonlinear elements have an operating-point-dependent ratio at nonzero current, but not generally constant resistance.

Derivation steps

The linear material model $J = \sigma{}E$ gives $U = RI$ for a uniform conductor; linearity remains an assumption to verify.

1Assume a uniform conductor with length l, constant cross-section A and constant conductivity σ.
2For the uniform conductor, substitute $J=\frac IA$ and $E=\frac U\ell$ into the local material model $J=\sigma E$ .
3Rearrange for voltage: $U=\frac{\ell I}{\sigma A}$ .
4Using $\rho=\frac1\sigma$ and $R=\rho\frac\ell A$ gives $U=RI$ . Resistance describes the component; resistivity describes its material state.
5Check constant operating conditions. Over the relevant range, the U–I characteristic must be a line through the origin.

Rearrangements

Current from voltage and resistance

I = \frac{U}{R}

A larger resistance means a smaller current at the same voltage.

Resistance from voltage and current

R = \frac{U}{I}

Use voltage across the component and current through that same component. At $I=0$ the ratio is undefined; one measured point does not establish an ohmic characteristic.

Task variant

A 9 V battery drives current through $R = 180 \Omega{}$ . Find I.

$I=\frac UR=\frac{9\,\mathrm V}{180\,\Omega}=0.05\,\mathrm A=50\,\mathrm{mA}$ .

A component carries $I = 0.3 A$ at $U = 12 V$ . What is its resistance?

$R=\frac UI=\frac{12\,\mathrm V}{0.3\,\mathrm A}=40\,\Omega$ .

Common mistakes

Substituting milliamperes directly without converting to amperes.

Divide mA by 1000 first: $50 mA = 0.05 A$ .

Applying the law to incandescent lamps or diodes.

Constant R requires a line through the origin under unchanged conditions. With nonlinear behaviour, $R=\frac UI$ is an operating-point value at the particular nonzero current.

Mixing up U, R and I when rearranging.

Use the formula triangle: U on top, R and I below; cover the unknown.

Exam context

Typically the entry step in circuit problems: first partial currents or voltages with $U = R\cdot{}I$ , then power or total resistance.

Formula cluster

DC circuit

Forms the backbone of circuit analysis together with power and equivalent resistance.

P = U·I

Electric Power

Rges = R₁+R₂ ; 1/Rges = 1/R₁+1/R₂

Resistors in Series and Parallel

C = Q/U

Capacitance of a Capacitor

R = ρ·l/A

Resistivity (Conductor Resistance)

Worked example

A current of $I = 0.05 A$ flows through a resistor $R = 220 \Omega{}$ . The applied voltage: $U = 220 \Omega{} \times{} 0.05 A = 11 V$ .

Applications

Circuit design, series resistors for LEDs, measurement technology (multimeters), troubleshooting electric circuits

How do you calculate with Ohm's Law?

Here is how to work through a typical Ohm's Law (U = R·I) task step by step:

1
Task
A 9 V battery drives current through $R = 180 \Omega{}$ . Find I.
Solution path
$I=\frac UR=\frac{9\,\mathrm V}{180\,\Omega}=0.05\,\mathrm A=50\,\mathrm{mA}$ .
2
Task
A component carries $I = 0.3 A$ at $U = 12 V$ . What is its resistance?
Solution path
$R=\frac UI=\frac{12\,\mathrm V}{0.3\,\mathrm A}=40\,\Omega$ .

Understand and apply the equation

Why does doubling voltage sometimes double current and sometimes not? Ohm’s law describes a component whose resistance stays constant under unchanged conditions. First identify the matching voltage and current, then read a characteristic curve and finally check the calculation against the circuit.

Ohm’s law: how do you calculate U, R and I?

For the same ohmic component under unchanged operating conditions, $U=RI$ . The rearrangements for current and resistance appear below; their denominators must be nonzero. Voltage is a difference between two terminals; current is charge flow through a cross-section. A voltmeter therefore connects to both terminals of the component, while the matching current must pass through that same component. Prerequisites are volts, amperes, ohms and simple rearrangement of fractions. You do not need microscopic theory to check measured data for proportionality.[2]

U is the voltage across the resistor and I the current through it. For an ohmic component their ratio is constant under unchanged operating conditions. Doubling voltage then doubles current. English textbooks often use V for voltage; this represents the same quantity.[5]

U=RI,\qquad I=\frac{U}{R},\qquad R=\frac{U}{I}

Voltage, current and resistance isolated in turn; the quotients require nonzero denominators.

U in volts (V), I in amperes (A), R in ohms (Ω). One ohm equals one volt per ampere.
1 mA = 10⁻³ A and 1 kΩ = 10³ Ω. Consequently V/kΩ also gives mA.
Signs must follow the chosen reference directions. Here current and voltage drop across the passive resistor use the same direction.

[5]

How does a characteristic curve identify an ohmic resistor?

With I on the horizontal axis and U on the vertical axis, an idealised ohmic resistor has a straight characteristic through the origin. Its slope is R. Swapping the axes makes the slope $\frac1R$ . A single measurement pair cannot establish ohmic behaviour.[2]

R=\frac{\Delta U}{\Delta I}=\frac{U_2-U_1}{I_2-I_1}

Determine the slope of a linear U–I characteristic using two distinct current values.

For a measurement series, change voltage in steps. Measure voltage in parallel with the component and current in series. At constant temperature, points approximately following a line through the origin support the ohmic model over the measured range.[2]

A line through the origin at fixed resistance

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The model has constant 220 Ω. With the mA axis, read slope in volts per milliampere and convert it to resistance units.[2]

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Connect the law to material and geometry

The key observation is not merely that more voltage produces more current. It is the same factor: doubling current must also double voltage. Compare several points under the same temperature conditions and additionally check the origin. A single ratio gives resistance only at that operating point.[2]

Begin with the material model J = σE, which assumes current density J is proportional to electric field E. For a uniform wire of length l and constant cross-section A, approximate $E=\frac Ul$ and $J=\frac IA$ . These assumptions do not replace experimental evidence of material linearity.[3]

\frac{I}{A}=\sigma\frac{U}{l}\quad\Longrightarrow\quad U=\frac{l}{\sigma A}I

Substitute current density and field, then multiply by $\frac l\sigma$ .

\rho=\frac1\sigma,\qquad R=\rho\frac{l}{A}\quad\Longrightarrow\quad U=RI

Resistivity ρ combines material and geometry into the component resistance R.

ρ and R differ: ρ is resistivity, a material property in Ω·m; R describes a component in Ω. Doubling length at fixed material and area doubles R. Doubling area halves R. For a circular cross-section A = πr², so doubling radius quadruples area.[3]

Measurement points, units and voltmeter loading

Find current and dissipated power

Given: U = 9.0 V across R = 470 Ω. Resistance is assumed constant.
Find: Find I in mA and dissipated power P.

Approach: Model the resistor as ohmic and constant under the stated conditions. The same applied DC voltage determines current and dissipated power.
$\begin{aligned}I&=\frac UR\\{}P&=UI=\frac{U^2}R\end{aligned}$
Divide U = RI by R.
$I=\frac{U}{R}$
Substitute volts and ohms; the quotient gives amperes.
$I=\frac{9.0\,\mathrm V}{470\,\Omega}\approx0.01915\,\mathrm A=19.15\,\mathrm{mA}$
Calculate absorbed power using P = UI.
$P=\frac{U^2}{R}=\frac{81}{470}\,\mathrm W\approx0.1723\,\mathrm W$

Result: I ≈ 19 mA and P ≈ 0.17 W. A component rated for 0.125 W would exceed that rating in the assumed steady operation.

Check: Reverse check: 470 Ω × 0.01915 A ≈ 9.0 V. Checking current alone is insufficient: compare power with the actual component data sheet.

Resistance from two measurement points

Given: A linear characteristic through the origin contains (I₁ = 10 mA; U₁ = 2.2 V) and (I₂ = 25 mA; U₂ = 5.5 V).
Find: Find slope R and compare it with both $\frac UI$ ratios.

Approach: The given line passes through the origin. Its slope and both individual ratios must therefore yield the same resistance; convert currents to a common unit first.
$\begin{aligned}R&=\frac{U_2-U_1}{I_2-I_1}\\{}R&=\frac{U_1}{I_1}=\frac{U_2}{I_2}\end{aligned}$
Find differences: ΔU = 3.3 V and ΔI = 15 mA = 0.015 A.
Divide voltage difference by current difference.
$R=\frac{3.3\,\mathrm V}{0.015\,\mathrm A}=220\,\Omega$
Check the individual ratios: $\frac{2.2}{0.010}=220$ and $\frac{5.5}{0.025}=220$ .

Result: R = 220 Ω. At 6.6 V, this constant resistance predicts a current of 30 mA.

Check: Equal slope and a line through the origin are stated assumptions here. Without them, extrapolation beyond the measurements would require validation.

A meter can change the circuit. The following original model deliberately gives the voltmeter finite input resistance. The node balance then explains why source and load currents must not be interchanged. Its loading is calculated using the series and parallel combination of the resistors.[7][8]

A voltmeter loads a voltage divider

Given: An ideal 6 V source supplies a 100 Ω series resistor and a 200 Ω load. A model voltmeter with 200 Ω input resistance is connected across the load; all resistances are constant and ohmic.
Find: Find load voltage and current before and after meter loading.

Approach: after connection, current splits at the load node. Load and meter share voltage but have separate currents.
$\begin{aligned}R_p&=\frac{R_LR_M}{R_L+R_M}\\I_s&=\frac{U_s}{R_s+R_p}\\U_L&=I_sR_p\\I_L&=\frac{U_L}{R_L}\end{aligned}$
Without the meter, the circuit is unbranched.
$\begin{aligned}I_0&=\frac{6\,\mathrm V}{300\,\Omega}=0.020\,\mathrm A\\U_{L,0}&=0.020\cdot200\,\mathrm V=4.0\,\mathrm V\end{aligned}$
After connection, find the parallel equivalent and recover load current.
$\begin{aligned}R_p&=100\,\Omega\\I_s&=\frac{6\,\mathrm V}{200\,\Omega}=0.030\,\mathrm A\\U_L&=3.0\,\mathrm V\\I_L&=I_M=\frac{3\,\mathrm V}{200\,\Omega}=0.015\,\mathrm A\end{aligned}$

Result: Load voltage falls from 4 V to 3 V and load current from 20 mA to 15 mA, while source current rises to 30 mA.

Check: Check: branch currents add to 30 mA and voltage drops add to 6 V. Load voltage divided by source current would give 100 Ω but does not describe the individual load. Its matching ratio remains $\frac{3\,\mathrm V}{0.015\,\mathrm A}=200\,\Omega$ .

Why power belongs in the calculation

Electrical power states how quickly a resistor converts electrical energy into heat. Combining P = UI with U = RI gives two useful forms. At fixed R, doubling voltage doubles current but quadruples power. A correct current calculation can therefore still describe an unsuitable thermal operating point.[4]

P=UI=I^2R=\frac{U^2}{R}

Equivalent power expressions for the ohmic resistor considered here.

For constant power during Δt, energy is W = PΔt. Watts multiplied by seconds give joules. Power and energy are different quantities: a resistor dissipating 0.20 W converts 0.20 × 300 = 60 J over five minutes.[6]

Five exercises with solutions

Calculate · Foundation

1. A 2.2 kΩ resistor has 5.0 V across it. Find current in mA.

Show hint

V/kΩ gives mA directly; alternatively convert to ohms first.

Learning goal: Determine current with consistent units.

Show solution and calculation

Approach: assume constant ohmic resistance and rearrange $U=RI$ for current.
Current is approximately 2.273 mA.
Substitution into the resistor relation restores the specified 5 V.

\begin{aligned}I&=\frac UR\\&=\frac{5.0\,\mathrm V}{2200\,\Omega}\approx2.273\,\mathrm{mA}\end{aligned}

Transfer · Application

2. A constant resistor dissipates 0.090 W at 3.0 V. What is its power at 6.0 V?

Show hint

At fixed R, P is proportional to U².

Learning goal: Use quadratic voltage-power dependence.

Show solution and calculation

Approach: at fixed ohmic resistance, power is proportional to voltage squared.
Resistance is 100 Ω.
Doubling voltage quadruples power to 0.36 W, provided resistance remains unchanged.

\frac{P_2}{P_1}=\left(\frac{U_2}{U_1}\right)^2=4

Interpret · Application

3. You measure 2.0 mA at 1.0 V and 3.0 mA at 2.0 V. Are these exact values consistent with a constant ohmic resistance?

Show hint

Compare both $\frac UI$ ratios.

Learning goal: Check constant resistance against measured pairs.

Show solution and calculation

Approach: compare both resistance ratios rather than only the increasing current readings.
No: R₁ = 500 Ω and R₂ ≈ 667 Ω differ.
For real measurements, assess uncertainty and possible heating first; the stated exact values do not fit a common line through the origin.

\begin{aligned}R_1&=\frac{1.0\,\mathrm V}{0.0020\,\mathrm A}=500\,\Omega\\R_2&=\frac{2.0\,\mathrm V}{0.0030\,\mathrm A}\approx666.7\,\Omega\end{aligned}

Interpret · Application

At constant external conditions a component gives exactly 4 mA at 2 V and 2 mA at 4 V. Interpret both points on an I-versus-U plot. Can one positive constant-resistance line pass through both? Compare powers.

Show hint

An ohmic line through the origin has positive slope; power is the product of both coordinates.

Learning goal: Distinguish a measurement pair with decreasing current from ohmic proportionality.

Show solution and calculation

Approach: Check resistance ratios and curve direction separately from power.
The resistances are 500 Ω and 2000 Ω; the points do not lie on one positive line through the origin.
Both powers nevertheless equal 8 mW.
Equal power does not establish constant resistance, and two points alone do not establish the cause.

\begin{aligned}R_1&=\frac{2\,\mathrm V}{0.004\,\mathrm A}=500\,\Omega\\{}R_2&=\frac{4\,\mathrm V}{0.002\,\mathrm A}=2000\,\Omega\\{}P_1&=P_2=0.008\,\mathrm W\end{aligned}

[2][4]

Error analysis · Application

In the specified range a curve follows exactly $U=1\,\mathrm V+(200\,\Omega)I$ . Someone calls its 200 Ω slope the constant ohmic resistance of the whole component. Check at 5 mA and 10 mA; the range excludes the origin.

Show hint

Slope and ratio differ for a line with an offset.

Learning goal: Compare secant slope with static resistance.

Show solution and calculation

Approach: Obtain voltages from the curve, then calculate each static ratio.
Voltages are 2 V and 3 V, with ratios 400 Ω and 300 Ω.
Secant slope stays 200 Ω but the ratio is not constant.
A local linear relation with an offset is not an ohmic origin-line model; extrapolation beyond the stated range is unjustified.

\begin{aligned}R_{\mathrm{static}}&=\frac UI=\frac{1\,\mathrm V}{I}+200\,\Omega\\{}R(5\,\mathrm{mA})&=400\,\Omega\\{}R(10\,\mathrm{mA})&=300\,\Omega\end{aligned}

[2]

Lamp, diode and AC: when is constant R insufficient?

A filament lamp’s resistance changes with filament temperature. Ohm’s law with constant R is a material and component model, not a universal law for every electrical device. Diodes have nonlinear characteristics. Heating can change a metal wire’s resistance too. A $\frac UI$ ratio at one operating point does not establish a constant characteristic.[2][3]

An ideal purely resistive component also satisfies u(t) = Ri(t) for instantaneous alternating quantities; voltage and current are in phase. Capacitors and inductors require additional relationships. For sinusoidal excitation, their response is described using complex impedance Z. For a series combination of resistor, inductor and capacitor, the impedance of the series RLC circuit is the appropriate terminal model.[1][9]

Use voltage across the same component through which the specified current flows.
Do not substitute milliamperes as amperes: that creates a factor-of-1 000 error.
A larger slope means greater resistance only when plotting U against I.
$R=\frac UI$ cannot be evaluated at I = 0. In the ideal ohmic model, determine R from the characteristic slope instead.

In brief

Match voltage and current to the same component. Use constant resistance only when the characteristic and operating conditions justify it. After calculating, check units, reverse substitution and, for a real load, dissipated power.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made the ohmic component model and rearrangements more direct; clarified the characteristic, meter loading, resistivity and AC limits and linked relevant follow-on topics.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. Ohm's Law. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/ohmsches-gesetz

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Frequently asked questions about Ohm's Law

How do you calculate with Ohm’s law

U = R\cdot{}I

First choose an ohmic component under unchanged operating conditions. Voltage is measured between its terminals, while the matching current flows through that same component. For the retained values $R=220\,\Omega$ and $I=0.05\,\mathrm A$ , $U=RI=11\,\mathrm V$ . Current follows from $I=\frac UR$ , and resistance at nonzero current from $R=\frac UI$ . Convert prefixes consistently before substitution. Afterwards check that volts, amperes and ohms match and that reverse substitution restores the original value. The formula triangle can help rearrangement, but does not replace choosing the correct component or checking constant resistance.

For which conductors does Ohm’s law actually hold?+

A conductor is ohmic over the range considered when voltage remains proportional to current and its characteristic passes through the origin. Appropriate unchanged operating conditions are required. Many metal wires approximately meet this condition at constant temperature. An operating filament lamp may heat up, however, bending its characteristic; a diode is another counterexample. The actual curve matters rather than a blanket statement about every material class. At nonzero current, $R=\frac UI$ can also be calculated for a nonlinear element. That operating-point value does not justify predicting other voltages with the same resistance. Compare several measurements under the same conditions first.

How do you rearrange

U = R\cdot{}I

for I or R?+

Dividing $U=RI$ by nonzero resistance gives $I=\frac UR$ . At nonzero current, division by I gives $R=\frac UI$ . An ideal 9 V source and constant 180 Ω resistor give $I=50\,\mathrm{mA}$ . The check $180\,\Omega\cdot0.050\,\mathrm A=9\,\mathrm V$ restores the source voltage, so resistance times current is the appropriate reverse calculation. Units help too: volts divided by ohms gives amperes, and volts divided by amperes gives ohms. Rearrangement is algebraic; the component model still determines whether constant resistance is physically justified. At zero current the ratio cannot be evaluated.

What is the difference between voltage, current and resistance?+

Voltage is a potential difference between two terminals and describes electrical energy transfer per charge. Current is directed charge per time through a cross-section. Resistance relates the two for the component considered. Units show the difference: a volt is a joule per coulomb, an ampere a coulomb per second, and an ohm a volt per ampere. An open connection can have voltage while carrying no steady current. Conversely, the complete closed circuit determines current. A water analogy can illustrate pressure difference and flow but remains limited. Match voltage across a component to current through that same component; source current in a branched network is not automatically load current.

Why is the current sometimes small despite a high voltage?+

In a constant ohmic model, the ratio of voltage to resistance determines current. The retained model value of 230 V across 46 kΩ gives $I=\frac{230\,\mathrm V}{46000\,\Omega}=5\,\mathrm{mA}$ . The same voltage magnitude can produce very different currents at other resistances. This calculation describes the specified two-terminal element, not a human body or a fully designed protective measure. A real short circuit also includes source and wiring resistance or current limiting. An ideal nonzero voltage source directly across ideal zero resistance instead has no finite consistent DC solution. Check source, current path and model domain together rather than deriving current from voltage alone.

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Subject references and further reading

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[1]OpenStax, University Physics Vol. 2, §9.4 Ohm’s Law

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Symbol

Meaning

Unit

U

Electric voltage

R

Electrical resistance

I

Electric current

Derivation & background

Exam blueprint

Validity range

Derivation steps

The linear material model $J = \sigma{}E$ gives $U = RI$ for a uniform conductor; linearity remains an assumption to verify.

1Assume a uniform conductor with length l, constant cross-section A and constant conductivity σ.
2For the uniform conductor, substitute $J=\frac IA$ and $E=\frac U\ell$ into the local material model $J=\sigma E$ .
3Rearrange for voltage: $U=\frac{\ell I}{\sigma A}$ .
4Using $\rho=\frac1\sigma$ and $R=\rho\frac\ell A$ gives $U=RI$ . Resistance describes the component; resistivity describes its material state.
5Check constant operating conditions. Over the relevant range, the U–I characteristic must be a line through the origin.

How do you calculate with Ohm's Law?

Here is how to work through a typical Ohm's Law (U = R·I) task step by step:

Task

A 9 V battery drives current through $R = 180 \Omega{}$ . Find I.

Solution path

$I=\frac UR=\frac{9\,\mathrm V}{180\,\Omega}=0.05\,\mathrm A=50\,\mathrm{mA}$ .

Task

A component carries $I = 0.3 A$ at $U = 12 V$ . What is its resistance?

Solution path

$R=\frac UI=\frac{12\,\mathrm V}{0.3\,\mathrm A}=40\,\Omega$ .

Understand and apply the equation

Ohm’s law: how do you calculate U, R and I?

U=RI,\qquad I=\frac{U}{R},\qquad R=\frac{U}{I}

Voltage, current and resistance isolated in turn; the quotients require nonzero denominators.

U in volts (V), I in amperes (A), R in ohms (Ω). One ohm equals one volt per ampere.
1 mA = 10⁻³ A and 1 kΩ = 10³ Ω. Consequently V/kΩ also gives mA.
Signs must follow the chosen reference directions. Here current and voltage drop across the passive resistor use the same direction.

[5]

How does a characteristic curve identify an ohmic resistor?

R=\frac{\Delta U}{\Delta I}=\frac{U_2-U_1}{I_2-I_1}

Determine the slope of a linear U–I characteristic using two distinct current values.

A line through the origin at fixed resistance

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The model has constant 220 Ω. With the mA axis, read slope in volts per milliampere and convert it to resistance units.[2]

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Connect the law to material and geometry

\frac{I}{A}=\sigma\frac{U}{l}\quad\Longrightarrow\quad U=\frac{l}{\sigma A}I

Substitute current density and field, then multiply by $\frac l\sigma$ .

\rho=\frac1\sigma,\qquad R=\rho\frac{l}{A}\quad\Longrightarrow\quad U=RI

Resistivity ρ combines material and geometry into the component resistance R.

Measurement points, units and voltmeter loading

Find current and dissipated power

Given: U = 9.0 V across R = 470 Ω. Resistance is assumed constant.
Find: Find I in mA and dissipated power P.

Approach: Model the resistor as ohmic and constant under the stated conditions. The same applied DC voltage determines current and dissipated power.
$\begin{aligned}I&=\frac UR\\{}P&=UI=\frac{U^2}R\end{aligned}$
Divide U = RI by R.
$I=\frac{U}{R}$
Substitute volts and ohms; the quotient gives amperes.
$I=\frac{9.0\,\mathrm V}{470\,\Omega}\approx0.01915\,\mathrm A=19.15\,\mathrm{mA}$
Calculate absorbed power using P = UI.
$P=\frac{U^2}{R}=\frac{81}{470}\,\mathrm W\approx0.1723\,\mathrm W$

Result: I ≈ 19 mA and P ≈ 0.17 W. A component rated for 0.125 W would exceed that rating in the assumed steady operation.

Check: Reverse check: 470 Ω × 0.01915 A ≈ 9.0 V. Checking current alone is insufficient: compare power with the actual component data sheet.

Resistance from two measurement points

Given: A linear characteristic through the origin contains (I₁ = 10 mA; U₁ = 2.2 V) and (I₂ = 25 mA; U₂ = 5.5 V).
Find: Find slope R and compare it with both $\frac UI$ ratios.

Approach: The given line passes through the origin. Its slope and both individual ratios must therefore yield the same resistance; convert currents to a common unit first.
$\begin{aligned}R&=\frac{U_2-U_1}{I_2-I_1}\\{}R&=\frac{U_1}{I_1}=\frac{U_2}{I_2}\end{aligned}$
Find differences: ΔU = 3.3 V and ΔI = 15 mA = 0.015 A.
Divide voltage difference by current difference.
$R=\frac{3.3\,\mathrm V}{0.015\,\mathrm A}=220\,\Omega$
Check the individual ratios: $\frac{2.2}{0.010}=220$ and $\frac{5.5}{0.025}=220$ .

Result: R = 220 Ω. At 6.6 V, this constant resistance predicts a current of 30 mA.

Check: Equal slope and a line through the origin are stated assumptions here. Without them, extrapolation beyond the measurements would require validation.

A voltmeter loads a voltage divider

Given: An ideal 6 V source supplies a 100 Ω series resistor and a 200 Ω load. A model voltmeter with 200 Ω input resistance is connected across the load; all resistances are constant and ohmic.
Find: Find load voltage and current before and after meter loading.

Approach: after connection, current splits at the load node. Load and meter share voltage but have separate currents.
$\begin{aligned}R_p&=\frac{R_LR_M}{R_L+R_M}\\I_s&=\frac{U_s}{R_s+R_p}\\U_L&=I_sR_p\\I_L&=\frac{U_L}{R_L}\end{aligned}$
Without the meter, the circuit is unbranched.
$\begin{aligned}I_0&=\frac{6\,\mathrm V}{300\,\Omega}=0.020\,\mathrm A\\U_{L,0}&=0.020\cdot200\,\mathrm V=4.0\,\mathrm V\end{aligned}$
After connection, find the parallel equivalent and recover load current.
$\begin{aligned}R_p&=100\,\Omega\\I_s&=\frac{6\,\mathrm V}{200\,\Omega}=0.030\,\mathrm A\\U_L&=3.0\,\mathrm V\\I_L&=I_M=\frac{3\,\mathrm V}{200\,\Omega}=0.015\,\mathrm A\end{aligned}$

Result: Load voltage falls from 4 V to 3 V and load current from 20 mA to 15 mA, while source current rises to 30 mA.

Why power belongs in the calculation

P=UI=I^2R=\frac{U^2}{R}

Equivalent power expressions for the ohmic resistor considered here.

Five exercises with solutions

Calculate · Foundation

1. A 2.2 kΩ resistor has 5.0 V across it. Find current in mA.

Show hint

V/kΩ gives mA directly; alternatively convert to ohms first.

Learning goal: Determine current with consistent units.

Show solution and calculation

Approach: assume constant ohmic resistance and rearrange $U=RI$ for current.
Current is approximately 2.273 mA.
Substitution into the resistor relation restores the specified 5 V.

\begin{aligned}I&=\frac UR\\&=\frac{5.0\,\mathrm V}{2200\,\Omega}\approx2.273\,\mathrm{mA}\end{aligned}

Transfer · Application

2. A constant resistor dissipates 0.090 W at 3.0 V. What is its power at 6.0 V?

Show hint

At fixed R, P is proportional to U².

Learning goal: Use quadratic voltage-power dependence.

Show solution and calculation

Approach: at fixed ohmic resistance, power is proportional to voltage squared.
Resistance is 100 Ω.
Doubling voltage quadruples power to 0.36 W, provided resistance remains unchanged.

\frac{P_2}{P_1}=\left(\frac{U_2}{U_1}\right)^2=4

Interpret · Application

3. You measure 2.0 mA at 1.0 V and 3.0 mA at 2.0 V. Are these exact values consistent with a constant ohmic resistance?

Show hint

Compare both $\frac UI$ ratios.

Learning goal: Check constant resistance against measured pairs.

Show solution and calculation

Approach: compare both resistance ratios rather than only the increasing current readings.
No: R₁ = 500 Ω and R₂ ≈ 667 Ω differ.
For real measurements, assess uncertainty and possible heating first; the stated exact values do not fit a common line through the origin.

\begin{aligned}R_1&=\frac{1.0\,\mathrm V}{0.0020\,\mathrm A}=500\,\Omega\\R_2&=\frac{2.0\,\mathrm V}{0.0030\,\mathrm A}\approx666.7\,\Omega\end{aligned}

Interpret · Application

At constant external conditions a component gives exactly 4 mA at 2 V and 2 mA at 4 V. Interpret both points on an I-versus-U plot. Can one positive constant-resistance line pass through both? Compare powers.

Show hint

An ohmic line through the origin has positive slope; power is the product of both coordinates.

Learning goal: Distinguish a measurement pair with decreasing current from ohmic proportionality.

Show solution and calculation

Approach: Check resistance ratios and curve direction separately from power.
The resistances are 500 Ω and 2000 Ω; the points do not lie on one positive line through the origin.
Both powers nevertheless equal 8 mW.
Equal power does not establish constant resistance, and two points alone do not establish the cause.

\begin{aligned}R_1&=\frac{2\,\mathrm V}{0.004\,\mathrm A}=500\,\Omega\\{}R_2&=\frac{4\,\mathrm V}{0.002\,\mathrm A}=2000\,\Omega\\{}P_1&=P_2=0.008\,\mathrm W\end{aligned}

[2][4]

Error analysis · Application

In the specified range a curve follows exactly $U=1\,\mathrm V+(200\,\Omega)I$ . Someone calls its 200 Ω slope the constant ohmic resistance of the whole component. Check at 5 mA and 10 mA; the range excludes the origin.

Show hint

Slope and ratio differ for a line with an offset.

Learning goal: Compare secant slope with static resistance.

Show solution and calculation

Approach: Obtain voltages from the curve, then calculate each static ratio.
Voltages are 2 V and 3 V, with ratios 400 Ω and 300 Ω.
Secant slope stays 200 Ω but the ratio is not constant.
A local linear relation with an offset is not an ohmic origin-line model; extrapolation beyond the stated range is unjustified.

\begin{aligned}R_{\mathrm{static}}&=\frac UI=\frac{1\,\mathrm V}{I}+200\,\Omega\\{}R(5\,\mathrm{mA})&=400\,\Omega\\{}R(10\,\mathrm{mA})&=300\,\Omega\end{aligned}

[2]

Lamp, diode and AC: when is constant R insufficient?

Use voltage across the same component through which the specified current flows.
Do not substitute milliamperes as amperes: that creates a factor-of-1 000 error.
A larger slope means greater resistance only when plotting U against I.
$R=\frac UI$ cannot be evaluated at I = 0. In the ideal ohmic model, determine R from the characteristic slope instead.

In brief

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made the ohmic component model and rearrangements more direct; clarified the characteristic, meter loading, resistivity and AC limits and linked relevant follow-on topics.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. Ohm's Law. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/ohmsches-gesetz

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Frequently asked questions about Ohm's Law

How do you calculate with Ohm’s law

U = R\cdot{}I

For which conductors does Ohm’s law actually hold?+

How do you rearrange

U = R\cdot{}I

for I or R?+

What is the difference between voltage, current and resistance?+

Why is the current sometimes small despite a high voltage?+

Ohm’s law

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Ohm's law calculator

Quick tips

Derivation & background

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Derivation steps

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Task variant

Common mistakes

Exam context

DC circuit

Worked example

Applications

How do you calculate with Ohm's Law?

Ohm’s law: how do you calculate U, R and I?

How does a characteristic curve identify an ohmic resistor?

A line through the origin at fixed resistance

Connect the law to material and geometry

Measurement points, units and voltmeter loading

Find current and dissipated power

Resistance from two measurement points

A voltmeter loads a voltage divider

Why power belongs in the calculation

Five exercises with solutions

1. A 2.2 kΩ resistor has 5.0 V across it. Find current in mA.

2. A constant resistor dissipates 0.090 W at 3.0 V. What is its power at 6.0 V?

3. You measure 2.0 mA at 1.0 V and 3.0 mA at 2.0 V. Are these exact values consistent with a constant ohmic resistance?

At constant external conditions a component gives exactly 4 mA at 2 V and 2 mA at 4 V. Interpret both points on an I-versus-U plot. Can one positive constant-resistance line pass through both? Compare powers.

In the specified range a curve follows exactly U=1 V+(200 Ω)IU=1\,\mathrm V+(200\,\Omega)IU=1V+(200Ω)I. Someone calls its 200 Ω slope the constant ohmic resistance of the whole component. Check at 5 mA and 10 mA; the range excludes the origin.

Lamp, diode and AC: when is constant R insufficient?

In brief

Sources for this explanation

Frequently asked questions about Ohm's Law

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Ohm's Law for exams

Ohm’s law

Variables & units

Ohm's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

DC circuit

Worked example

Applications

How do you calculate with Ohm's Law?

Ohm’s law: how do you calculate U, R and I?

How does a characteristic curve identify an ohmic resistor?

A line through the origin at fixed resistance

Connect the law to material and geometry

Measurement points, units and voltmeter loading

Find current and dissipated power

Resistance from two measurement points

A voltmeter loads a voltage divider

Why power belongs in the calculation

Five exercises with solutions

1. A 2.2 kΩ resistor has 5.0 V across it. Find current in mA.

2. A constant resistor dissipates 0.090 W at 3.0 V. What is its power at 6.0 V?

3. You measure 2.0 mA at 1.0 V and 3.0 mA at 2.0 V. Are these exact values consistent with a constant ohmic resistance?

At constant external conditions a component gives exactly 4 mA at 2 V and 2 mA at 4 V. Interpret both points on an I-versus-U plot. Can one positive constant-resistance line pass through both? Compare powers.

In the specified range a curve follows exactly U=1 V+(200 Ω)IU=1\,\mathrm V+(200\,\Omega)IU=1V+(200Ω)I. Someone calls its 200 Ω slope the constant ohmic resistance of the whole component. Check at 5 mA and 10 mA; the range excludes the origin.

Lamp, diode and AC: when is constant R insufficient?

In brief

Sources for this explanation

Frequently asked questions about Ohm's Law

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Ohm's Law for exams

In the specified range a curve follows exactly $U=1\,\mathrm V+(200\,\Omega)I$ . Someone calls its 200 Ω slope the constant ohmic resistance of the whole component. Check at 5 mA and 10 mA; the range excludes the origin.

In the specified range a curve follows exactly $U=1\,\mathrm V+(200\,\Omega)I$ . Someone calls its 200 Ω slope the constant ohmic resistance of the whole component. Check at 5 mA and 10 mA; the range excludes the origin.