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Ohm’s law, power and circuit formulas

Find voltage, current, resistance and power using Ohm’s law and series or parallel resistor rules.

52 entries

Rearrange the basic formulas

Ohm's law for voltage: V = I × R
Multiply current in amperes by resistance in ohms. [1]
Solve current: I = V / R
Divide applied voltage by ohmic resistance. [1]
Solve resistance: R = V / I
Divide voltage by current when current is nonzero. [1]
Voltage unit: 1 V = 1 A × 1 Ω
A one-ohm resistor at one ampere drops one volt. [1]
Current unit: 1 A = 1 C / s
Ampere is coulombs of charge per second. [1]
Resistance unit: 1 Ω = 1 V / A
A ratio between voltage and current in the ohmic regime. [1]
Milliamps to amps: 1 mA = 0.001 A
Divide milliampere readings by one thousand. [1]
Microamps to amps: 1 µA = 0.000001 A
Divide microampere readings by one million. [1]
Kilohms to ohms: 1 kΩ = 1,000 Ω
Scale resistance by a factor of one thousand. [1]
Megohms to ohms: 1 MΩ = 1,000,000 Ω
Scale resistance by a factor of one million. [1]
Voltage from milliampere: V = (I_mA / 1000) × R_Ω
Convert milliamps to amperes before multiplying ohms. [1]
Current in milliamps: I_mA = 1000 × V / R_Ω
Convert a computed current from amps into milliamps. [1]

Power and energy

Power from voltage and current: P = V × I
DC electrical power in watts from volts and amps. [1]
Power from current and resistance: P = I² × R
Resistive heating power for a steady ohmic element. [1]
Power from voltage and resistance: P = V² / R
Equivalent expression when the current obeys Ohm's law. [1]
Current from power and voltage: I = P / V
Rearrangement applies for nonzero voltage in this DC model. [1]
Resistance from voltage and power: R = V² / P
Rearranged dissipated power relationship. [1]
Resistance from current and power: R = P / I²
Rearrangement applies for nonzero current. [1]
Voltage from power and resistance: V = √(P × R)
Positive magnitude across a resistor at a known dissipation. [1]
Current from power and resistance: I = √(P / R)
Positive magnitude for an ohmic resistor. [1]
Watt definition: 1 W = 1 V × 1 A
Watt is a joule per second of power. [1]
Millwatts to watts: 1 mW = 0.001 W
Use watts in formulas unless all units are consistent. [1]
Energy from steady power: E = P × t
Watt-seconds are joules for constant power over t seconds. [1]
Watt-hours to joules: 1 Wh = 3,600 J
Multiply watt-hours by 3,600 to get joules. [1]

Series and parallel connections

Two resistors in series: R_total = R1 + R2
Individual resistances add in series. [2][1]
Three resistors in series: R_total = R1 + R2 + R3
Series resistance increases with each added resistor. [2][1]
Equal resistors in series: R_total = N × R
N equal resistors sum to N times one resistor. [2][1]
Same series current: I1 = I2 = I_total
A series branch has one common current. [2][1]
Series voltage division: V1 = V_total × R1 / (R1 + R2)
For two series resistors driven by a fixed total voltage. [2][1]
Series voltage sum: V_total = V1 + V2
Voltage drops around simple series path sum to source voltage. [2][1]
Two resistors in parallel: R_total = (R1 × R2) / (R1 + R2)
Two-resistor simplification of reciprocal resistance addition. [2][1]
General parallel resistance: 1/R_total = 1/R1 + 1/R2 + ...
Add conductances and then invert for total resistance. [2][1]
Equal resistors in parallel: R_total = R / N
N equal resistors lower equivalent resistance by factor N. [2][1]
Same parallel voltage: V1 = V2 = V_total
Parallel branches share the applied node-to-node voltage. [2][1]
Current sum at a junction: I_total = I1 + I2
The source current is the sum of parallel branch currents. [2][1]
Branch current: I_branch = V_total / R_branch
Current depends on each branch's resistance at shared voltage. [2][1]
Series power sum: P_total = P1 + P2
Individual resistive dissipations add to total power. [2][1]
Parallel limit check: R_total < smallest R_branch
With positive resistors, a parallel equivalent is below any branch value. [2][1]

Numerical examples and checks

Find 5 V across 1 kΩ: 5 V / 1,000 Ω = 0.005 A = 5 mA
A small current flows at five volts across one kilo-ohm. [1][2]
Find 12 V across 4 Ω: 12 V / 4 Ω = 3 A
Ohm's law gives three amperes in ideal DC steady state. [1][2]
Find drop from 0.2 A and 10 Ω: 0.2 A × 10 Ω = 2 V
The resistor drops two volts at that current. [1][2]
Find resistance from 9 V and 3 mA: 9 V / 0.003 A = 3 kΩ
Convert milliamps into amps before dividing. [1][2]
Power at 5 V and 0.2 A: 5 V × 0.2 A = 1 W
Multiply measured DC voltage by current. [1][2]
Power at 12 V across 6 Ω: 12² / 6 = 24 W
The ideal element would dissipate 24 watts. [1][2]
Power at 0.1 A through 100 Ω: 0.1² × 100 = 1 W
Squaring current is essential in I²R. [1][2]
Two 100 Ω series: 100 + 100 = 200 Ω
Equal resistors in series double resistance. [1][2]
Two 100 Ω parallel: (100 × 100) / (100 + 100) = 50 Ω
Parallel equal resistors halve resistance. [1][2]
Two parallel 100 Ω and 200 Ω: (100 × 200) / 300 ≈ 66.7 Ω
Parallel result is below both branch values. [1][2]
Three 1 kΩ parallel: 1,000 / 3 ≈ 333.3 Ω
Three identical branches give one-third equivalent resistance. [1][2]
Series divider from 10 V across 1k/1k: V_each = 10 × 1000 / 2000 = 5 V
Equal series resistances split ideal voltage equally. [1][2]
Current at 9 V across 3 kΩ: 9 / 3000 = 0.003 A = 3 mA
Correct dimensional conversion yields milliamps. [1][2]
Check resistor energy: 2 W × 10 s = 20 J
Ten seconds of steady 2 W dissipation means twenty joules. [1][2]

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Updated 2026-10-20SI electrical units · ideal DC resistors

Sources

cheatsheetatlas.com/sheets/ohms-law/ · Updated 2026-10-20 · SI electrical units · ideal DC resistors

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