Ohm’s Law Calculator

Lei de Ohm

A Lei de Ohm relaciona tensão (V), corrente (I) e resistência (R) em um circuito: V = I × R. A potência elétrica é P = V × I. Combinando as duas, obtemos P = I² × R = V² / R.

• Tensão (V) — diferença de potencial em volts.
• Corrente (I) — fluxo de elétrons em amperes.
• Resistência (R) — oposição à corrente em ohms.
• Potência (P) — energia consumida em watts.

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Ohm's law: voltage, current and resistance

Ohm's law states that the current through a conductor is proportional to the voltage across it: V = R · I, where V is voltage in volts, R is resistance in ohms (Ω) and I is current in amperes. Rearranged: R = V / I and I = V / R. Electrical power follows from P = V · I = R · I² = V² / R, measured in watts. Published by Georg Ohm in 1827 (Die galvanische Kette mathematisch bearbeitet), the law holds for ohmic (linear) resistors only — it does not apply to diodes, transistors, or incandescent bulbs at high temperatures, where resistance changes with current. In Brazil, residential voltage is 127 V (some states) or 220 V, at 60 Hz (versus 50 Hz in Europe). Example: a 60 W bulb on a 127 V outlet draws I = 60 / 127 ≈ 0.47 A, with resistance R = 127² / 60 ≈ 268 Ω.

Applications: wire sizing, fuses and electronics

Ohm's law sizes wires by current capacity (Brazilian NBR 5410 allows roughly 10 A/mm² for residential copper), selects circuit breakers so they trip before wires overheat, and dimensions current-limiting resistors for LEDs (R = (Vsource − VLED) / ILED). It also underpins voltage dividers, pull-up/pull-down resistors in microcontrollers, and power dissipation calculations to avoid burning components.

FAQ

Does Ohm's law apply to AC circuits? Yes for purely resistive loads. For capacitive or inductive loads (motors, transformers), use complex impedance Z instead of R: V = Z · I.

Why doesn't the law work for diodes? Diodes are non-ohmic: current rises exponentially with voltage past the threshold (~0.7 V for silicon), so R is not constant.

How do I pick an LED resistor? Subtract the LED forward voltage from the supply, then divide by the desired current. Example: 5 V supply, 2 V red LED at 20 mA → R = (5 − 2) / 0.02 = 150 Ω.

Is 127 V or 220 V more efficient? Higher voltage means lower current for the same power, so less loss in cables (P_loss = R · I²). That's why long-distance transmission uses hundreds of kilovolts.