Use consistent units and identify the component or circuit path before applying an equation. The worked examples below use steady DC conditions and idealized values. These calculations explain electrical relationships; selecting wiring and protection also requires actual equipment specifications and installation requirements.

Voltage Drop Calculation

Vdrop = I × Rloop, where current I is in amperes and total loop resistance is in ohms. Include both the outgoing and return paths.

For two equal-length conductors with the same resistance per unit length: Vdrop = (2 × L × I × r) ÷ 1000. Here L is the one-way length in feet and r is the wire resistance in ohms per 1,000 feet (Ω/kft). The divisor converts feet to thousands of feet; omit it only when resistance is already expressed per foot. Do not double a length that already includes both conductors.

Voltage drop equals 2 times one-way length times current times resistance per thousand feet, divided by 1000
The existing formula image uses R for resistance per thousand feet; the text uses r to distinguish it from total circuit resistance.

Arithmetic example: L = 10 ft, I = 10 A and hypothetical r = 1 Ω/kft gives 0.20 V drop. At a 12 V supply, (0.20 ÷ 12) × 100 is about 1.67%.

Victron’s Wiring Unlimited theory chapter explains cable and connection resistance. This equal-conductor calculation excludes extra losses at terminals, switches and other components; use the actual complete path when assessing a circuit.

Ohm’s Law

Defines the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit.

V=I×R

  • Voltage (V) is measured in volts (V).
  • Current (I) is measured in amperes (A).
  • Resistance (R) is measured in ohms (Ω).

Power Equation

  • P=V×I
    • Power (P) is measured in watts (W).

Relates electrical power (P) with voltage (V) and current (I). For these steady DC examples, use values at the same operating point; AC average-power calculations require the appropriate waveform and power-factor treatment.

Power In Terms Of Resistance And Current

Derived from Ohm's Law and the Power equation.

  • P=I² × R

Power In Terms Of Voltage And Resistance

P = V² ÷ R for power dissipated in a resistor. Divide by resistance; do not multiply. Use voltage across that resistor and its resistance in ohms.

Worked check: 12 V across a 6 Ω resistor gives I = 12 ÷ 6 = 2 A. Each route gives 24 W: V × I = 12 × 2; I² × R = 2² × 6; V² ÷ R = 12² ÷ 6.

OpenStax’s electrical-energy and power chapter derives these relationships. The resistance-based forms describe resistor dissipation.

Kirchhoff’s Voltage Law (KVL)

States that the sum of all electrical voltages around any closed network is zero.

  • ∑V=0
    • This principle is used to calculate unknown voltages in a circuit loop.

Kirchhoff’s Current Law (KCL)

States that the total current entering a junction must equal the total current leaving the junction.

  • ∑I in = ∑I out

Capacitance And Charge

Describes the relationship between the charge (Q) stored in a capacitor, its capacitance (C), and the voltage (V) across it.

  • Q=C×V
    • Charge (Q) is measured in coulombs (C).
    • Capacitance (C) is measured in farads (F).

Inductive Reactance

Determines the inductive reactance (X_L) for a sinusoidal signal at frequency f. Here L means inductance, not the cable length used in the voltage-drop section.

  • XL   = 2 π f L
    • XL  is measured in ohms (Ω).
    • Frequency (f) is measured in hertz (Hz).
    • Inductance (L) is measured in henrys (H).

Record your units, resistance source, complete cable path and load conditions alongside the calculation. Continue with the DC wire-sizing reference and breaker comparison before preparing an installation design.