One tap fills in a typical airfield case. The two CCR examples treat kVA as an equivalent resistive load.
Known Values
Values must be greater than 0. The CCR step table for the same resistance is on the next tab.
Report details
Same Resistance at Each CCR Step
| Step | Current | Voltage | Power | Power bar | % of full power |
|---|
The twelve Ohm's law formulas
Two of the four quantities fix the other two. Each row below shows the three ways to find one quantity, depending on what you already know.
Formula table
| To find | From V and I | From V and R | From I and R |
|---|---|---|---|
| Voltage V | — | — | I × R |
| Current I | — | V ÷ R | — |
| Resistance R | V ÷ I | — | — |
| Power P | V × I | V² ÷ R | I² × R |
| To find | From P and V | From P and I | From P and R |
|---|---|---|---|
| Voltage V | — | P ÷ I | √(P × R) |
| Current I | P ÷ V | — | √(P ÷ R) |
| Resistance R | V² ÷ P | P ÷ I² | — |
Why it matters on an airfield
A series circuit carries the same current through every fixture, set by the CCR step. The voltage the CCR must supply is that current times the total resistance, and the heat lost in the cable is I² × R. Because loss follows the square of current, it falls to about 18 % of the full-brightness value at 2.8 A on a 6.6 A circuit.
Worked example
A 4 km primary loop of #8 AWG has about 8.24 Ω of resistance at 20 °C. It carries 6.6 A at full brightness.
| Item | Value |
|---|---|
| Known | I = 6.6 A, R = 8.24 Ω |
| Voltage drop, V = I × R | 54.38 V |
| Cable loss, P = I² × R | 358.9 W |
| Same cable at 2.8 A (B1) | 23.07 V, 64.6 W |
| Loss at B1 vs full brightness | 18.0 % |
What this tool does not do
It treats the circuit as purely resistive. It does not handle reactive loads, power factor or temperature change of resistance. For those, use the Series Circuit Impedance and Circuit Power Factor calculators.