NEC Article 690.8(A)(3) & 690.8(B)Worked ExampleElectrical

Solar PV Inverter Output Conductor Sizing: 7.6 kW 240V Continuous Output

Calculate minimum conductor size and overcurrent protection for a 7.6 kW 240V grid-tied solar inverter under NEC 690.8 continuous duty rules.

Direct Answer
39.58 A Conductor Ampacity (8 AWG Copper with 40A Breaker)

Calculated strictly in accordance with NEC Article 690.8(A)(3) & 690.8(B).

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Given Scenario Parameters

wattage7600
voltage240
continuousContinuous (125%)

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Interactive Electric Load Calculator

Pre-loaded with NEC 220.51 & 424.3(B) parameters

Nameplate wattage of the heating unit or load

Line-to-line or line-to-neutral

NEC 424.3(B) mandates that fixed electric space heating equipment branch circuits must be sized at 125% of the total load current.

Calculated OutputLive Recalculation
Base Running Current31.67 A7600W ÷ 240V
Required Ampacity (125%)39.58 A31.67A × 1.25
Minimum Circuit Breaker:40 Ampere
Recommended Conductor (75°C Cu):8 AWG Copper
Overcurrent Protection Rule:NEC 240.4 & 424.3(B)
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Step-by-Step Solution

Formal engineering derivation citing applicable National Electrical Code articles.

Governing Code Equation
Conductor Ampacity = (Inverter Max Current) * 1.25
Citing NEC Article 690.8(A)(3) & 690.8(B)
1

Calculate maximum continuous inverter output current: 7,600 W / 240 V = 31.67 Amperes.

2

Per NEC 690.8(B), multiply maximum inverter output current by 125% for continuous duty: 31.67 A * 1.25 = 39.58 Amperes.

3

Select conductor from Table 310.16 at 75°C: 8 AWG copper (rated 50A) or 8 AWG THHN.

4

Select overcurrent protective device: 40 Ampere double-pole circuit breaker.

Conclusion: 39.58 A Conductor Ampacity (8 AWG Copper with 40A Breaker)
NEC Article 690.8(A)(3) & 690.8(B)