Maximum short-circuit current

Calculate AC and DC current in amps from voltage, power (watts/kW), power factor, or resistance. Supports single-phase, two-phase (2-wire), three-phase, and Ohm’s Law. Ideal for electrical engineers, electricians, students, and DIY projects. Includes motor load estimation Supported Systems & Key Formulas System Type Formula Notes DC I = P / V I = V / R No power factor; pure resistive load Single-Phase AC I = P / (V × cosφ) I = S / V cosφ = power factor; S = apparent power Two-Phase (2-Wire) AC I = P / (V × cosφ) × √2 Important: This mode is specifically designed for two-phase, two-wire systems where the phase difference is 90°. The factor √2 arises because each phase contributes independently, and the total power is distributed across two conductors with orthogonal phase angles. Common in older industrial installations and some specialized motor drives. Three-Phase AC I = P / (√3 × V × cosφ) I = S / (√3 × V) √3 ≈ 1.732; used for balanced three-phase systems Ohm’s Law I = V / R I = V / Z R = resistance (Ω); Z = impedance (Ω) Example Calculation Three-phase system: 400V, 10kW, PF = 0.85 I = 10,000 / (√3 × 400 × 0.85) I ≈ 10,000 / (1.732 × 400 × 0.85) I ≈ 10,000 / 586.4 ≈ 17.0 A Typical Equipment Current Reference Equipment Power Voltage Power Factor Estimated Current Household Refrigerator 150 W 230 V 0.95 ≈ 0.69 A LED Lighting (Total Load) 200 W 230 V 0.98 ≈ 0.89 A Electric Kettle 2.2 kW 230 V 1.0 ≈ 9.6 A Air Conditioner (1.5 HP) 1.1 kW 230 V 0.85 ≈ 5.6 A Three-Phase Water Pump 7.5 kW 400 V 0.88 ≈ 12.3 A Industrial Motor (15 kW) 15 kW 400 V 0.9 ≈ 24.1 A Solar Inverter (DC Output) 5 kW 48 V — ≈ 104 A Frequently Asked Questions (FAQ) Can this calculator handle both AC and DC circuits? Yes! It supports: DC circuits using I = P/V or I = V/R AC single-phase with power factor AC two-phase (2-wire) — specifically designed for 90° phase-shifted systems AC three-phase balanced systems All formulas are dynamically applied based on your selection. What if I don’t know the power factor? Use typical values: Resistive loads (heaters, incandescent lamps): PF = 1.0 Inductive loads (motors, transformers): PF = 0.8–0.9 Capacitive loads (some electronics): PF ≈ 0.95 Applications Home Electrical Projects: Size breakers, wiring, and outlets based on appliance current draw. Solar & Battery Systems: Calculate DC current for inverters, charge controllers, and battery banks. Industrial Maintenance: Estimate motor or pump operating current before installation. Education & Training: Visualize how voltage, power, and power factor affect current in real time. Energy Audits: Convert equipment nameplate data (kW, V, PF) into actual current for load analysis. Who Should Use This Tool? Electrical engineers and designers Electricians and maintenance technicians Students of physics or electrical engineering DIY homeowners installing EV chargers, solar panels, or workshops Renewable energy professionals

Calculate neutral current in 3-phase systems with unbalanced loads using phase A, B, C currents. No harmonics assumed. Compliant with NEC 220.61 & IEC 60364. Calculate the neutral current in a three-phase, four-wire system based on unbalanced phase currents. This tool uses vector summation principles for linear loads only , assuming no harmonic distortion. When You Need This Calculation Determining whether your neutral conductor is properly sized for mixed single-phase loads Verifying compliance with NEC Article 220.61 or IEC 60364-5-52 for neutral load calculation Preventing overheating due to excessive neutral current in panelboards Designing feeders for commercial buildings with unbalanced lighting and receptacle loads Performing preliminary load analysis during electrical inspections Underlying Principle The neutral current is calculated using the following formula: I N = √(I A 2 + I B 2 + I C 2 - I A ×I B - I B ×I C - I C ×I A ) Where: I A , I B , I C : Phase currents in amperes (A) I N : Resultant neutral current (A) Note : This formula applies to sinusoidal, linear loads with 120° phase separation. It does not account for harmonic currents, which can significantly increase neutral current. Practical Example Scenario : A commercial building has the following phase currents: Phase A: 10 A Phase B: 5 A Phase C: 2 A Calculation : I_N = √(10² + 5² + 2² - 10×5 - 5×2 - 2×10) = √(100 + 25 + 4 - 50 - 10 - 20) = √(59) ≈ 7.68 A → Rounded to 7.7 A This means the neutral conductor must carry ~7.7 A continuously. If not sized correctly, it may overheat or trip protective devices. Key Limitations & Assumptions No harmonics : Assumes purely resistive/reactive loads (e.g., motors, heaters). Does not model triplen harmonics from VFDs, computers, or LED drivers. Linear loads only : Not suitable for non-linear loads where current waveforms are distorted. Phase angle assumption : Assumes standard 120° phase difference between A-B-C. Not valid for unbalanced systems with phase shifts. Single frequency : Only applicable to 50/60 Hz AC systems. Not for DC : This is an AC phasor sum; does not apply to DC systems. Use Cases by Industry Industry Application Commercial Buildings Panelboard neutral sizing for lighting and outlets Industrial Facilities Feeder design for mixed motor and control loads Data Centers Assessing neutral current in UPS distribution Residential Verifying service entrance neutral capacity Education Teaching three-phase system behavior For Professionals Who design or inspect electrical systems per NEC or IEC 60364 size neutral conductors to prevent overheating troubleshoot neutral-related faults in panelboards perform load calculations for new construction or retrofits teach power system fundamentals to students Compliance Context NEC Article 220.61 : Requires neutral load calculation based on unbalanced current IEC 60364-5-52 : Specifies rules for final circuit and socket-outlet installations IEEE 141 : Recommends practices for electric power distribution UL 486A/B : Wiring device standards including neutral handling Frequently Asked Questions How do you calculate neutral current in a 3-phase system? For unbalanced linear loads without harmonics, use: I_N = √(I_A² + I_B² + I_C² - I_A×I_B - I_B×I_C - I_C×I_A). This accounts for phase angles and vector addition. In balanced systems, I_N = 0. Why is neutral current important? Neutral current carries the imbalance between phases. If too high, it can overheat conductors, trip breakers, or cause voltage drop—especially in lighting or IT systems where loads are often unbalanced. What happens if neutral is not connected? Without a neutral connection, phase voltages become unbalanced, potentially causing equipment damage or malfunction. The neutral ensures stable voltage across single-phase loads. Does this calculator include harmonic effects? No. This tool assumes linear loads with no harmonics. For non-linear loads (e.g., VFDs, LED drivers), triplen harmonics (3rd, 9th, etc.) add in phase on the neutral—can cause up to 1.7× total load.

Calculate the minimum short-circuit current (fault current) at the end of a low-voltage circuit — essential for verifying protective device sensitivity and ensuring safe electrical system design. This tool helps engineers, electricians, and designers determine how much current flows during a short circuit in single-phase, two-phase, or three-phase systems using copper or aluminum conductors. Supports mm²/AWG units, meters/feet/yards length units, and parallel conductors. Key Features: Single-phase, two-phase, three-phase systems support Copper/Aluminum conductors with adjustable cross-sections mm²/AWG unit compatibility for international use Parallel conductors support for complex circuits Accurate calculations based on line resistance and reactance Formula Used The calculation is based on the following formula: I sc,min = U / (√3 × (R L + X L )) Where: U: System voltage R L : Line resistance X L : Line reactance Example For a three-phase system with a voltage of 400 V, a length of 50 m, and a copper conductor of 16 mm², the calculated minimum short-circuit current is approximately 8.5 kA. Applications Residential wiring projects: Ensuring circuit breakers can handle potential faults. Commercial buildings: Verifying protection coordination for safety compliance. Industrial installations: Designing robust electrical systems that meet rigorous standards. Educational purposes: Teaching students about electrical engineering principles and fault analysis. Frequently Asked Questions (FAQ) Q: What is a short circuit? A: A short circuit occurs when there is an abnormal low-resistance connection between two points of an electrical circuit that are normally at different voltages. Q: How do I find a short circuit? A: Use diagnostic tools like multimeters or specialized equipment to detect unusual voltage drops or high currents indicating a fault. Q: Does a short circuit always trip a breaker? A: Not necessarily. If the fault current is too low due to long wire runs or high impedance, the breaker may not trip. This is why calculating minimum fault current is crucial. Q: How to calculate current in a short circuit? A: The calculator automates this process using the standard formula provided above, taking into account system voltage, conductor material, length, and size. Why Choose Our Calculator? Our Minimum Short-Circuit Current Calculator provides accurate results for professionals needing to ensure the reliability and safety of their electrical designs. With support for multiple system types, conductor materials, and units, it’s a versatile tool for any project.

Calculate the operating current (I b ) of electrical equipment based on active power, voltage, power factor, and coincidence factor—essential for sizing cables, breakers, and protection devices in AC systems. This tool helps engineers, electricians, and designers determine the actual running current under real-world conditions, including multiple loads with varying usage patterns. Supports: Single-phase, two-phase, and three-phase AC systems Standard voltages (e.g., 230V L-N, 400V L-L) Custom voltage input Adjustable power factor (cosφ) Adjustable coincidence factor (F c ) for group load estimation Input via active power (P) or apparent power (S) Key Formulas Single-phase: I b = (F c × P) / (U 0 × cosφ) I b = (F c × S) / U 0 Two-phase: I b = (F c × P) / (U × cosφ) I b = (F c × S) / U Three-phase: I b = (F c × P) / (√3 × U × cosφ) I b = (F c × S) / (√3 × U) Apparent Power: S = √(P² + Q²) Variables: I b : Operating current (A) F c : Coincidence factor (0–1) P: Active power (W) Q: Reactive power (VAr) S: Apparent power (VA) U 0 : Line-to-neutral voltage (V) U: Line-to-line voltage (V) cosφ: Power factor (0.1–1.0) Example Three-phase system, 400 V (L-L), 10 kW, PF = 0.85, F c = 1.0 → Operating current ≈ 19.5 A Frequently Asked Questions (FAQ) What is operating current? Operating current (I b ) is the actual current drawn by equipment during normal operation, adjusted for load diversity using the coincidence factor. Why include coincidence factor? Because not all loads operate at full capacity simultaneously. For example, a factory may have 10 motors, but only 7 run at once → use F c = 0.7 to avoid over-sizing circuits. Can I use this for DC systems? No—this calculator is designed for AC systems only. For DC, use I = P / V (no √3 or power factor). How do I find the power factor if I don’t know it? Use typical values: 0.8–0.9 for motors, 0.95–1.0 for resistive loads (heaters, lighting). You can adjust it to see how current changes. Applications Electrical Design: Size conductors, circuit breakers, and fuses based on actual load demand. Panel Load Calculation: Estimate total current for multiple devices using coincidence factor. Energy Audits: Convert nameplate data into real-world current draw. Renewable Integration: Calculate inverter output current with partial load factors. Educational Use: Teach students about AC power relationships and load diversity. Who Should Use This Tool? Electrical engineers designing LV distribution systems Electricians installing motors, pumps, or HVAC systems Maintenance technicians troubleshooting overloads Facility managers planning upgrades or expansions Students learning AC power fundamentals