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 / VI = 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 / RI = 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: IN = √(IA2 + IB2 + IC2 - IA×IB - IB×IC - IC×IA) Where: IA, IB, IC: Phase currents in amperes (A) IN: 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 RL: Line resistance XL: 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 (Ib) 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 (Fc) 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: Ib: Operating current (A) Fc: Coincidence factor (0–1) P: Active power (W) Q: Reactive power (VAr) S: Apparent power (VA) U0: 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 (Ib) 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 Fc = 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

Calculate maximum prospective short-circuit current (Isc) at any point in a low-voltage AC circuit per IEC 60909. Supports 3-phase, L-L, L-N, and L-PE faults based on source impedance and conductor data. Who Uses This Tool? Designed for electrical engineers, solar designers, electricians, and facility managers who need to verify equipment safety and code compliance. Whether you're sizing breakers for a commercial panel, designing a PV system, or performing an arc flash study, this calculator helps ensure your installation can safely withstand fault conditions. Calculation Principles The maximum short-circuit current is determined by the total impedance between the source and the fault location, including: Available short-circuit current at the origin (kA) System voltage (V) – with 5% overvoltage factor (c = 1.05) Line length (m/ft/yd) Conductor material (Copper or Aluminum) Conductor cross-section (mm² or AWG) Cable type (Unipolar, Multicore, etc.) affecting reactance Number of parallel conductors per phase/neutral/PE Type of fault: Three-phase (L1-L2-L3) Phase-to-phase (L-L) Phase-to-neutral (L-N) Phase-to-protective earth (L-PE) Longer lines, smaller cross-sections, aluminum conductors, or higher-resistance installations reduce the short-circuit current at the load end due to increased line impedance (R + jX). Underlying Formulas (IEC-based) Three-phase fault: I SC = (1.05 × U) / (√3 × Z L) Phase-to-phase fault: I SC = (1.05 × U) / (2 × Z L) Phase-to-neutral fault: I SC = (1.05 × U) / (√3 × (Z L + Z N)) Phase-to-PE fault: I SC = (1.05 × U) / (√3 × (Z L + Z PE)) Where ZL, ZN, and ZPE are the total impedances of phase, neutral, and protective earth conductors, respectively—adjusted for length, material, cross-section, and parallel paths. Key Applications Circuit breaker & fuse selection: Verify that the device’s rated breaking capacity (Icn or Icu) exceeds the calculated ISC at its installation point—essential for meeting NEC and IEC SCCR requirements. Protection coordination: Ensure selective tripping between upstream and downstream devices by comparing actual fault currents against time-current curves. Arc flash risk assessment: Estimate incident energy levels to determine if arc-resistant switchgear or PPE is required. Conductor thermal withstand: Check that cables can endure short-circuit heating without damage (using I²t or adiabatic equation). Solar PV system design: Confirm that MPPT trackers and inverters are rated for the maximum short-circuit current from your PV strings—critical for avoiding equipment damage under fault conditions. Typical Input Example • Source short-circuit current: 10 kA • System voltage: 220 V (single-phase) or 400 V (three-phase) • Conductor material: Copper • Phase size: 1.5 mm² (≈ 16 AWG) • Neutral & PE size: 1.5 mm² • Line length: 10 meters • Fault type: Phase-to-earth (L-PE) Common Questions Answered How do you calculate maximum short-circuit current? It’s derived from system voltage and total impedance: ISC = (1.05 × U) / Ztotal, where Ztotal includes both source and line impedances. This tool automates the full calculation—including conductor resistance, reactance, parallel paths, and fault type. What is SCCR, and why does it matter? SSCR (Short-Circuit Current Rating) is the maximum fault current a device can safely interrupt. Your calculated ISC must be ≤ the SCCR of all downstream equipment—otherwise, catastrophic failure can occur during a fault. Does the NEC 125% rule apply here? Yes—for continuous loads, NEC Article 210.20 requires overcurrent devices to be rated at least 125% of the load current. While this rule governs normal operation, your short-circuit analysis ensures the same device can also handle abnormal fault conditions. Can this be used for solar systems? Absolutely. Solar designers use this tool to verify that the short-circuit current from PV strings does not exceed the input rating of inverters or MPPT trackers—a common requirement in UL 1741 and IEC 62109.