Motor full-load current

This tool calculates the efficiency of an electric motor as the ratio between shaft output power and electrical input power. Typical efficiency ranges from 70% to 96%. Input motor parameters to automatically calculate: Electrical input power (kW) Motor efficiency (%) Supports single-, two-, and three-phase systems Real-time bidirectional calculation Key Formulas Electrical Input Power: Single-phase: P_in = V × I × PF Two-phase: P_in = √2 × V × I × PF Three-phase: P_in = √3 × V × I × PF Efficiency: % = (P_out / P_in) × 100% Example Calculations Example 1: Three-phase motor, 400V, 10A, PF=0.85, P_out=5.5kW → P_in = √3 × 400 × 10 × 0.85 ≈ 5.95 kW Efficiency = (5.5 / 5.95) × 100% ≈ 92.4% Example 2: Single-phase motor, 230V, 5A, PF=0.8, P_out=1.1kW → P_in = 230 × 5 × 0.8 = 0.92 kW Efficiency = (1.1 / 0.92) × 100% ≈ 119.6% (Invalid!) Important Notes Input data must be accurate Efficiency cannot exceed 100% Use high-precision instruments Efficiency varies with load

A tool to calculate the slip of an AC induction motor, which is the difference between the stator magnetic field speed and the rotor speed. Slip is a key parameter affecting torque, efficiency, and starting performance. This calculator supports: Input synchronous and rotor speed → automatically calculate slip Input slip and synchronous speed → automatically calculate rotor speed Input frequency and pole pairs → automatically calculate synchronous speed Real-time bidirectional calculation Key Formulas Synchronous Speed: N_s = (120 × f) / P Slip (%): Slip = (N_s - N_r) / N_s × 100% Rotor Speed: N_r = N_s × (1 - Slip) Example Calculations Example 1: 4-pole motor, 50 Hz, rotor speed = 2850 RPM → N_s = (120 × 50) / 2 = 3000 RPM Slip = (3000 - 2850) / 3000 × 100% = 5% Example 2: Slip = 4%, N_s = 3000 RPM → N_r = 3000 × (1 - 0.04) = 2880 RPM Example 3: 6-pole motor (P=3), 60 Hz, slip = 5% → N_s = (120 × 60) / 3 = 2400 RPM N_r = 2400 × (1 - 0.05) = 2280 RPM Use Cases Motor selection and performance evaluation Industrial motor monitoring and fault diagnosis Teaching: Induction motor operation principles VFD control strategy analysis Motor efficiency and power factor study

This tool calculates the running and starting capacitor values needed to operate a three-phase induction motor on single-phase power. Suitable for small motors (< 1.5 kW), with output power reduced to 60–70%. Input motor rated power, single-phase voltage, and frequency to automatically calculate: Running capacitor (μF) Starting capacitor (μF) Supports kW and hp units Real-time bidirectional calculation Key Formulas Running Capacitor: C_run = (2800 × P) / (V² × f) Starting Capacitor: C_start = 2.5 × C_run Where: P: Motor power (kW) V: Single-phase voltage (V) f: Frequency (Hz) Example Calculations Example 1: 1.1 kW motor, 230 V, 50 Hz → C_run = (2800 × 1.1) / (230² × 50) ≈ 11.65 μF C_start = 2.5 × 11.65 ≈ 29.1 μF Example 2: 0.75 kW motor, 110 V, 60 Hz → C_run = (2800 × 0.75) / (110² × 60) ≈ 2.9 μF C_start = 2.5 × 2.9 ≈ 7.25 μF Important Notes Only suitable for small motors (< 1.5 kW) Output power drops to 60–70% of original Use capacitors rated at 400V AC or higher Starting capacitor must be automatically disconnected Motor should be connected in "Y" configuration

This tool calculates the power factor (PF) of an electric motor as the ratio between active power and apparent power. Typical values range from 0.7 to 0.95. Input motor parameters to automatically calculate: Power Factor (PF) Apparent Power (kVA) Reactive Power (kVAR) Phase Angle (φ) Supports single-, two-, and three-phase systems Key Formulas Apparent Power: Single-phase: S = V × I Two-phase: S = √2 × V × I Three-phase: S = √3 × V × I Power Factor: PF = P / S Reactive Power: Q = √(S² - P²) Phase Angle: φ = arccos(PF) Example Calculations Example 1: Three-phase motor, 400V, 10A, P=5.5kW → S = √3 × 400 × 10 = 6.928 kVA PF = 5.5 / 6.928 ≈ 0.80 φ = arccos(0.80) ≈ 36.9° Example 2: Single-phase motor, 230V, 5A, P=0.92kW → S = 230 × 5 = 1.15 kVA PF = 0.92 / 1.15 ≈ 0.80 Important Notes Input data must be accurate PF cannot exceed 1 Use high-precision instruments PF varies with load