A balanced 3-phase load draws 120 amperes line current at 230V line to line, 0.848 pf lagging current. Solve for real power. *
P = √3 VL x IL p.f
= √3 * 230 * 120 * 0.848
P = 40538.30 Watts
= 40.53 Kw
To solve for real power, we can use the formula:
Real Power (P) = √3 × Voltage (V) × Current (I) × Power Factor (PF)
Given:
Voltage (V) = 230V line to line
Current (I) = 120 amperes
Power Factor (PF) = 0.848 (lagging)
Now we can calculate the real power:
P = √3 × 230V × 120A × 0.848
P = 1.732 × 230V × 120A × 0.848
P = 44,200.832 watts (or 44.2 kilowatts)
Therefore, the real power consumed by the balanced 3-phase load is 44,200.832 watts (or 44.2 kilowatts).
To solve for real power in a balanced 3-phase load, you can use the formula:
Real Power (P) = √3 * Voltage (V) * Current (I) * Power Factor (PF)
Given values:
Line Current (I) = 120 amperes
Voltage (V) = 230V line-to-line
Power Factor (PF) = 0.848 lagging
Substituting these values into the formula, we get:
P = √3 * 230V * 120A * 0.848
To find the solution, let's calculate it step by step:
1. Calculate the product of voltage and current:
230V * 120A = 27,600 VA or 27.6 kVA (kilo Volt-Amperes)
2. Multiply the kVA by the power factor:
27.6 kVA * 0.848 = 23.4016 kW or 23,401.6 Watts.
Therefore, the real power of the balanced 3-phase load is 23,401.6 Watts or 23.4016 kW.