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Capacitor kVAR to bring cable current to rating

Short answer

Compute real power (150.95 kW) and reactive power (154.0 kVAr). Permissible apparent power at rated current is 168.92 kVA, giving reactive power 75.82 kVAr. Required capacitor is 78.2 kVAr.

A 415 V conductor cable is rated at 235 Amperes but is carrying a load of 300A at 0.7 power factor. What KVAR of capacitor is required to reduce the current to its normal rated value?

Given: V_L = 415 V, cable rating 235 A, present load 300 A at pf 0.7 (lag). Real power stays the same; capacitors reduce only the reactive part.

Present load

S₁ = √3 × 415 × 300 = 215.64 kVA
P  = S₁ × 0.7       = 150.95 kW
Q₁ = S₁ × sin(cos⁻¹0.7) = 215.64 × 0.7141
   = 154.00 kVAr

Allowed load at rated current

S₂ = √3 × 415 × 235 = 168.92 kVA
Q₂ = √(S₂² − P²) = √(168.92² − 150.95²)
   = 75.82 kVAr
New pf = P / S₂ = 150.95 / 168.92 = 0.894 lag

Capacitor required

Qc = Q₁ − Q₂ = 154.00 − 75.82 = 78.18 kVAr

Answer: about 78.2 kVAr of capacitors (3-phase); the pf rises to about 0.894 lag and the current falls to 235 A. In practice the next standard size, 80 kVAr, is chosen.

✓ Answer checked by Claude Opus 5.5 (high effort)

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