---
title: "Capacitor kVAR to bring cable current to rating"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:capacitor-kvar-415v-cable"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/capacitor-kvar-415v-cable
site: BCT Engineering
---

# Capacitor kVAR to bring cable current to rating

Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry

- Asked in: 2074 Bhadra, 4 marks; 2072 Asoj, 4 marks (asked 2 times)
- Video: [APFC Panel KVAR Calculation | Required KVAR Calculation Methods | APFC Panel Designing](https://www.youtube.com/watch?v=GwKykjCUvdA) — Electrical Infinity
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/capacitor-kvar-415v-cable
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10

## Question

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?

## 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.

## Answer

**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
```text
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
```text
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
```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When capacitors are added to reduce the cable current, which quantity remains unchanged?
   A. Reactive power
   B. Real power P
   C. Power factor
   D. Apparent power
   Answer: B. Capacitors only reduce reactive power; the real power drawn by the load stays the same. (Key point: What stays constant)

2. For 300 A at 415 V, 3-phase, pf 0.7, what are S and P?
   A. S = 215.64 kVA, P = 150.95 kW
   B. S = 124.5 kVA, P = 87.2 kW
   C. S = 215.64 kVA, P = 215.64 kW
   D. S = 150.95 kVA, P = 215.64 kW
   Answer: A. S = sqrt3 x 415 x 300 = 215.64 kVA, and P = 0.7 x 215.64 = 150.95 kW. (Key point: Present load)

3. How is the allowed reactive power at 235 A found?
   A. Q2 = S2 x 0.7
   B. Q2 = sqrt(S2^2 - P^2) with S2 = sqrt3 x 415 x 235
   C. Q2 = P x 235
   D. Q2 = Q1 x 235/300 x 0.7
   Answer: B. At 235 A, S2 = 168.92 kVA, and with P fixed, Q2 = sqrt(168.92^2 - 150.95^2) = 75.82 kVAr. (Key point: Reactive power at rated current)

4. What capacitor rating is required?
   A. 154.00 kVAr
   B. 75.82 kVAr
   C. 230 kVAr
   D. About 78.2 kVAr
   Answer: D. Qc = Q1 - Q2 = 154.00 - 75.82 = 78.18 kVAr, giving a new pf of about 0.894 lag. (Key point: Capacitor rating)

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