---
title: "Line pf of 35 kW induction motor with capacitors"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:induction-motor-35kw-capacitor-line-pf"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-motor-35kw-capacitor-line-pf
site: BCT Engineering
---

# Line pf of 35 kW induction motor with capacitors

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2081 Bhadra, 8 marks; 2073 Shrawan; 2070 Chaitra (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-motor-35kw-capacitor-line-pf
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6

## Question

A 35 kW induction motor has pf 0.9 and efficiency 0.9 at full load, pf 0.6 and efficiency 0.7 at half load. At no load, the current is 25% of full load current and pf 0.1. Capacitors are supplied to make the line pf 0.8 at half load. With these capacitors in circuit, find the line pf at (i) full load (ii) no load.

## Short answer

The capacitor bank rating is fixed at 14.58 kVAR based on the half-load condition. With this capacitor, the line power factor becomes 0.994 lagging at full load and 0.271 leading at no load.

## Answer

The capacitor kVAR is fixed by the half-load condition and then stays connected at full load and at no load.

#### Input power and reactive power of the motor
**Full load:** output 35 kW, $\eta = 0.9$, pf 0.9
$$
\begin{aligned}
P_{FL} &= \frac{35}{0.9} = 38.89\ \text{kW} \\
Q_{FL} &= P_{FL}\tan\phi = 38.89 \times \tan(\cos^{-1}0.9) = 38.89 \times 0.4843 = 18.83\ \text{kVAR}
\end{aligned}
$$

**Half load:** output 17.5 kW, $\eta = 0.7$, pf 0.6
$$
\begin{aligned}
P_{HL} &= \frac{17.5}{0.7} = 25\ \text{kW} \\
Q_{HL} &= 25 \times \tan(\cos^{-1}0.6) = 25 \times 1.3333 = 33.33\ \text{kVAR}
\end{aligned}
$$

**No load:** current is 25% of full-load current, pf 0.1
$$
\begin{aligned}
S_{FL} &= \frac{38.89}{0.9} = 43.21\ \text{kVA}, \quad S_{NL} = 0.25 \times 43.21 = 10.80\ \text{kVA} \\
P_{NL} &= 10.80 \times 0.1 = 1.080\ \text{kW}, \quad Q_{NL} = 10.80 \times \sin(\cos^{-1}0.1) = 10.75\ \text{kVAR}
\end{aligned}
$$

#### Capacitor rating (line pf 0.8 at half load)
$$
\begin{aligned}
Q_C &= P_{HL}(\tan\phi_1 - \tan\phi_2) = 25(1.3333 - 0.75) = 14.58\ \text{kVAR}
\end{aligned}
$$

#### (i) Line pf at full load
$$
\begin{aligned}
Q' &= 18.83 - 14.58 = 4.25\ \text{kVAR (lagging)} \\
\cos\phi &= \frac{38.89}{\sqrt{38.89^2 + 4.25^2}} = 0.994\ \text{lagging}
\end{aligned}
$$

#### (ii) Line pf at no load
$$
\begin{aligned}
Q' &= 10.75 - 14.58 = -3.84\ \text{kVAR (leading)} \\
\cos\phi &= \frac{1.080}{\sqrt{1.080^2 + 3.84^2}} = 0.271\ \text{leading}
\end{aligned}
$$

| Load | P (kW) | Q motor (kVAR) | Q net (kVAR) | Line pf |
|---|---|---|---|---|
| Full | 38.89 | 18.83 | 4.25 | 0.994 lag |
| Half | 25.00 | 33.33 | 18.75 | 0.800 lag |
| No load | 1.08 | 10.75 | −3.84 | 0.271 lead |

**Answer:** Capacitor $= 14.58$ kVAR; line pf (i) full load $\approx 0.994$ lagging, (ii) no load $\approx 0.271$ leading.

## Check yourself

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

1. At which load is the capacitor kVAR fixed, and what happens at other loads?
   A. Fixed by the half-load pf of 0.8; the same capacitor stays connected at full and no load
   B. Fixed by full-load pf; it is switched out at no load
   C. Fixed at no load; it is doubled at full load
   D. Recalculated separately for each load
   Answer: A. The question states capacitors give line pf 0.8 at half load, so Q_C is found there and remains constant. (Key point: Capacitor sizing condition)

2. Which expression gives the capacitor rating at half load?
   A. Q_C = 35(tan phi1 - tan phi2)
   B. Q_C = P_HL(tan phi1 - tan phi2) = 25(1.3333 - 0.75) = 14.58 kVAR
   C. Q_C = 38.89 x 0.4843
   D. Q_C = 25 x (0.6 - 0.8)
   Answer: B. Half-load input is 17.5/0.7 = 25 kW; tan(cos^-1 0.6) = 1.3333 and tan(cos^-1 0.8) = 0.75. (Key point: Capacitor kVAR formula)

3. With the capacitor connected, what is the full-load line pf?
   A. About 0.9 lagging, unchanged
   B. About 0.271 leading
   C. About 0.994 lagging, since net Q = 18.83 - 14.58 = 4.25 kVAR
   D. About 0.8 lagging
   Answer: C. P = 38.89 kW and net Q = 4.25 kVAR give cos phi = 0.994 lagging. (Key point: Full-load line pf)

4. Why is the no-load line pf leading?
   A. No-load P is larger than Q
   B. The motor generates power at no load
   C. The capacitor is disconnected at no load
   D. Capacitor kVAR (14.58) exceeds the motor no-load kVAR (10.75), leaving 3.84 kVAR leading
   Answer: D. Net Q = 10.75 - 14.58 = -3.84 kVAR, with P = 1.08 kW, giving pf 0.271 leading. (Key point: No-load line pf)

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