---
title: "Minimum current and emf of 10 HP motor"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:motor-10hp-minimum-current"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-10hp-minimum-current
site: BCT Engineering
---

# Minimum current and emf of 10 HP motor

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2070 Asar, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-10hp-minimum-current
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 400 V, 10 HP, 3-phase synchronous motor has negligible armature resistance and synchronous reactance of 10 Ω/phase. Determine the minimum current and the corresponding induced emf for full load conditions. Assume an efficiency of 85%.

## Short answer

At minimum current, pf = 1. Input power = 8776.5 W, so I_min = 8776.5/(√3×400) = 12.67 A. Induced emf Ef = √(V_ph² + (I Xs)²) = 263.4 V per phase (456 V line).

## Answer

For a given load, armature current is **minimum at unity power factor**, because then the whole current is active current. **Assumptions:** star connection, 1 hp = 746 W.

#### Minimum current

$$
\begin{aligned}
P_{out} &= 10 \times 746 = 7460\ \text{W} \\
P_{in} &= \frac{7460}{0.85} = 8776.5\ \text{W} \\
I_{min} &= \frac{P_{in}}{\sqrt3\,V_L \times 1} = \frac{8776.5}{\sqrt3 \times 400} = 12.67\ \text{A}
\end{aligned}
$$

#### Induced emf

At unity pf, $\vec E_f = \vec V - jX_s\vec I_a$ with $I_a$ in phase with $V$:

$$
\begin{aligned}
V &= \frac{400}{\sqrt3} = 230.94\ \text{V},\quad I_aX_s = 12.67 \times 10 = 126.68\ \text{V} \\
E_f &= \sqrt{230.94^2 + 126.68^2} = 263.40\ \text{V per phase} \\
\delta &= \tan^{-1}\frac{126.68}{230.94} = 28.75^\circ
\end{aligned}
$$

```text
 V = 230.94 ------------------>|
   \                           | IXs = 126.68
    \ delta                    |
     '------------------------>'
          Ef = 263.40   (Ia along V)
```

Line value $= \sqrt3 \times 263.40 = 456.2$ V.

**Answer:** Minimum current $= 12.67$ A (at unity pf); induced emf $= 263.4$ V per phase $= 456$ V line.

## Check yourself

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

1. At what power factor is the armature current minimum for a given load?
   A. Zero lagging
   B. Unity
   C. 0.8 leading
   D. 0.5 lagging
   Answer: B. At unity pf the whole current is active, so for a given power the current is minimum. (Key point: Minimum current condition)

2. How is the minimum current found for 10 hp output at 85 percent efficiency?
   A. Pin = 7460/0.85 = 8776 W; I = Pin/(sqrt3 x 400) = 12.67 A
   B. I = 7460/(sqrt3 x 400) = 10.77 A
   C. I = 7460 x 0.85/(sqrt3 x 400)
   D. I = 7460/400 = 18.65 A
   Answer: A. Input is output divided by efficiency; then I = Pin/(sqrt3 VL) at unity pf. (Key point: Input power and current)

3. How is Ef computed at unity pf?
   A. Ef = V + Ia Xs
   B. Ef = sqrt(V^2 + (Ia Xs)^2) = 263.4 V per phase
   C. Ef = V - Ia Xs
   D. Ef = Ia Xs
   Answer: B. Ia is in phase with V and jXs Ia is perpendicular, so Ef = sqrt(230.94^2 + 126.68^2) = 263.4 V per phase. (Key point: Induced emf at unity pf)

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