---
title: "Short shunt vs long shunt generated voltage (150 A, 230 V)"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:short-shunt-150a-230v-long-shunt"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-150a-230v-long-shunt
site: BCT Engineering
---

# Short shunt vs long shunt generated voltage (150 A, 230 V)

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2076 Baisakh, 10 marks
- Video: [Short Shunt and Long Shunt Generator Solved Example](https://www.youtube.com/watch?v=SXV7QFc8sjQ) — Vishnu Online Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-150a-230v-long-shunt
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A short shunt compound dc generator supplies a load current of 150 A at 230 V. The generator has following winding resistances: Armature resistance = 0.15 ohm, Series field resistance = 0.1 ohm, Shunt field resistance = 100 ohm. Calculate (a) emf generated if the carbon brush drop is 2 V per brush, and also (b) calculate the ratio of voltage generated if the same generator is connected as long shunt generator to the original short shunt compound generator.

## Short answer

Short-shunt: Vsh=245 V, Ish=2.45 A, Ia=152.45 A, Eg=271.87 V (with 4 V brush drop). Long-shunt: Ish=2.3 A, Ia=152.3 A, Eg=272.08 V; ratio long/short = 1.00076.

## Answer

**Given:** $I_L = 150$ A, $V = 230$ V, $R_a = 0.15\ \Omega$, $R_{se} = 0.1\ \Omega$, $R_{sh} = 100\ \Omega$, brush drop 2 V per brush, so total brush drop $= 2 \times 2 = 4$ V (one positive and one negative brush in the current path).

#### (a) Short-shunt connection
```text
     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 230 V
     |      |                 |
     +------+-----------------+
```

$$
\begin{aligned}
V_{sh} &= V + I_L R_{se} = 230 + 150 \times 0.1 = 245\ \text{V} \\
I_{sh} &= \frac{245}{100} = 2.45\ \text{A} \\
I_a &= I_L + I_{sh} = 150 + 2.45 = 152.45\ \text{A} \\
E_{g,s} &= V_{sh} + I_a R_a + V_{b} \\
&= 245 + 152.45 \times 0.15 + 4 \\
&= 245 + 22.87 + 4 = 271.87\ \text{V}
\end{aligned}
$$

#### (b) Same machine as long shunt (same load: 150 A at 230 V)
```text
     +--Ra--Rse--+------------+
     |           |            |
    (Eg)        Rsh         Load 230 V
     |           |            |
     +-----------+------------+
```

Now the shunt field is across the terminals and the series field carries $I_a$:

$$
\begin{aligned}
I_{sh} &= \frac{230}{100} = 2.3\ \text{A} \\
I_a &= 150 + 2.3 = 152.3\ \text{A} \\
E_{g,l} &= V + I_a (R_a + R_{se}) + V_b \\
&= 230 + 152.3 \times 0.25 + 4 \\
&= 230 + 38.075 + 4 = 272.075\ \text{V}
\end{aligned}
$$

Ratio of generated voltages:

$$
\frac{E_{g,l}}{E_{g,s}} = \frac{272.075}{271.8675} = 1.00076
$$

**Answer:** (a) $E_g = 271.87$ V (short shunt). (b) $E_g = 272.08$ V as long shunt; ratio long : short $= 1.00076$ (about 1.0008 : 1). The long-shunt machine needs a slightly higher emf because the series field now carries the larger armature current.

## Check yourself

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

1. With 2 V per brush, what total brush drop is used?
   A. 2 V
   B. 1 V
   C. 4 V, as one positive and one negative brush are in the path
   D. 8 V
   Answer: C. Current passes through one positive and one negative brush, so total drop is 2 x 2 = 4 V. (Key point: Total brush drop)

2. What is E_g for the short-shunt connection?
   A. 230 + 150 x 0.25 + 4 = 271.5 V
   B. 245 + 152.45 x 0.15 + 4 = 271.87 V
   C. 230 + 152.3 x 0.15 + 4
   D. 245 + 150 x 0.15 + 4
   Answer: B. V_sh = 245 V, I_sh = 2.45 A, I_a = 152.45 A, so E_g = 245 + 22.87 + 4 = 271.87 V. (Key point: Short shunt emf)

3. In the long-shunt connection, what are I_sh and I_a?
   A. I_sh = 2.45 A, I_a = 152.45 A
   B. I_sh = 2.3 A, I_a = 147.7 A
   C. I_sh = 2.3 A, I_a = 152.3 A
   D. I_sh = 2.3 A, I_a = 150 A
   Answer: C. The shunt field is across the terminals: I_sh = 230/100 = 2.3 A, and I_a = 150 + 2.3 = 152.3 A. (Key point: Long shunt currents)

4. Why is the long-shunt emf slightly larger, giving a ratio of about 1.0008?
   A. The series field carries the larger I_a and the full 230 V is across the shunt field
   B. The shunt field resistance is higher in long shunt
   C. Brush drop doubles in long shunt
   D. The armature resistance changes
   Answer: A. In long shunt the drop is I_a(R_a + R_se) = 38.075 V, more than the 22.87 V + 15 V of short shunt, so E_g is 272.08 V. (Key point: Long shunt emf and ratio)

## Related in Electrical Machines I

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