---
title: "Short circuit at load B (ohmic data)"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:short-circuit-load-b-ohmic"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/short-circuit-load-b-ohmic
site: BCT Engineering
---

# Short circuit at load B (ohmic data)

Power System Analysis II · Chapter 3: Power System Fault Calculation

- Asked in: 2075 Asoj, 10 marks
- Video: [Short Circuit MVA - Symmetrical Fault Analysis - Power System 2](https://www.youtube.com/watch?v=Jn2UbN_nNt4) — Ekeeda
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/short-circuit-load-b-ohmic
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

Calculate short circuit MVA and short circuit current when three phase fault occur at load B of the following single line diagram. [Figure: G1 20 MVA, 6.6 kV, X = 0.655 ohm and G2 10 MVA, 6.6 kV, X = 1.31 ohm on a common 6.6 kV bus which also supplies Load A (15 MVA, 6.6 kV, pf = 0.9 lag); transformer Tr-1 30 MVA, 6.6/66 kV, X = 14.52 ohm (referred to HV side); transmission line x = 17.4 ohm; transformer Tr-2 30 MVA, 66/3.81 kV, X = 14.52 ohm (referred to HV side); 3.81 kV bus with generator G3 (30 MVA, 3.81 kV, X = 0.1452 ohm) and Load B (30 MW, 3.81 kV, pf = 0.9 lag)]

## Short answer

On 30 MVA, 3.81 kV base, the short-circuit MVA is 148.3 MVA and the short-circuit current is 22.48 kA at the load B bus.

## Answer

Base: 30 MVA, with base voltages 6.6 kV, 66 kV and 3.81 kV in the three zones. Loads are neglected and the pre-fault voltage is 1.0 pu (usual for fault level), so all generator EMFs are 1 pu. The fault is on the 3.81 kV bus of load B.

#### Base impedances
$$
\begin{aligned}
Z_{b,6.6} &= \frac{6.6^2}{30} = 1.452\ \Omega \\
Z_{b,66} &= \frac{66^2}{30} = 145.2\ \Omega \\
Z_{b,3.81} &= \frac{3.81^2}{30} = 0.48387\ \Omega
\end{aligned}
$$

#### Per-unit reactances
| Element | Ohms | Zone base (Ω) | pu |
|---|---|---|---|
| G1 | 0.655 | 1.452 | 0.4511 |
| G2 | 1.31 | 1.452 | 0.9022 |
| Tr-1 (HV) | 14.52 | 145.2 | 0.1000 |
| Line | 17.4 | 145.2 | 0.1198 |
| Tr-2 (HV) | 14.52 | 145.2 | 0.1000 |
| G3 | 0.1452 | 0.48387 | 0.3001 |

```text
 G1--+
     +--Tr1--line--Tr2--+-- 3.81 kV bus (F, load B)
 G2--+                  |
                        G3
```

#### Thevenin reactance at load B
$$
\begin{aligned}
X_{G1\|G2} &= \frac{0.4511 \times 0.9022}{0.4511 + 0.9022} = 0.3007 \\
X_{left} &= 0.3007 + 0.1000 + 0.1198 + 0.1000 = 0.6206 \\
X_{th} &= \frac{0.6206 \times 0.3001}{0.6206 + 0.3001} = 0.2023\ \text{pu}
\end{aligned}
$$

#### Short-circuit MVA and current
$$
\begin{aligned}
\text{SC MVA} &= \frac{30}{0.2023} = 148.32\ \text{MVA} \\
I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 3.81 \times 10^3} = 4546.1\ \text{A} \\
I_f &= \frac{1}{0.2023} \times 4546.1 = 4.9439 \times 4546.1 = 22475\ \text{A}
\end{aligned}
$$

**Answer:** short-circuit MVA $\approx 148.3$ MVA, short-circuit current $\approx 22.48$ kA at 3.81 kV.

## Check yourself

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

1. Why are three different base impedances (1.452, 145.2, 0.48387 ohm) used?
   A. Each voltage zone has its own base voltage, so Z_base = kV^2/MVA differs per zone
   B. The generators have different MVA ratings
   C. Because the loads have different power factors
   D. Because the line is long
   Answer: A. Transformers split the system into zones with base voltages 6.6, 66 and 3.81 kV; the same 30 MVA base gives a different Z_base in each. (Key point: Zone base impedances)

2. G1 has X = 0.655 ohm on the 6.6 kV bus. Its per-unit value is
   A. 0.655 x 1.452 = 0.951
   B. 0.655/145.2 = 0.0045
   C. 0.655/1.452 = 0.4511
   D. 0.655/0.48387 = 1.354
   Answer: C. G1 is in the 6.6 kV zone, so divide by Z_base = 6.6^2/30 = 1.452 ohm. (Key point: Per-unit from ohms)

3. How is X_th at load B (3.81 kV bus) formed?
   A. (G1||G2) + Tr1 + line + Tr2 = 0.6206, in parallel with G3 (0.3001)
   B. All six reactances added in series
   C. G1, G2 and G3 all in series
   D. 0.6206 added in series to G3
   Answer: A. The left source chain gives 0.6206 pu, and G3 sits directly on the faulted bus, so it is in parallel: 0.2023 pu. (Key point: Thevenin reduction)

4. Which base current converts the pu fault current to amperes?
   A. The 6.6 kV base current
   B. The 66 kV base current
   C. The average of all three
   D. The 3.81 kV base current (4546 A), since the fault is on that bus
   Answer: D. Amperes are found with the base current of the faulted zone: I_f = 4.944 x 4546 = 22.5 kA. (Key point: Fault current in amperes)

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