---
title: "Current I using star-delta transformation"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:star-delta-current-i-19v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-delta-current-i-19v
site: BCT Engineering
---

# Current I using star-delta transformation

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2068 Chaitra, 4 marks
- Video: [Delta to Wye and Wye to Delta Transformations | Engineering Circuit Analysis | (Solved Examples)](https://www.youtube.com/watch?v=XtciQU4KMSg) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-delta-current-i-19v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Find the current I as shown in figure below using star - delta transformation. [Figure: 19 V source; 10 Ω shunt; 5 Ω in the top branch; bridge with 2 Ω, 2 Ω, 2 Ω and 10 Ω resistors.]

## Short answer

Convert three 2 Ω star to 6 Ω delta, then parallel combinations: T-U 2.727 Ω, T-G and U-G 3.75 Ω each. Total resistance 2.375 Ω. Source current I = 19 V / 2.375 Ω = 8 A.

## Answer

#### Reading the figure
The 19 V source (+ on top) feeds node T. The current *I* is the current leaving the source.
- 10 Ω from T to the bottom rail (G).
- 5 Ω from T to the top right node U.
- A star of three 2 Ω resistors: T to O, U to O, O to G.
- 10 Ω from U to G.

#### Step 1: Star (2, 2, 2 Ω at O) to delta
For equal resistors, R_Δ = 3 R_Y = 6 Ω. The delta joins T, U and G, with 6 Ω between each pair.

```text
  T ---[5 || 6]--- U          T-U : 5 Ω ∥ 6 Ω
  |                |          T-G : 10 Ω ∥ 6 Ω
  [10 || 6]    [10 || 6]      U-G : 10 Ω ∥ 6 Ω
  |                |
  G ---------------G
```

#### Step 2: Combine in parallel
- T–U: 5 × 6/(5 + 6) = 2.727 Ω
- T–G: 10 × 6/(10 + 6) = 3.75 Ω
- U–G: 10 × 6/(10 + 6) = 3.75 Ω

#### Step 3: Total resistance across the source
Path T–U–G = 2.727 + 3.75 = 6.477 Ω, in parallel with T–G = 3.75 Ω:

R_total = 3.75 × 6.477/(3.75 + 6.477) = 2.375 Ω

#### Step 4: Source current
I = 19/2.375 = 8.0 A

(Check: 19/3.75 = 5.067 A plus 19/6.477 = 2.933 A gives 8.0 A.)

**Answer: I = 8 A**

## Check yourself

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

1. For three equal 2 ohm resistors in star, what is each delta resistor?
   A. 6 ohm (3 times R_Y)
   B. 2/3 ohm
   C. 2 ohm
   D. 4 ohm
   Answer: A. For equal resistors R_delta = 3 R_Y. (Key point: Star to delta formula)

2. After conversion, what happens to the 5 ohm and 10 ohm resistors?
   A. They are removed
   B. They combine in parallel with the 6 ohm delta resistors across the same node pairs
   C. They are added in series with the delta
   D. They are shorted
   Answer: B. Each delta resistor lies between the same two nodes as an existing resistor: 5||6 for T-U and 10||6 for T-G and U-G. (Key point: After conversion)

3. What is I?
   A. 19 A
   B. 4 A
   C. 8 A, since R_total = 2.375 ohm and I = 19/2.375
   D. 2.375 A
   Answer: C. Total resistance across the source is 2.375 ohm so I = 8 A. (Key point: Final current)

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