---
title: "Current in 20Ω using star-delta transformation"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:star-delta-current-20-ohm-2v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-delta-current-20-ohm-2v
site: BCT Engineering
---

# Current in 20Ω using star-delta transformation

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2070 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-20-ohm-2v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Determine the current in 20Ω resistor of the network shown in figure below using Star Delta Transformation. [Figure: bridge network with 6 Ω and 12 Ω in the upper arms, 3 Ω and 10 Ω in the lower arms, 20 Ω in the middle bridge arm, 2 V source at the bottom.]

## Short answer

Convert the delta (6, 20, 3 Ω) to star; total resistance is 7.526 Ω, source current 0.2658 A. Back in the original bridge, voltage difference across the 20 Ω yields a current of 0.0078 A from Bm to T.

## Answer

#### Reading the figure
Bridge nodes: T (top), Lf (left), Rt (right), Bm (bottom). Arms: Lf–T = 6 Ω, T–Rt = 12 Ω, Lf–Bm = 3 Ω, Bm–Rt = 10 Ω, and the 20 Ω between T and Bm. The 2 V source is connected across Lf and Rt. The 2 V battery sits in the bottom wire with its + terminal towards the left node Lf, so Lf is the positive end.

#### Step 1: Delta (6, 20, 3) to star
The delta Lf–T–Bm has R_LT = 6, R_TB = 20, R_BL = 3. Sum = 29 Ω.

- Leg at Lf: 6 × 3/29 = 0.621 Ω
- Leg at T: 6 × 20/29 = 4.138 Ω
- Leg at Bm: 20 × 3/29 = 2.069 Ω

#### Step 2: Reduce
- Path through T: 4.138 + 12 = 16.138 Ω
- Path through Bm: 2.069 + 10 = 12.069 Ω
- Parallel: 16.138 × 12.069/28.207 = 6.905 Ω

Total resistance = 0.621 + 6.905 = 7.526 Ω

Source current I = 2/7.526 = 0.2658 A

#### Step 3: Go back to the original bridge
Voltage across the parallel paths = 0.2658 × 6.905 = 1.835 V

- Current through the 12 Ω: 1.835/16.138 = 0.1137 A, so V_T (above Rt) = 0.1137 × 12 = 1.3645 V
- Current through the 10 Ω: 1.835/12.069 = 0.1520 A, so V_Bm = 0.1520 × 10 = 1.5205 V

#### Step 4: Current in the 20 Ω
I₂₀ = (V_T − V_Bm)/20 = (1.3645 − 1.5205)/20 = −0.0078 A

**Answer: Current in the 20 Ω ≈ 7.8 mA, flowing from the bottom node Bm to the top node T**

## Check yourself

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

1. Which delta is converted to star?
   A. Lf-T-Rt
   B. T-Rt-Bm
   C. All of them together
   D. Lf-T-Bm (6, 20, 3 Ω)
   Answer: D. The delta containing the 20 Ω gives a series-parallel network. (Key point: Delta chosen)

2. The star leg at Lf is
   A. 6 + 3 = 9 Ω
   B. 20 x 3/29 = 2.07 Ω
   C. 6 x 3/29 = 0.621 Ω
   D. 29/9 Ω
   Answer: C. Product of the two adjacent delta arms over the sum. (Key point: Star leg)

3. After solving the star circuit, we go back to the original bridge because
   A. the star circuit is wrong
   B. the 20 Ω no longer exists in the star circuit
   C. to find the supply voltage
   D. to change the source
   Answer: B. The 20 Ω current needs node voltages of the original bridge. (Key point: Return to bridge)

4. The current in the 20 Ω is
   A. about 7.8 mA from Bm to T
   B. about 0.27 A from T to Bm
   C. zero
   D. 1.8 A
   Answer: A. (1.3645 - 1.5205)/20 = -0.0078 A. (Key point: Result)

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