Current in 20Ω using star-delta transformation

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