Star–Delta Transformation

Tip

Exam Focus

Directly asked in NTC 2082 [4]. Memorise both directions and the balanced shortcut. $\Delta\!\to\!Y$: “product of the two adjacent delta arms over the sum of all three”. $Y\!\to\!\Delta$: “sum of star pair-products over the opposite star arm”. Balanced: $R_\Delta=3R_Y$.

Three resistances can meet at a common node as a star (Y) or form a closed loop as a delta ($\Delta$). Conversion keeps the resistance seen between every pair of terminals unchanged, letting us reduce bridge networks that are neither series nor parallel.

Equivalent star and delta three-terminal networks.

Caution

Key Formula — Delta $\to$ Star

Let $S_\Delta=R_{12}+R_{23}+R_{31}$. Each star arm = product of the two adjacent delta arms $\div\,S_\Delta$:

$$R_1=\frac{R_{12}R_{31}}{S_\Delta},\quad R_2=\frac{R_{12}R_{23}}{S_\Delta},\quad R_3=\frac{R_{23}R_{31}}{S_\Delta}.$$

[!IMPORTANT]

Key Formula — Star $\to$ Delta

Let $S_Y=R_1R_2+R_2R_3+R_3R_1$. Each delta arm = $S_Y\,\div$ the opposite star arm:

$$R_{12}=\frac{S_Y}{R_3},\quad R_{23}=\frac{S_Y}{R_1},\quad R_{31}=\frac{S_Y}{R_2}.$$

[!NOTE]

Key Point

For a balanced network (all arms equal): $\;R_\Delta=3R_Y\;$ or $\;R_Y=R_\Delta/3$.

[!NOTE]

Example — Delta to star

A balanced delta with $R_{12}=R_{23}=R_{31}=9\,\Omega$ converts to a star with $R_Y=R_\Delta/3=3\,\Omega$ per arm. Check with the general formula: $S_\Delta=27\,\Omega$, $R_1=\dfrac{9\times9}{27}=3\,\Omega$.