Both **delta** and **star** (Y) **circuits** are circuits with three resistors connected.

Using the **Δ-Y (delta-star) conversion** formula, you can calculate the combination of resistor values that make these circuits equivalent.

Read this article to understand **the principle of the Delta-Y conversion formula and its proof**.

## Δ-Y (delta-star) conversion

### Δ circuit & Y circuit

Circuits such as those in Fig. 1 are called **Δ (delta) circuits** and those in Fig. 2 are called **Y (star) circuits**.

The two can be rewritten as mutually equivalent circuits.

### Conversion from Δ to Y

The Δ circuit can be rewritten as a Y circuit by setting \(r_a,r_b,r_c\) as follows.

$$r_a=\frac{R_b R_c}{R_a+R_b+R_c}$$

$$r_b=\frac{R_c R_a}{R_a+R_b+R_c}$$

$$r_c=\frac{R_a R_b}{R_a+R_b+R_c}$$

In other words, the sum of the resistances of the **Δ circuits** is the denominator and **the product of the two resistances of the Δ circuits adjacent to the resistance of the Y circuit** is the numerator.

### Conversion from Y to Δ

The Y circuit can be rewritten as a Δ circuit by setting \(R_a,R_b,r_R\) as follows.

$$R_a=\frac{r_a r_b+r_b r_c+r_c r_a}{r_a}$$

$$R_b=\frac{r_a r_b+r_b r_c+r_c r_a}{r_b}$$

$$R_c=\frac{r_a r_b+r_b r_c+r_c r_a}{r_c}$$

That is, **the resistance of the Y circuit on the opposite side of the resistance of the Δ circuit** is the denominator and **the sum of the products of the adjacent resistances of the Y circuit** is the numerator.

### Proof

If the Δ and Y circuits are equivalent, they are also equivalent if parts of them are extracted. If only the A-B connection is extracted as in Fig. 3, the corresponding part of the Δ-circuit is shown in Fig. 4 below (no special attention is paid to terminal C).

Let the composite resistance of A-B in the Y circuit be \(r_a+r_b\) and the composite resistance in the Δ circuit be \(R^D_{AB}\),

$$\frac{1}{R^D_{AB}}=\frac{1}{R_c}+\frac{1}{R_a+R_b}=\frac{R_a+R_b+R_c}{R_c(R_a+R_b)}$$

then

$$R^D_{AB}=\frac{R_c(R_a+R_b)}{R_a+R_b+R_c}$$

Since these are equivalent,

$$r_a+r_b=\frac{R_c(R_a+R_b)}{R_a+R_b+R_c}\tag{1.1}$$

holds, and similarly between B-C and C-A,

$$r_b+r_c=\frac{R_a(R_b+R_c)}{R_a+R_b+R_c}\tag{1.2}$$

$$r_c+r_a=\frac{R_b(R_c+R_a)}{R_a+R_b+R_c}\tag{1.3}$$

are obtained.

If you add up these sides, then

$$2(r_a+r_b+r_c)=\frac{2(R_aR_b+R_bR_c+R_cR_a)}{R_a+R_b+R_c}$$

$$r_a+r_b+r_c=\frac{R_aR_b+R_bR_c+R_cR_a}{R_a+R_b+R_c}$$

subtracting the sides of equation \((1.1), (1.2), (1.3)\) respectively, from which

$$r_a=\frac{R_b R_c}{R_a+R_b+R_c}\tag{2.1}$$

$$r_b=\frac{R_c R_a}{R_a+R_b+R_c}\tag{2.2}$$

$$r_c=\frac{R_a R_b}{R_a+R_b+R_c}\tag{2.3}$$

is obtained.

Also, multiplying the sides of the expression \((1.3), (1.1)\) gives

$$(r_c+r_a)(r_a+r_b)=\frac{R_b(R_c+R_a)R_c(R_a+R_b)}{(R_a+R_b+R_c)^2}$$

$$r_ar_b+r_br_c+r_cr_a+r^2_a=\frac{R_bR_c(R_c+R_a)(R_a+R_b)}{(R_a+R_b+R_c)^2}$$

and dividing the sides by the expression \((2.1)\) gives

$$\frac{r_ar_b+r_br_c+r_cr_a}{r_a}+r_a=\frac{(R_c+R_a)(R_a+R_b)}{R_a+R_b+R_c}$$

$$\frac{r_ar_b+r_br_c+r_cr_a}{r_a}+r_a=\frac{R_a(R_a+R_b+R_c)+R_bR_c}{R_a+R_b+R_c}$$

$$\frac{r_ar_b+r_br_c+r_cr_a}{r_a}+r_a=R_a+\frac{R_bR_c}{R_a+R_b+R_c}$$

Furthermore, subtracting the sides of equation \((2.1)\), we get

$$\frac{r_ar_b+r_br_c+r_cr_a}{r_a}=R_a$$

Similar calculations for the pairs of equations \((1.1)-(1.2)-(2.2), (1.2)-(1.3)-(2.3)\) lead to

$$\frac{r_ar_b+r_br_c+r_cr_a}{r_b}=R_b$$

$$\frac{r_ar_b+r_br_c+r_cr_a}{r_c}=R_c$$

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