When networks having a large number of branches are to be solved by the use of Kirchhoff s law, a: great difficulty is experienced in solving several simultaneous equations. Such complicated networks, however, can be simplified by successively replacing delta meshes by equivalent star systems and vice versa.

Consider the two circuits shown in· the Figs.- 102 and 103. They will be equivalent if the resistance measured between any two of the terminals 1, 2 and 3 is the same in the two cases.

R_{1} + R_{2} = R_{12} (R_{23} + R_{31}) / R_{12} + R_{23} + R_{31}

Similarly R_{2} + R_{3} = R_{23} (R_{31} + R_{12}) / R_{12} + R_{23} + R_{31}

R_{3} + R_{1} = R_{31} (R_{12} + R_{23} ) / R_{12} + R_{23} + R_{31 }

Solving (24), (2b) and (26) simultaneously, we get

R_{1} = R_{12} R_{31} / R_{12} + R_{23} + R_{31}

R_{2} = R_{23} R_{12} / R_{12} + R_{23} + R_{31}

R_{3} = R_{31} R_{23} / R_{12} + R_{23} + R_{31}

From above it may be noted that resistance of each arm of the star is given by the product of the resistances of the two delta sides that meet at its end divided by the sum of the three delta resistances.

From eqns. (24) to (26), eqns. for star to delta conversion can also be obtained. These are as follows:

R_{12} = R1 R_{2} + R_{2} R_{3} + R_{3} R1 / R_{3}

R_{23} = R1 R_{2} + R_{2} R_{3} + R_{3} R1 / R_{1}

R_{31} = R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1} / R_{2}

in electronics, star and delta circuits are generally referred to as ‘t and 1t circuits respectively.

**Example 24.** Fig. 104 shows a number of resistances connected in delta and star. Using star/delta conversion method complete the network resistance measured between (i) Land M (ii) M and N and (iii) N and L.

**Solution.** Three resistances 12 n, 6 Q and Q are star connected. Transform them into delta with ends at the same points as before.

Refer Fig. 105. R_{12 }= R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1} / R_{3 } = 12 x 6 + 6 x 8 + 8 x 12 / 8 = 27 Q

R_{23} = R_{1} R_{2} + R_{2}R_{3} + R_{3}R_{1} / R_{1} = 12 x 6 + 6 x 8 x 12 / 12 = 18 Q

R_{31} = 12 x 6 + 6 x 8 x 12 / 6 = 36 Q

Similarly

Fig. 106 shows this transformed circuit connected to original delta connected resistances in the circuit 18 n, 3 Q and 2 Q.

Here 18 Q and 36 Q are in parallel;

3 Q and 27 Q are in parallel, and

2 Q and 18 Q are in parallel.

These resistances are equivalent to:

18 x 36 / 18 + 36 = 12 Q ; 3 x 27 / 3 + 27 = 2.7 Q and 2 x 18 / 2 + 18 = 1.8 Q

This is shown in Fig. 107.

(i) Resistance between L and M,

R_{LM} = 12 x (2.7 + 1.8) / 12 + (2.7 + 1.8) = 3.27 Q

(ii) Resistance between M and N,

R_{MN} = 1.8 x (12 + 1.8) / 1.8 + 12 + 2.7 = 1.6 Q

(iii) Resistance between N and L,

R_{NL} = 2.7 x (12 + 1.8) / 2.7 + (12 + 1.8) = 2.25

**Example 25.** In the circuit shown in Fig. 108, find the resistance between M and N.

**Solution.** Connecting the 1 2 3 delta [Fig. 109 (i)] to equivalent star [Fig. 109 (ii)]

R_{1} = R_{12} R_{31} / R_{12} + R_{23} + R_{31} = 5 x 3 / 5 + 2 + 3 = 1.5

R_{2} = R_{23} R_{12} / R_{12} + R_{23} + R_{31} = 2 x 5 / 5 + 2 + 3 = 1

R_{3} = R_{31} R_{23} / R_{12} + R_{23} + R_{31} = 3 x 2 / 5 + 2 + 3 = 0.6

Thus the original circuit reduces to that shown in Fig. 110 which further reduces to the circuit shown in Fig. 111.

Now, the inner star circuit of Fig. 110 shown as Fig. 111 is equivalent to the delta circuit

shown in Fig. 112 (ii) as appears from calculations given below :

R_{12 }= R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1} / R_{3 } = 8 x 10 + 10 x 10 + 10 x 8 / 10 = 26 Q

R_{23} = R_{1} R_{2} + R_{2}R_{3} + R_{3}R_{1} / R_{1} = 8 x 10 + 10 x 10 x 10/ 8 = 32.5 Q

R_{31} = 8 x 10 + 10 x 10 x 10 x 8 / 10 = 26 Q

The given circuit thus ultimately reduces to the circuit shown in Fig. 113 which in turn is equivalent to the circuit given in Fig. 114.

In Fig. 113, the observe that :

30 Q and 26 Q are in parallel and are equivalent to :

30 x 26 / 30 + 26 = 13.9

30 and 26 n are in parallel and are equivalent to : 13.9 n (as above)

32.5 Q and 30 Q are in parallel and are equivalent to :

32.5 x 30 / 32.5 + 30 = 15.6

Hence total resistance between M and N,

R_{MN} = 15.6 x (13.9 + 13.9) / 15.6 + (13.9 + 13.9)

= 433.69 / 43.4 = 9.99

**Example 26.** Find the current I supplied by the battery for Fig. 115, using delta/star transformation.

**Solution.** Delta connected resistances 25 n, 10 n and 15 Q are transformed to equivalent star as given below : (See Fig. 116)

R_{1} = R_{12} R_{31} / R_{12} + R_{23} + R_{31} = 10 x 25 / 10 + 15 + 25 = 5

R_{2} = R_{23} R_{12} / R_{12} + R_{23} + R_{31} = 15 x 10 / 10 + 15 + 25 = 3

R_{3} = R_{31} R_{23} / R_{12} + R_{23} + R_{31} = 25 x 15 / 10 + 15 + 25 = 7.5

The given circuit thus reduces to the circuit shown in Fig. 117.

The equivalent resistance of

(20 + 5) Q || (10 + 7.5) Q = 25 x 17.5 / 25 + 17.5 = 10.29 Q

Thus the given circuit ultimately reduces to the circuit shown in Fig. 118.

Total resistance = 10.29 + 3 + 2.5 = 15.79 Q

Hence current through the battery,

I = 15 / 15.79 = 0.95 A

**Example 27.** Using Delta-star conversion find resistance between terminals AB in Fig. 124.

**Solution.**

• Applying delta-star conversion to CGD and EGF of Fig. 124, we get Fig. 125 (i).

• Further simplification reduces the circuit to its equivalents as in Fig. 125 (ii) and later as in Fig. 125 (iii ).

• Convert CHJ to equivalent star as in B Fig.125 (iv).

• Using series-parallel combinations, we have

R_{AB} = (10 + 0.66) + [(0.752 + 1.6) || (2.64 + 5.6)]

= 10.66 + 2.352 x 8.24 / (2.352 + 8.24 = 12.49 Q

**Example 28.** Find the resistance at the A-B terminals in the electric circuit of Fig. 126, using Δ-Y transformations.

**Solution.** Converting CEF into star connections, we get (Refer Fig. 127) :

CN = 60 x 40 / 60 + 40 + 100 = 24 Q

NE = 40 x 100 / 60 + 40 + 100 = 20 Q

NF = 60 x 100 / 60 + 40 + 100 = 30 Q

R_{ND} = (100 + 20) x (90 x 30) / 120 + 120 = 60 Q

R_{AB} = 12 + 60 = 72 Q