Question

(a) With the aid of delta and star connection diagrams, state the basic equation from which the delta-star and star-delta conversion equation can be derived. (6)

(b) Three batteries A, B,and C have their negative terminals connected together, between the positive terminals of A and B there is a resistor of 0.5 ohm and between B and C there is a resistor of 0.3 ohm (10)

Battery A 105 V, Internal resistance 0.25 ohm

Battery B 100 V, Internal resistance 0.2 ohm

Battery C 95 V, Internal resistance 0.25 ohm

Determine the current values in the two resistors and the power dissipated by them. (10)

Answer 1

Part (a)

Delta-star and star-delta conversion equations:

The basic equation is that the resistance between any two terminals must be the same in both the star and delta networks.
For a delta network with resistances R12 (between terminals 1-2), R23 (2-3), R31 (3-1), and a star network with resistances R1, R2, R3 (each connected to a terminal and a common centre point):
Resistance between terminals 1 and 2: in delta it is R12 in parallel with (R23 + R31); in star it is R1 + R2. Equating:

R1 + R2 = R12 (R23 + R31) / (R12 + R23 + R31)

R2 + R3 = R23 (R31 + R12) / (R12 + R23 + R31)

R3 + R1 = R31 (R12 + R23) / (R12 + R23 + R31)

Solving these gives the delta-to-star conversion:

R1 = R12 R31 / (R12 + R23 + R31)

R2 = R12 R23 / (R12 + R23 + R31)

R3 = R23 R31 / (R12 + R23 + R31)

And the star-to-delta conversion:

R12 = (R1 R2 + R2 R3 + R3 R1) / R3

R23 = (R1 R2 + R2 R3 + R3 R1) / R1

R31 = (R1 R2 + R2 R3 + R3 R1) / R2

For equal resistances: delta R = 3 x star r (R = 3r), and star r = R/3.

Part (b)

Three batteries A, B, C with negative terminals common. Resistor 0.5 ohm between A and B, 0.3 ohm between B and C.

Battery A: 105 V, internal 0.25 ohm. Battery B: 100 V, internal 0.2 ohm. Battery C: 95 V, internal 0.25 ohm.
Let the node voltages at the positive terminals be Va, Vb, Vc (common negative = 0).
Current from A into the 0.5 ohm resistor: (105 - Va)/0.25 = (Va - Vb)/0.5.

105 - Va = 0.5(Va - Vb) -> 1.5 Va - 0.5 Vb = 105. (1)

Current from C into the 0.3 ohm resistor: (95 - Vc)/0.25 = (Vb - Vc)/0.3.

95 - Vc = 0.8333(Vb - Vc) -> 5 Vb + Vc = 570. (2)

At node B: (Va - Vb)/0.5 + (100 - Vb)/0.2 = (Vb - Vc)/0.3.

2(Va - Vb) + 5(100 - Vb) = 3.333(Vb - Vc)

2 Va - 10.333 Vb + 3.333 Vc = -500. (3)

From (1): Va = 70 + 0.3333 Vb. From (2): Vc = 570 - 5 Vb.
Substitute into (3): 2(70 + 0.3333 Vb) - 10.333 Vb + 3.333(570 - 5 Vb) = -500

140 + 0.6667 Vb - 10.333 Vb + 1900 - 16.667 Vb = -500

2040 - 26.333 Vb = -500 -> Vb = 2540/26.333 = 96.46 V.

Va = 70 + 0.3333 x 96.46 = 70 + 32.15 = 102.15 V.
Vc = 570 - 5 x 96.46 = 570 - 482.3 = 87.7 V.
Current in 0.5 ohm resistor: I_AB = (Va - Vb)/0.5 = (102.15 - 96.46)/0.5 = 5.69/0.5 = 11.38 A (from A to B).
Current in 0.3 ohm resistor: I_BC = (Vb - Vc)/0.3 = (96.46 - 87.7)/0.3 = 8.76/0.3 = 29.2 A (from B to C).
Power in 0.5 ohm: P = I^2 R = 11.38^2 x 0.5 = 129.5 x 0.5 = 64.8 W.
Power in 0.3 ohm: P = 29.2^2 x 0.3 = 852.6 x 0.3 = 255.8 W.

So the 0.5 ohm resistor carries 11.4 A (64.8 W) and the 0.3 ohm resistor carries 29.2 A (255.8 W).

✓ Verified Model Answer (Text Solution)

Delta-star and star-delta conversion equations:

Three batteries A, B, C with negative terminals common. Resistor 0.5 ohm between A and B, 0.3 ohm between B and C.