Question

A 2-wire ring distributor ABC is supplied at A at 400 V. Point loads of 20 A at a p.f. of 0.8 lagging and 30 A at a p.f. 0.6 lagging are tapped off at B and C respectively. Both the power factors refer to the voltage at A. The respective go-and-return impedances of sections AB,BC and CA are (1 + j2) ohm, (2 + j3) ohm and (1 + j3) ohm. Calculate the current flowing through each section and the potentials at B and C. Use Superposition theorem.

Answer 1

The distributor circuit is shown in Fig.(a). Currents in various sections are as shown. First, consider the load at point B acting alone as in Fig.(b). The input current at point A divides in the inverse ratio of the impedances of the two paths A B and ACB.

Let the currents be I₁′ and I₂′

Now, consider the load at point C to act alone as shown in Fig. (c). Let the currents now be I₁″ and I₂″.

As per Superposition theorem, the current I₁ in section A B is the vector sum of I₁′ and I₁″

Potential of B = 400 – drop over A B

= 400 − (19.5 − j16.5)(1 + j2) = 347.5 − j22.5 = 348 ∠ −3.7°V

Potential of C = 400 − (14.5 − j19.5)(1 + j3) = 327 − j24 = 328 ∠−4.2°V