View the step-by-step solution to:

The 3-wire d. system supplies a load of 4 resistance across +ve wire and the neutral wire and a load of 6 resistance across ve outer and the neutral...

The 3-wire d.c. system supplies a load of 4 Ω resistance across +ve wire and
the neutral wire and a load of 6 Ω resistance across −ve outer and the neutral at the far end of the
distributor. The resistance of each conductor is 0.15 Ω and voltage across each outer and neutral is
240 V at the load end. Determine the load current and load voltages when there is a break in the
(i) neutral wire (ii) positive outer (iii) negative outer. Assume that the load resistances and the
feeding end voltages remain the same.

Top Answer

Current in the positive outer, I1 = 240/4 = 60 A Current in the negative outer, I2 = 240/6 = 40 A Current in the neutral wire... View the full answer

1 comment
  • Current in the positive outer, I1 = 240/4 = 60 A Current in the negative outer, I2 = 240/6 = 40 A Current in the neutral wire = I1 − I2 = 60 − 40 = 20 A Voltage between +ve outer and neutral at feeding point is V1 = VEL + I1RAE + (I1 − I2) RNL = 240 + 60 × 0·15 + 20 × 0·15 = 252 V Voltage between −ve outer and neutral at feeding point is V2 = VLC − (I1 − I2) RNL + I2 RBC = 240 − 20 × 0·15 + 40 × 0·15 = 243 V (i) When neutral breaks. When there is a break in the neutral, the system is equivalent to 2- wire d.c. system having load resistance = 4 + 6 = 10 Ω and p.d. = 252 + 243 = 495 V at the feeding end. If I is the load current, then, Total circuit resistance = 10 + 0·15 + 0·15 = 10·3 Ω ∴ Load current, I = 495/10·3 = 48·06 A Voltage across 4 Ω resistance = I × 4 = 48·06 × 4 = 192·24 V Voltage across 6 Ω resistance = I × 6 = 48·06 × 6 = 288·36 V (ii) When +ve outer breaks. When there is a break in the +ve outer, there will be no current in 4 Ω load. The circuit is again 2-wire d.c. system but now load is 6 Ω and p.d. at the feeding point is 243 V. Total circuit resistance = 6 + 0·15 + 0·15 = 6·3 Ω If I′ is the load current, then, I′ = 243/6·3 = 38·57 A Voltage across 6 Ω = I′ × 6 = 38·57 × 6 = 231·42 V (iii) When −ve outer breaks. When there is a break in the negative outer, there will be no current in 6 Ω load. The circuit is again 2-wire d.c. system but now load is 4 Ω and p.d. at the feeding point is 252 V. Total circuit resistance = 4 + 0·15 + 0·15 = 4·3 Ω If I ″ is the load current, then, I″ = 252/4·3 = 58·6 A Voltage across 4 Ω = I″ × 4 = 58·6 × 4 = 234·42 V
    • ayura190012
    • Jul 19, 2016 at 9:21pm

Sign up to view the full answer

Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

-

Educational Resources
  • -

    Study Documents

    Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

    Browse Documents
  • -

    Question & Answers

    Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed!

    Ask a Question
Ask a homework question - tutors are online