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A distribution system shown in the figure below.
The point of minimum potential along the tract from A will be _______ m
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- 3.67 m
- 367 m
- 36.7 m
- 0.367 m
Correct Option: A
Potential of P = 575 – IA 0.04 x
= 590 – (600 – IA) 0.04 (8 – x)
| Simplifying, we have x = | (177 - 0.32IA) | |
| 24 |
| and IA = | ||
| 0.32 |
then, VP = 575 - IA 0.04x
| = 575 – | 0.04x | |
| 0.32 |
= 575 – 22.01 x – 3x²
| ∴ | = - 22.01 - 6x - 3x² = 0 | |
| dx |
or x = 3.67 m
