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The total reactance and total susceptance of a lossless overhead EHV line, operating at 50 Hz, are given by 0.045 pu and 1.2 pu respectively. If the velocity of wave propagation is 3 × 105 km/s, then approximate length of the line is
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- 122 km
- 185 km
- 222 km
- 272 km
Correct Option: B
Assume that ZB is the base impedance,
then, X(in Ω) 0.045 ZB
and, Y(inΩ) = 1.2/ZB
If l is length of line, then, = ωLl and Y = ωCl
| or L = | = | ..............(A) | ||
| ωl | ωl |
| and C = | = | ..............(B) | ||
| ωl | ωlZB |
Now, velocity of propagation
| = | |
| √LC |
| ⇒ vc = | = | ||
| √(0.045ZB/ωl)(1.2/ωlZB) | √0.045 × 1.2 |
| ⇒ 3 × 108 = | |
| √0.045 × 1.2 |
⇒ l ≌ 222 × 10³m = 222 km.
