Question

A 3-phase line having negligible resistance and 22 ohm inductive reactance per phase operates with 110 kV sending-end voltage and 100 kV receiving-end voltage. The maximum power that this line can transmit is

Answer 1

Answer:

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Answer 2

Answer:

The maximum power that the 3-phase line can transmit is 500MW.

Explanation:

Given:

Reactance ($X$) = 22Ω, sending-end voltage ($V_{1}$) = 110kV, receiving-end voltage ($V_{2}$) = 100kV

To find:

Maximum power transmitted ($P_{max}$)

Formula:

$P_{max} = \frac{V_{1}V_{2} }{X}$

Solution:

Substituting the given values in the above formula, we get

$P_{max} = \frac{V_{1}V_{2} }{X}$

$P_{max}= \frac{100*10^{3} *110*10^{3} }{22}$

$P_{max}= \frac{11000*10^{6} }{22}$

$P_{max}=500*10^{6}$

$P_{max}=500MW$

Therefore, the maximum power transmitted is 500 M watts.

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