Engineering Solve AC Circuit Analysis: Calculate Current, Real & Reactive Power

AI Thread Summary
The discussion focuses on calculating the current, real power, and reactive power in an AC circuit with a given impedance and voltage levels. The calculated current in the line is 763.45 ∠ -8.38°, derived from the voltage difference and impedance. The real power consumed is calculated as 291.5 kW, while the reactive power is significantly higher at 700 kVAr. Concerns are raised about the reactive power being larger than the real power, which is noted as unusual. Overall, the calculations appear correct, and the results are confirmed by other participants.
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Homework Statement


Two nodes in a network are connected by a line with an impedance of 0.5 + 1.2j Ω. The voltage at the sending end is 33KV and at the receiving end is 32.5 KV with a lag of 1.5° compared to the sending end..

Calculate the current in the line and the real and reactive power consumed by the line itself.


Homework Equations


Standard AC circuit analysis equations.


The Attempt at a Solution



I think I've got this just want to make sure that I'm not doing anything silly with the angles etc:

ISR = VS - VR / 0.5 + j1.2 = (33) - (32.5 ∠ -1.5°) / 0.5 + j1.2

= 763.45 ∠ -8.38°

Real Power consumed = I2*Re(Z) = (763.45)^2 * 1/2 = 291.5 KW
Reactive = I2Im(Z) = (763.45)^2 * 1.2 = 700KVAr

Have I done anything silly? I feel to ask because it seems unusual to me that the reactive power is larger than the real power consumed by the line.

Many thanks !
 
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Looks okay.
 
Great, thanks a lot !
 

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