AC Circuit Qs: Power & Current w/ 200V RMS

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Discussion Overview

The discussion revolves around calculating the rms current drawn from an AC source with given active and reactive power values. Participants explore different methods and approaches to arrive at the correct current value, while addressing discrepancies in their calculations.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents a scenario with a 200 V rms source supplying 600 W of active power and 800 VAR of reactive power, noting a discrepancy in their calculated rms current.
  • Another participant suggests using resistive and reactive components in series to verify the calculations, providing their own calculations that yield a different current value.
  • A formula involving the square root of 3 is introduced by a participant, which raises questions about its applicability in the context of single-phase versus three-phase systems.
  • A later reply acknowledges the mistake regarding the use of the square root of 3, suggesting it should be disregarded in this case.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct calculation method or the resulting current value, with multiple competing views and methods presented throughout the discussion.

Contextual Notes

There are unresolved assumptions regarding the application of formulas and the context of the power calculations, particularly concerning the phase of the system being analyzed.

dhruv.tara
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Now this is not a homework question, just a practise question and my answer doesn't match.

An AC source of 200 V rms supplies active power of 600 W and reactive power of 800 VAR. The rms current drawn from the source is ? (the answer is 3.75 while mine is 5)

I did was squared and added the powers to get V^2*I^2, plugged in the value of V and got I as 5... Any help?
 
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I like your answer.

Try it with 24 ohms resistive and 32 ohms reactive in series.
I squared R:
5 * 5 * 24 = 600 watts
I squared X
5 * 5 * 32 = 800 VAR

resultant of r + x:
SQ rt of (24 * 24 + 32 * 32) = 40 ohms
200 V / 40 ohms = 5 amps
 
Last edited:
p=sqrt(3)*V*I*p.f
 
from where did that sqrt(3) factor come into? I have seen this factor before, but as far as I remember that was in 3 phase systems... And I am not dealing with such problem yet I think...
 
sorry, my mistake, you're right dump the sqrt(3).
 
hmm okiess... thanks all
 

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