Understanding Phasor Representation and Calculating Current in AC Circuits

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The discussion focuses on calculating current in an AC circuit using phasor representation, specifically addressing discrepancies in results. The user initially calculated the total impedance as 210+j232.5 but later corrected it to 310 ohms. A key point is the importance of multiplying by the complex conjugate of the impedance to simplify the expression correctly. The correct final current calculation should yield I=1176-j1302, emphasizing the need to avoid mistakes like taking the square root of the denominator. Proper handling of complex numbers is crucial for accurate results in phasor calculations.
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http://i47.tinypic.com/2mg2iyf.jpg
i ant to find the current in phasor representation.
i made found the Z total 210+j232.5
the current sourse is 2170
I=\frac{2170}{310+j232.5}=1454.56-j1610.4

but the calculation igives another result
why?
 
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The first problem that comes to mind is that the 210 ohms (real part) should actually be 310 ohms.

Also, do you know how to express a complex number in such a way that the imaginary part appears only in the numerator (i.e. there is no j on the bottom)?
 
sorry on paper i wrote 310
and i did multiply the numerator and denominator by the "opposite"
of the compex number that is on the denominator

and i got a totaly different result

who is correct?
 
The solution in the jpg image you posted is correct. You have to multiply the expression in your first post (numerator and denominator) by the complex conjugate of the impedance and then simplify.
 
<br /> I=\frac{2170(310-j232.5)}{387.5}=1176-j1302<br />

i did that as you see
why i still get the wrong result ?
 
It looks like you took the square root of the bottom. You should have zz*=|z|2 there, not |z|.
 
thanks:)
 

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