Sum of branch currents more than total current?

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SUMMARY

The discussion centers on the calculation of branch currents in a parallel circuit with impedances of 7-2jΩ and 4+2jΩ, supplied by a 97.5A current source. Using the current-divider rule, the calculated branch currents are I1=39.54A and I2=64.53A, which sum to 104.07A, exceeding the supplied current. The discrepancy arises from the misunderstanding that the magnitudes of the complex currents do not equate to the total supplied current, emphasizing the importance of considering complex numbers in circuit analysis.

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Homework Statement


An impedance of 7-2jΩ and an impedance of 4+2jΩ and a current source of 97.5A are all connected in parallel. Find the individual branch currents.

Homework Equations



Current-divider rule:

I1=I*Z1/(Z1+Z2) (only have to consider the magnitudes of the impedances, not the angle)

The Attempt at a Solution



I've obtained a value of I1=39.54A and I2=64.53A. However, these currents sum up to 104.07A, whereas the current source is supplying a current of only 97.5A. Why is there a discrepancy?
 
Last edited:
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For complex current I1 and I2 it's true that I1 + I2 = (97.5 + j*0) A, but it doesn't follow that |I1| + |I2| = |(97.5 + j*0)| A
 
Oh, I see. My bad.

Thank you very much, milesyoung!
 

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