Engineering Finding the line current in an ac circuit

AI Thread Summary
The load impedance is calculated as 18 + j15 ohms, resulting in a magnitude of 23.43 ohms at an angle of 39.8 degrees. The line voltage Vab is determined to be 173.21 volts at an angle of 30 degrees. Using these values, the line current Iab is found to be 7.39 amps at an angle of -9.8 degrees. Finally, the phase current Ia is calculated to be 12.80 amps at an angle of -39.8 degrees, indicating that the current leads the voltage by 39.8 degrees. The final result confirms the line current in the AC circuit.
james99
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Homework Statement


Given the load impedence is 18 + j15 ohms, what is the line current. Follow this link to see the circuit and the variables. http://www.flickr.com/photos/61312863@N07/5584165017/


Homework Equations


Vab = Van*root[3] angle 30 degrees
Iab = Vab/Z
Ia = Iab*root[3] angle -30 degrees


The Attempt at a Solution


Load impedance = 18 + j15 = 23.43 angle 39.8

Vab = root[3]*100 angle 30 degrees = 173.21 angle 30 degrees

Iab = 173.21<30/23.43<39.8 = 7.39<-9.8

Ia = 7.39<-9.8 * root [3]<30 = 12.80<-39.8
 
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Therefore, the line current in the circuit is 12.80 amps, with a phase angle of -39.8 degrees. This means that the current is leading the voltage by 39.8 degrees.
 

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