Let a star three-phase balanced circuit with line voltage of 230 V, and load \( Z = 16 + 12j \) Calculate the line currents:
so we have this circuit :
Considering \( V_a \) to be the origin of phase-voltage so \( V_{ab} \)
We have : \( V_{ab} = V_a \sqrt{3} (30°) \\
V_{bc} = V_b \sqrt{3} (-90°) \\
V_{ca} = V_c \sqrt{3} (150°)\)
and we have \( I_a = \frac{V_a}{|Z|} (-arg(Z)) \\
I_b = \frac{V_b}{|Z|} (-arg(Z)) \\
I_c = \frac{V_c}{|Z|} (-arg(Z)) \)
Thus:
\( I_a = \frac{V_{ab}}{\sqrt{3}|Z|} (-30-arg(Z)) \\
I_b = \frac{V_{bc}}{|Z|} (90-arg(Z)) \\
I_c = \frac{V_{ca}}{|Z|} (-150-arg(Z)) \)
with \( V_{ab} = 230 V = V_{bc} = V_{ca} \)

 

Let a star three-phase balanced circuit with line voltage of 230 V, and load \( Z = 16 + 12j \) Calculate the line currents:
so we have this circuit : View attachment 309940
Considering \( V_a \) to be the origin of phase-voltage so \( V_{ab} \)
We have : \( V_{ab} = V_a \sqrt{3} (30°) \\
V_{bc} = V_b \sqrt{3} (-90°) \\
V_{ca} = V_c \sqrt{3} (150°)\)
and we have \( I_a = \frac{V_a}{|Z|} (-arg(Z)) \\
I_b = \frac{V_b}{|Z|} (-arg(Z)) \\
I_c = \frac{V_c}{|Z|} (-arg(Z)) \)
Thus:
\( I_a = \frac{V_{ab}}{\sqrt{3}|Z|} (-30-arg(Z)) \\
I_b = \frac{V_{bc}}{|Z|} (90-arg(Z)) \\
I_c = \frac{V_{ca}}{|Z|} (-150-arg(Z)) \)
with \( V_{ab} = 230 V = V_{bc} = V_{ca} \)

Hi,

What is it you are having a problem with; do you not know how to calculate the numerical values of the currents from the given information?

 

Hi,

What is it you are having a problem with; do you not know how to calculate the numerical values of the currents from the given information?

no i just someone to verify my solution thats it, because i had been told to post my solution and eventually someone will answer if its valid or not, infinite thanks !

 

no i just someone to verify my solution thats it, because i had been told to post my solution and eventually someone will answer if its valid or not, infinite thanks !

Hi,

Oh ok, so you mean the other text there was your own text as those are your results. It looks almost correct so far.

Then I have to ask, why is your solution for 'Ia' so different from the other two 'Ib' and 'Ic'?
Why does 'Ia' have a sqrt(3) in it when the others do not, or stated another way, why do the other two have no sqrt(3) while the first 'Ia' does?

 

Hi,

Oh ok, so you mean the other text there was your own text as those are your results. It looks almost correct so far.

Then I have to ask, why is your solution for 'Ia' so different from the other two 'Ib' and 'Ic'?
Why does 'Ia' have a sqrt(3) in it when the others do not, or stated another way, why do the other two have no sqrt(3) while the first 'Ia' does?

ah yes my bad they all need to have that sqrt(3) in the denominator coming from the line voltage right?

 

ah yes my bad they all need to have that sqrt(3) in the denominator coming from the line voltage right?

Yes going from line-to-line voltage to line voltage you have to divide by sqrt(3), and that goes for any of the phases.
You might take solace in the fact that a balanced three phase system is the simplest and easiest to analyze.