A balanced Y-Y four-wire system has phase voltages Van=120 0, Vbn=120 -120, Vcn=120 -240V The load impedance per phase is 10+j19 , and the line impedance per phase is 1+j2 . Solve for the line currents and neural current
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To solve for the line currents and neutral current in the given balanced Y-Y four-wire system, we will follow these steps: Show more…
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A balanced Y-Y four-wire system has phase voltages \[ \begin{array}{c} \mathbf{V}_{a n}=120 / 0^{\circ}, \quad \mathbf{V}_{b n}=120 /-120^{\circ} \\ \mathbf{V}_{c n}=120 / 120^{\circ} \mathrm{V} \end{array} \] The load impedance per phase is $19+j 13 \Omega$, and the line impedance per phase is $1+j 2 \Omega$. Solve for the line currents and neutral current
The voltages at the terminals of a balanced load consisting of three 10-ohm resistors connected in Y are: Vab = 100 angle 0 degrees Vbc = 80.8 angle -121.44 degrees Vca = 90 angle 130 degrees Assuming that there is no connection to the neutral of the load, find the line currents to the load. Notes: In the absence of neutral connection, there will be no zero sequence components to solve for. Considering phase displacements. Phase voltage sequence components: Van1 = [Vab1 / (sqrt 3)] angle -30 Van2 = [Vab2 / (sqrt 3)] angle 30 Phase currents to load: Ia = Van / R, and so on.
Sri K.
A four-wire wye-wye circuit has $$\begin{array}{c} \mathbf{V}_{a n}=120 / 120^{\circ}, \quad \mathbf{V}_{b n}=120 / 0^{\circ} \\ \mathbf{V}_{c n}=120 /-120^{\circ} \mathrm{V} \end{array}$$ If the impedances are \[ \begin{array}{c} \mathbf{Z}_{A N}=20 / 60^{\circ}, \quad \mathbf{Z}_{B N}=30 / 0^{\circ} \\ \mathbf{Z}_{c n}=40 / 30^{\circ} \Omega \end{array} \] find the current in the neutral line.
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