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In a three phase system, what is the instantaneos voltage of phase 2 when phase 1 is...
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In a three phase system all three phases are separated by equal angles, the sum of all 3 angles being 360 deg.
`3*theta =360 degree`
it means the angle separation between two consecutive phases is
`theta =360/3 =120 degree`
Therefore the values of the instantaneous voltage for all 3 phases are
`V1(t) = V_max*sin(omega*t +0)) "Volts"`
`V2(t) = V_max*sin(omega*t +120)) "Volts"`
`V3(t) = V_max*sin(omega*t +240)) "Volts"`
To compute the value of V2 we need to know first the values of Vmax and angle `(omega*t)`
From the text we have
`-85 = V_max*sin(omega*t)` (1)
`116 = V_max*sin(omega*t +240) ` (2)
Second equation can be written as
`116 = V_max*sin(omega*t)*cos(240) +V_max*cos(omega*t)*sin(240)` `116 = -85*(-0.5) + V_max*(-0.866)*cos(omega*t)`
`73.5 =-0.866*V_max*cos(omega*t)` (3)
And by dividing equation (1) with equation (3) we get
`1.1547*tan(omega*t) = 1.1565`
or equivalent `(omega*t) =45.044 degree =45 degree`
and from equation (1) we have
`V_max = -85/sin(45) =-120.2 "Volts"`
Now we can compute the value of the instantaneous voltage of the second phase
`V2 = -120.2*sin(45+120) = -31.11 "Volts"`
Posted by valentin68 on September 10, 2013 at 10:41 AM (Answer #1)
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