Considering ABC sequence, if Vbn = \(10\angle 40^{\theta}\) then Vcb is:
Take Van as a reference phasor
1
\(10\angle 70^{\theta}\)
2
\(10\sqrt {3}\angle 250^{\theta}\)
3
\(10\sqrt {3}\angle 70^{\theta}\)
4
\(10\angle 290^{\theta}\)
Considering ABC sequence, if Vbn = \(10\angle 40^{\theta}\) then Vcb is:
Take Van as a reference phasor