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Three equal impedances (10 + j10) Ω are connected in delta across a three-phase balanced supply, the angle between the line current IR and line voltage VRY is given by
- A45°
- B75°
- C15°
- D0°
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Impedance = (10 + j10) Ω
In a delta connected balanced supply, the angle between the line current IR and line voltage VRY is given by
? = (30 + θ)
Where θ is the impedance angle.
$(\theta = {\tan ^{ - 1}}\frac{X}{R} = 45^\circ )$
? = 30 + 45 = 75°
In a delta connected balanced supply, the angle between the line current IR and line voltage VRY is given by
? = (30 + θ)
Where θ is the impedance angle.
$(\theta = {\tan ^{ - 1}}\frac{X}{R} = 45^\circ )$
? = 30 + 45 = 75°