In a three-phase system with a line-to-line voltage of 400 V and a line current of 10 A, what is the total real power in Watt, delivered to the load if the power factor is 0.9?
3600√3 W
The total real power in a three-phase system can be calculated using the following formula:
\(P = \sqrt{3} V_{L} I_{L} \cos(\phi)\)
Where:
- \(P\) is the total real power in Watts
- \(V_{L}\) is the line-to-line voltage (400 V)
- \(I_{L}\) is the line current (10 A)
- \(\cos(\phi)\) is the power factor (0.9)
Substituting the given values into the formula:
\(P = \sqrt{3} \times 400 V \times 10 A \times 0.9\)
\(P = 3600\sqrt{3} W\)
Therefore, the total real power delivered to the load is \(3600\sqrt{3} W\).
Similar Questions
- In a balanced 3 phase circuit, neutral current is always ________.
Important Questions from Three Phase Circuits
The direction of rotation of a three-phase induction motor is determined by its ________.
The real power taken by three-phase load is given by -
A phase sequence indicator rotates clockwise for phase sequence of RYB. If the phase sequence is changed to BRY, it will _______.
If a phase sequence indicator rotates clockwise for a phase sequence of RYB, then what happens when the phase sequence is changed to BRY?
In a star-connected system, the phase angle difference between line and phase voltage is:
