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A 3-phase star connected balanced load is supplied from a 3 phase, 418 V supply. The line current is 16 A and the power taken by the load is 11000 W. Find (i) impedance in each branch (ii) power factor and (iii) power consumed if the same load is connected in delta. Impedance Power factor Power when connected in delta

User Fedtuck
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To find the impedance in each branch, power factor, and power consumed when the load is connected in delta, we can use the following formulas and equations:

(i) Impedance in each branch:

The impedance in each branch can be calculated using the formula:

Z = V_line / I_line

Where:

V_line = Line voltage = 418 V

I_line = Line current = 16 A

Impedance in each branch = Z = 418 V / 16 A = 26.125 Ω (approximately)

(ii) Power factor:

The power factor can be calculated using the formula:

Power Factor (PF) = P / (V_line * I_line * sqrt(3))

Where:

P = Power taken by the load = 11000 W

V_line = Line voltage = 418 V

I_line = Line current = 16 A

Power Factor (PF) = 11000 W / (418 V * 16 A * sqrt(3))

PF = 0.468 (approximately)

(iii) Power consumed when connected in delta:

When the load is connected in delta, the line current will be the same as the phase current. Therefore, we can use the same power taken by the load, which is 11000 W.

So, the power consumed when the load is connected in delta = 11000 W

To summarize:

(i) Impedance in each branch = 26.125 Ω (approximately)

(ii) Power factor = 0.468 (approximately)

(iii) Power consumed when connected in delta = 11000 W

User Veger
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