Assume 3 numbers of pure resistive loads each of value R.
When these loads are connected in delta the power in each resistor is V^2/R and the total power is 3×V^2/R.
When these loads are connected in star the voltage applied across each = V/(sqrt of 3). Power in each load becomes = V^2/(3×R) . and the total load is =(3×V^2)/(3×R)=V^2/R. You can see this is 3 times less than the total power when the loads are connected in delta.
One important point is that the loads should withstand higher voltage (line voltage) and higher current when connected in delta than when they are connected in star
so power in delta is 3 time power correct answer is 3P
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Assume 3 numbers of pure resistive loads each of value R.
When these loads are connected in delta the power in each resistor is V^2/R and the total power is 3×V^2/R.
When these loads are connected in star the voltage applied across each = V/(sqrt of 3). Power in each load becomes = V^2/(3×R) . and the total load is =(3×V^2)/(3×R)=V^2/R. You can see this is 3 times less than the total power when the loads are connected in delta.
One important point is that the loads should withstand higher voltage (line voltage) and higher current when connected in delta than when they are connected in star
so power in delta is 3 time power correct answer is 3P
welldone.. Very good question
Assume 3 numbers of pure resistive loads each of value R.
When these loads are connected in delta the power in each resistor is V^2/R and the total power is 3×V^2/R.
When these loads are connected in star the voltage applied across each = V/(sqrt of 3). Power in each load becomes = V^2/(3×R) . and the total load is =(3×V^2)/(3×R)=V^2/R. You can see this is 3 times less than the total power when the loads are connected in delta.
One important point is that the loads should withstand higher voltage (line voltage) and higher current when connected in delta than when they are connected in star
so power in delta is 3 time power correct answer is 3P
Thanks for explaining it. answer updated