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Line 81:
Line 81: − + Line 90:
Line 90: + − 10.3 x 20.45 x 0.03 = 6.32V+ − − − + − 6.8 x 20.45 x 0.03 = 4.17V+ − − −
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T<span lang="EN-US"></span>he voltage drop is then calculated from the table as :
T<span lang="EN-US"></span>he voltage drop is then calculated from the table as :
K x I<sub>B</sub> x L
'''K x I<sub>B</sub> x L'''
* K = Given by the table.
* K = Given by the table.
[[File:IEC60364-5-52 P6.png|center|thumb|573x573px]]
[[File:IEC60364-5-52 P6.png|center|thumb|573x573px]]
Again, following the same example with 4mm<sup>2</sup> cable selected, assuming the cable is to run for 30m (0.03km) connected to resistive load (cos ϕ = 1), then the voltage drop for copper can be calculated as:
Again, following the same example with 4mm<sup>2</sup> cable selected, assuming the cable is to run for 30m (0.03km) connected to resistive load (cos ϕ = 1), then the voltage drop for copper can be calculated as:
10.3 x 20.45 x 0.03 = 6.32 V
6.32V/220V x 100 = 2.87% voltage drop
6.32V/220V x 100 = 2.87% voltage drop
If this value exceeds the voltage drop limit in your country, then select the next bigger cable till the value drops to within the acceptable limits. For example, if the voltage drop limit is 2%, we then select 6mm<sup>2</sup> and recalculate the voltage drop as:
If this value exceeds the voltage drop limit in your country, then select the next bigger cable till the value drops to within the acceptable limits. For example, if the voltage drop limit is 2%, we then select 6mm<sup>2</sup> and recalculate the voltage drop as:
6.8 x 20.45 x 0.03 = 4.17 V
4.17V/220V x 100 = 1.89% voltage drop
4.17V/220V x 100 = 1.89% voltage drop
Therefore, the cable to be used for this circuit is 6mm<sup>2.</sup>
Therefore, the cable to be used for this circuit is 6mm<sup>2.</sup>