Current Flowing Through Parallel Connected Conductors

The first thing to know is the supply voltage.

The second thing to know is whether the cables are part of a 3-phase load or whether they are supplying a single phase load.

The third thing to know is why two cables of differing sizes have been used.

sir,

supply voltage is 415V, its 3phase load, feeding from msb to smsb, 3phase load but not symmetrical,all are lighting loads distributed, its existing, just i want to know how to calculate the current flowing through each conductor, i think it may required 70 sqmm for 200A, instead of that they may provide available sizes. my doubt is how the current is distributing by two cables.

thanks

3 Phase power is P = Sqrt(3) x V x I

The ratio of the current in the two conductors is same as of ratio of their conductor size (area), i.e , I1/ I2 = A1/A2

ok sir, if so,
50/35=I1/I2,
ie, 50I2=35I1, but here I1 or I2 not available, only available thing is I1+I2=200A

200(50/85) ≈ 117.6 A; 200(35/85) ≈ 82.4 A.

yes sir, it is something meaning ful, can u pls. tell where i can find the formuale for this.

That was the formula! (#5↑)

yes, its fine but any reference? to quote

The formula given you is correct, but if you realllllllly need an Official Version that you can quote, keep in mind Official Versions will substantiate their claims in ways only Official Versions can: in gruesome, gory detail. It's what makes them Official, after all. Not much different-looking than the word Offal. Mere coincidence? I think not.

So, if you like watching laws and sausages being made, you'll simply love this.

You asked!

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Probably you want to know how the formula come about. Here is the derivation. The two conductors having resistance R1 an R2 will have same voltage drop, because there are in parallel. Therefore V1=I1R1= V2=I2R2. Resistence =resistivity x length/ area. Therefore, when you put them together, will give you the ratio I1/I2= A1/A2

Here in North America, that is strictly forbidden. Wires of different diameters (and lengths for that matter) run in parallel to the same load will have different circuit resistance values. The larger wire will have less resistance, so more current will flow in that conductor until it heats up to the point where the positive coefficient of temperature of the conductor material makes the resistances equal. Only then will they share equally, but that usually means the current rating of the larger conductor has been exceeded and the insulation fails (unless of course the current in the circuit is below the rating of the single larger conductor alone). There is no way to accurately calculate how they will share, hence it being forbidden (at least here).

...which is why the installation needs review and assessment for improvement by a qualified electrician on safety grounds.

100kW at 380V = 151Amp total load current at unity power factor...if not unity power factor as usual the total amps can be more

50sq cable = 88amp of load

35sq cable = 62amp of load

your 200amp MCB should do the thing for the total load

greetings M & L