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@6, if your going to specify a percentage for volt drop you should let them know that this volt drop is the total from the intake point to the final circuit not just from a sub-board.
Put your email up and I’ll send you on a fully worked out example.
Cable sizing depends on the regulations for your location and on four factors 1. Length 2. Cross-sectional area 3. Temperature 4. Conductor material.
The cable must satisfy the following conditions: Ib ≤ In ≤ Iz, Iz = In / (Ca x Cg)
Where Ib = design current. In = protective devise rating. Iz = current carrying capacity of cable under specific conditions.
Ca = cable ambient temperature factor. Cg = cable grouping factor.
The volt drop for the run of cable then must be checked = mV x A x m
Where mV = volt drop per meter (for a perticular size of cable). A = Ib. m = length of run (meters).
The max overload must then be checked = I2 ≤ 1.45 (this is an Irish figure) x Iz
Where I2 = the current that will operate a protective device (obtained from the manufacturer)
The cable withstand energy for short-circuits must also be checked = I^2 x t ≤ K^2 x S^2
where I = fault current. t = time for protective device to operate. K = conductor materrial factor.
S = cable cross sectional area.2011/09/02 at 5:36 pm in reply to: what is the theoretical explanaton for the current lead in capacitor? #12455
When you first turn on the power in a capacitive circuit, the capacitor acts like a short circuit. Current rushes in at maximum rate, then gradually decreases as the capacitor charges.
While this is happening, the voltage across the capacitor slowly rises, from zero to max. At full charge, the capacitor acts like an open circuit. Current is zero and voltage is at its maximum.
So I can be said that voltage lags the current by 90′
I made a mistake at the end, it should be cos (x)^-1 not cos (x)^1, kW / kVA = cos (x), cos (x)^-1 = power factor.