Recent content by Babadag

The only thing that can be considered a “mathematical trick” it is the shape and dimensions of the measuring loop where the only thing precise is the distance on the tank, the rest is my fantasy.

Since Windows 11 up-dating it produces chaos -usually, I had to revise my calculation. The inductance formula is incomplete. Theoretically, the magnetic field around a conductor where a current flow is H=I/(2*pi()*x) [A/m] if x is the radius from conductor centre line up toany point outside...

In my opinion it is about the circuit reactance.

Suppose an observer measured the time since the stone began to fall and it was 10 seconds. The initial velocity of the stone in the opposite direction was 10 m/s and it stopped its ascent in 10/9.8 seconds. That means the rock flies up 9.8*(10/9.8)^2/2=5.1m From this point the stone falls in 10...

Something like this

The multimeter presents the voltage drop between its connecting points. That means, this voltage is proportional with the current flowing through the multimeter impedance-I think mainly through a reactance [inductive].However, the circuit passes through grounding resistance ,your body and the...

Skin effect and proximity effect it can be calculated according to IEC 60287-1-1 2.1.2 Skin effect factor ys and 2.1.4 Proximity effect factor yp for three-core cables and for three single-core cables

That's the part that looks like your case

So, this cable conductor is waterproof, round aluminium rope and it is not copper. Calculated, according IEC 60287-2 [and -1], at 30oC air -no wind, no sun-current capacity is 1279A per cable, and you have 4 cables in parallel per one phase. If distance is 128 mm centre-line to centre-line for...

First, I am not sure, I understood well your problem. You have 2 single core cables per phase, of 800 mm^2 copper conductor XLPE insulated [20 kV] each, for 4000 A. If the cables run in a rack at a distance of minimum one cable overall diameter, maximum current per one cable, up to 90oC, will be...

I forgot to say that 65 kW is evacuated in the air, while 11.39 MW still remain in the transformer because the exhaust begins after 5-10 seconds when the outside temperature of the transformer is high enough.

The apparent power [kVA or MVA] on the name plate it is not the power “consumed” in the transformer, but only the transferred power to the load. The power consumed in the transformer it is the transformer losses-electrical and magnetic. Let's take an example. ABB 33/11 kV 10 MVA transformer...

In order to calculate the power wasted in a transformer short-circuit case you need the following data: S=rated kVA, uk%-short-circuit voltage, kxr= the X on R ratio[X/R]. The transformer impedance [short-circuit impedance] Z=VLL^2/S*uk%/100 where Z=R+Xi and R=Z/sqrt(1+kxr^2) Then Isc=VLL/Z for...

According to (Endreny 1967) Endrenyi, J., “Analysis of transmission tower potentials during ground faults" for a transmission line R1∞=R/2+sqrt(R^2+R^2/4) See also: https://ace.ucv.ro/sintes11/Volume2/5%20ELECTRONICS/E17%20-%20VINTAN%20Maria.pdf

It seems that actually it is a little more complicate. The supply voltage is AC -as usual- and a driver AC/DC is provided in order to supply DC. Then 350 mA it could be the DC output [18-42 V].Let’s say Vdc=20 V Pdc=0.35*20=7 W. In the interval of 115 to 125 Vac the LED current is 350 mA, but...