current
I do not understand the answer of the following question :
A three-phase power in the lab have symmetrical three-phase voltages 400 / 230V and
terminals marked L1, L2, L3 and N. A 60W bulb is connected between phase L1
and N. Another 60W lamp is connected in the same way, but the phase L2
and N.
Calculate the current in the neutral conductor N.
The answer is 0,26 A.
Why it is 0,26 A , should not it be 0,26 A + 0,26 A (from the both lamps)?
Thanks
In your edit, what's missing is that the rate of cooling will depend on the temperature. In general, the cooling rate will increase as the temperature increases. When the temperature rises enough that the cooling rate matches the heating rate, the temperature will stabilize.
But the actual cooling rate is very difficult to calculate. It depends on what other materials the copper is in contact with (conductive cooling), the airflow around the conductor, etc.
As an added complication, the heating rate will also depend on temperature, because the resistance of the copper will increase at higher temperatures.
So without much more detailed information about your conductor and its environment, its not really possible to give a precise answer to your initial question, how hot will it get?.
As for the second question, how fast will it heat up if there's no cooling, you can calculate that from the heat capacity of copper, which Wikipedia gives as 0.385 J / (g K), or 3.45 J / (cm^3 K).
No guarantees as to correctness, but this shows the general method.
By convention, phase rotation is anti-clockwise. An inductive load will make the current lag behind the voltage. (Remember, inductors resist changes in voltage dv/dt.)
Diagrams below are not to scale.
Best Answer
To confirm, from \$ P = VI \$ that the current would be \$ I = \frac {P}{V} = \frac {60}{230} = 0.26~A \$. No problems there.
A very simple way to consider this problem is that if we connected a 60 W lamp from each phase to neutral then the neutral current would sum to zero.
Now consider what happens if we remove one of the three lamps: the neutral current must change by that amount, 0.26 A. That's the simple way to calculate for this problem.
The more general way would be to add the vectors. (And this is your problem: you forgot that they are not in phase.)
simulate this circuit – Schematic created using CircuitLab
Figure 1 (a) The phase A and B current vectors. (b) A and B vectors summed to give the resultant current.
Clearly from the vector diagram, since A and B were at 120° then in (b) they must be at 60°. Since they're the same size the triangle is equilateral. Therefore the sum must be 0.26 A.