Maximum switching current of this relay at 250 V is 10 A. Why is the maximum switching current at 30 V also 10 A? I wonder why it can tolerate 10 A at 250 V but not 80 A in 30 V because the power is the same?
P1 = P2 = 250 V x 10 A = 30 V x 83 A
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There is no 'power law' that applies to relay contacts.
When the contacts open carrying a current, an arc often forms in the gap between them, which damages the contacts.
AC current falls to zero 100 times per second (at 50Hz). This helps to extinguish the arc, which mitigates some of the effect of having a higher voltage.
DC current does not have this feature, less voltage is needed to maintain an arc.
Current carrying, with closed contacts, is limited by heating in the contact area. This will be a function of the contact resistance, which depends on contact area, pressure, material etc. This joule heating is insensitive to whether the current is AC or DC. Therefore it's not surprising to see the same current specification for AC and DC.
You will notice there are two AC specifications, 250V and 125V. Obviously both cannot be 'right'. They are ratings. The point about a rating is that it's a test done under well defined conditions. What that means is when the relay is tested with some specified circuit, for x thousand operations, some specified parameter, perhaps on resistance, will not have changed by more than y%. If you want the detail of the test circuit and pass/fail conditions, then you need to track down the specification that this has been rated against. But basically, more voltage means more energy available to heat contacts, and more voltage available to keep the arc going longer.
Different relays have different contact materials, pressures, masses, speed and distance of separation, and these will all have differing impacts on carrying current and breaking current and voltage. Some relays may well have different rated currents depending on whether AC or DC is used, this one doesn't.
The contact area and heat sink capabilities determines the maximum current irrespective of voltage.
AC voltage has zero current assisting normal arc extinction where DC must rely on sufficient contact separation to extinguish the arc. This is why maximum AC voltage is much higher than the maximum rated DC voltage.
Contacts can be used outside these parameters with snubber circuits but will be destroyed under very inductive loads, even at rated voltage and current.
Your calculations assume that the full voltage is dropped by the relay, which is certainly not the case, usually you have a very small voltage drop over the relay, as its contacts have a low resistance.
To understand the issue, it helps to look at the formula for the power dissipated in a resistance, which is:
$$P = I^2 R$$
As you can see, it is not dependent on the voltage. So if the dissipated power would be the limiting factor, only the current would influence that.
In this case power is probably the limiting factor (as it both specifies 10A for 230V AC and 10A for 125V AC). The copper conductors in the relay have a non-zero resistance, at a certain current the relay might get too hot and plastic parts might melt.
As @Neil_UK already stated, the reason for the different maximum voltage rating for AC and DC is because of the sparking behavior.
The ratings apply to the safety approvals above them. You should not use the relay above 125VAC if you need UL approval.
The ratings are determined by a number of factors- one of which is heating (the current passes through a flexure that must not get too hot or it could fail). The safety agencies also look askance on products that produce excessive amounts of smoke or burst into flames.
Contact life from erosion is another factor, contact gap (for inductive loads) and maximum surge current (for tungsten loads and motors). Ratings are not determined by any single factor.
Chances are if you tried to put 80A through that relay the contacts would instantly weld shut, and then it would burn up soon afterward no matter what you did to the coil.