How to calculate the percentage error for moving iron voltmeter?

the moving iron voltmeter has resistance of 5k ohm at 15c when it reads correcectly when connected to a supply of 200V.

and the copper coil's temperature coefficient at 15c is 0.004ohm/c.



so what's the percentage error in the reading when it's 50c?How to calculate the percentage error for moving iron voltmeter?
In a moving iron instrument, the deflection of the needle is proportional to the square of the current, but the linearity of the scale can be compensated somewhat by suitable design of the iron vanes. Since the scale is calibrated to indicate a value that is proportional to the current, I believe that the error can be calculated based on the change in current due to the change in resistance due to the temperature change.



At 15c, the current through the meter is 200/5000 = 0.04 A. If the temperature increases by 35c to 50c, the resistance increases by 0.004/c, the resistance at 50c is 5000 X 1.004 = 5020. The current will then be 200/5020 = .0398 A or .04/.0398 = 1.004 = 100.4% of the previous value. Therefore the error is 0.4%.



Correction:

I forgot to multiply 0.004/c X 35c = .14. The resistance at 50c = 5000 X 1.14 = 5700. The current will then be 200/5700 = 0.0351 A or .04/.0351 = 1.14 = 114% of the previous value. Therefore the error is 14%. Sorry.



Additional note:

I assume that the units for temperature coefficient are 1/c as published in various places.How to calculate the percentage error for moving iron voltmeter?
Get it calibrated.