Without any parameters, this is just a guess. if your inductor needs 5 A and has a resistance of R ohms then it requires 5 X R volts. If you then feed it from a voltage source higher then 5 X R volts, then the excess voltage must be dropped across a series resistor which is dropping (Vr), Vsupply- (5 X R) volts. So the secret is to supply |JUST enough voltage of 5 X R volts, then no series resistor is required.
If you are stuck with , say a supply of 24V and the resistance of the coil is say, 2 ohms, then it will drop 2 X 5 = 10V, so the resistor has to drop 24- 10 = 14 V at a current of 5 A so it will dissipate 5 X 14 = 70 V. A better way would be to use a BUCK converter, this is an electronic circuit that will convert your 24 V to 10V with a high efficiency so it would typically waste just a few watts.
Frank