An iron designed for use at 120 V and 5.0 A is connected to a source of 240 V. Calculate the current the iron will draw at the higher potential....

Even if we change the voltage the iron is connected to its resistance will still be the same. The resistance offered by a device can be related to the applied voltage and current, as per the following equation:

V = IR (Ohm's Law)

or R = V/I

where R is the resistance, V is the voltage and I is the current.

Since the iron has a rated voltage of 120 v and current of 5...

Even if we change the voltage the iron is connected to its resistance will still be the same. The resistance offered by a device can be related to the applied voltage and current, as per the following equation:

V = IR (Ohm's Law)

or R = V/I

where R is the resistance, V is the voltage and I is the current.

Since the iron has a rated voltage of 120 v and current of 5 A, its resistance is:

R = V/I = 120 v / 5 A = 24 ohm.

If the same iron is connected to twice the voltage, that is, 240 v, the current passing through it will be:

I = V/R = 240 v / 24 ohm = 10 A

Thus, a current of 10 A, or twice the design current, will pass through the iron.

When that happens, the device will stop working. The resistance wire will develop more heat than it normally does and will burn out. The device will short-circuit and you will have to get it repaired. (I actually connected a Wii, designed for 120 v to a 240 v supply and had a short circuit and had to get it repaired.)

This situation can be avoided by using devices rated over a broader voltage range (say 120-240 v, as is the case with most laptops these days) or using a voltage adapter. 

Hope this helps.