According to Ohm's Law, if the coil has 60 Ohms of resistance and can only support a 2 Amp load, the voltage must be decreased to

V = IR

V = 2A x 60 ohms = 120 Volts

Therefore, the series resistor must provide 120 Volts of drop from the 240 Volt source. Again, using ohms law we can solve for the size of the resistor that will cause this to happen:

V(drop) = IR

**R = V(drop)/I = 120V / 2 amp = 60 ohms.**

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