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clear//
//Variables
I = 250 * 10**-3 //Current (in Ampere)
R = 1.5 * 10**3 //Resistance (in ohm)
//Calculation
Vs = I * R //Source voltage (in volts)
I1 = 0.75 * I //New current (in Ampere)
R1 = Vs / I1 //New Resistance (in ohm)
R2 = R1 - R //Resistance to be added (in ohm)
//Result
printf("\n %0.3f ohm Resistance must be added in order to accomplish the reduction in current.",R2)
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