Question 211183
P = I^2R
(I/2)^2*(0.75R) = (I^2/4)*0.75R
= 0.1875(I^2R)
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The change is a reduction of 13/16 from the original.
13/16 = 0.8125 = 81.25%
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Another approach:
V = IR (volts = I*R)
P = VI (watts = volts * amps)
V = IR --> (0.5I)*(0.75R)
V --> 0.375IR
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P = 0.375IR*0.5I --> 0.1875I^2R