# SOLUTION: Not sure if this question belong here but it has to do with percentage change. The question is- the power of an electric circuit is given by P=I^2 R whre I= current and R= resist

Algebra ->  Algebra  -> Percentages: Solvers, Trainers, Word Problems and pie charts  -> Lessons -> SOLUTION: Not sure if this question belong here but it has to do with percentage change. The question is- the power of an electric circuit is given by P=I^2 R whre I= current and R= resist      Log On

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 Algebra: Percentage and Pie Charts Solvers Lessons Answers archive Quiz In Depth

 Question 211183: Not sure if this question belong here but it has to do with percentage change. The question is- the power of an electric circuit is given by P=I^2 R whre I= current and R= resistance Find the % change in Power if the current is halved and resistance decreases by 25%. Thank you.Found 2 solutions by rfer, Alan3354:Answer by rfer(12678)   (Show Source): You can put this solution on YOUR website!By putting numbers into the formula I come up with a 20% decrease in power. I used I=4 and R=10 Then I used I=2 and R=8 Answer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website!P = I^2R (I/2)^2*(0.75R) = (I^2/4)*0.75R = 0.1875(I^2R) --------------- The change is a reduction of 13/16 from the original. 13/16 = 0.8125 = 81.25% ----------- Another approach: V = IR (volts = I*R) P = VI (watts = volts * amps) V = IR --> (0.5I)*(0.75R) V --> 0.375IR ------- P = 0.375IR*0.5I --> 0.1875I^2R