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.Com
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(16322) (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(69443) (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%
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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
-------
P = 0.375IR*0.5I --> 0.1875I^2R
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