For two resistors R1 and R2 connected in a parallel , find % limiting error , if R1 = (200+_4) ohm and R2 = (100+- 2 ) ohm
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hello,
In parallel,
Rnet = R1*R2/(R1+R2) = 20000/300 = 200/3 ohm
For error : take partial derivative both sides but first take log to separate them out.
.i.e
log(1/Rnet) = log(1/R1) + log(1/R2)
drnet = 2 + 4 = 6ohm
error %=100%-(actual-error)/actual×100
100-(200/3-6)/200/3×100
=100-97
=3%
hence percent of error is 3%
hope this helps,if u like it please mark it as brainliest
In parallel,
Rnet = R1*R2/(R1+R2) = 20000/300 = 200/3 ohm
For error : take partial derivative both sides but first take log to separate them out.
.i.e
log(1/Rnet) = log(1/R1) + log(1/R2)
drnet = 2 + 4 = 6ohm
error %=100%-(actual-error)/actual×100
100-(200/3-6)/200/3×100
=100-97
=3%
hence percent of error is 3%
hope this helps,if u like it please mark it as brainliest
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