The values of two resistors are R1 = (6±0.3)kΩ and R2 = (10±0.0.2)kΩ. The percentage error in the equivalent resistors when they are connected in parallel is :

a) 5.125 %, b)2% , c) 3.125 %, d) 7%, e) 10.125%(answer)

How do you get this answer?

Thanks!

R= R1*R2/R1+R2, where R is equivalent resistance %error in R = % error in R1*R2 +%error in R1+R2 %error in R1*R2 =%error in R1 + %error R2 %error in R1= 0.3/6*100= 5%% error in R2 = 0.2/10 * 100 = 2%% error in R1+ R2= 5%+2% =7%%error in R1 +R2 = 0.3+0.2/16 *100 = 3.125%Total error in R = 7%+3.125% =10.125%