the value of two resistors are R1 =(6+_0.3) kohm and R2 =10+_0.2 kohm .the percentage error in equivalent resistance when they are connected in parallel is?

1/R = 1/R1+1/R2;
1/R = R1+R2/R1*R2;
R = R1*R2/R1+R2;
R = 10*6/10+6 = 60/16=3.75;

Now ,As We know;
Δ R/R = Δ R1/R1 + Δ R2/R2;
Δ R/3.75 = 0.3/6 + 0.2/10;
Δ R/3.75 = 0.05+0.02;
Δ R = 0.07*3.75;
Δ R = 0.2625 !!!

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