According to what I've learnt, in any expression of multiplication or division, the percentage errors of each term are added up to find the equivalent percentage error. That is, if
<span><span>y=<span>ABC
</span></span><span>
y=<span>ABC</span></span></span>then<span><span>% error in y= % error in A + % error in B + % error in C</span></span>
For the above problem, let <span><span>Rs</span></span> denote series combination. Then <span><span><span>Rs</span>=300±7</span></span> ohm.
Let <span><span>Rp</span></span> denote parallel combination.
<span><span>∴<span>Rp</span>=<span><span><span>R1</span><span>R2/</span></span><span><span>R1</span>+<span>R2 </span></span></span>= <span><span><span>R1</span><span>R2/</span></span><span>Rs</span></span></span></span>
Ignoring errors, we get <span><span><span>Rp</span>=<span>200/3 = </span></span></span><span><span>66.67</span></span> ohm
<span><span>%<span>error in R1</span>=3</span></span>, <span><span>%<span>error in R2</span>=2</span></span>, <span><span>%<span>error in Rs</span>=<span>7/3</span></span></span>
Hence, <span><span>%<span>error in Rp</span>=3+2+<span>7/3</span>=<span>22/3</span></span></span>
So, error in <span><span>Rp</span></span> will be <span><span><span>22/3</span>% of </span></span><span><span>200/3</span></span>, which is approximately <span>4.89.</span>
Hence, I got <span><span><span>Rp</span>=66.67±4.89</span></span> ohm.
However, the book used the formula described and proved here and arrived at the answer <span><span><span>Rp</span>=66.67±1.8</span></span>ohm.
