Two resistances are expressed as $R_1=\left(4\pm 0.5\right)\Omega$  and $R_2=\left(12\pm 0.5\right)\Omega$ . What is the net resistance when they are connected (i) in series and (iv) in parallel, with percentage error?

$R_3=R_1+R_2=16\Omega$

$R_P=\frac{R_1{R}_2}{R_1+R_2}=\frac{R_1{R}_2}{R_S}=3\Omega$

$\Delta R_S=\Delta R_1+\Delta R_2=1\Omega$

$\Rightarrow \frac{\Delta R_S}{R_S}\times 100=\frac{1}{16}\times 100\%$

$\Rightarrow \frac{\Delta R_S}{R_S}\times 100=6.25\%$

$\Rightarrow R_S=16\Omega \pm 6.25\%$

Similarly

$R_P=\frac{R_1{R}_2}{R_S}$

$\Rightarrow \frac{\Delta R_P}{R_P}=\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}+\frac{\Delta R_S}{R_S}$

$\Rightarrow \frac{\Delta R_P}{R_P}=\frac{0.5}{4}+\frac{0.5}{12}+\frac{1}{16}$

$\Rightarrow \frac{\Delta R_P}{R_P}=0.23$

$\Rightarrow \frac{\Delta R_P}{R_P}\times 100=23\%$

$\Rightarrow R_P=3\Omega \pm 23\%$