Two resistances are expressed as R1=(4±0.5)Ω  and R2=(12±0.5)Ω . What is the net resistance when they are connected (i) in series and (iv) in parallel, with percentage error?

R3=R1+R2=16Ω RP=R1R2R1+R2=R1R2RS=3Ω ΔRS=ΔR1+ΔR2=1Ω ⇒ΔRSRS×100=116×100% ⇒ΔRSRS×100=6.25% ⇒RS=16Ω±6.25% SimilarlyRP=R1R2RS ⇒ΔRPRP=ΔR1R1+ΔR2R2+ΔRSRS ⇒ΔRPRP=0.54+0.512+116 ⇒ΔRPRP=0.23 ⇒ΔRPRP×100=23% ⇒RP=3Ω±23%