Two identical resistors each of resistance 10 ohm are connected in (i) series, (ii) parallel to a battery of 6V. Calculate the ratio of power consumed by the combination of resistor in the two cases.

Explanation

`R_(1) = R_(2)=10 Omega`
(i) When connected in series
`R_(s)=R_(1)+R_(2)= 20 Omega`
(ii) When connected in parallel
`R_(p)= (R_(1)R_(2))/(R_(1)+R_(2))= (10xx10)/(10+10)=5 Omega`
`P_(p)= (V^(2))/(R_(p))=(6xx6)/(5) = 7.2 ` W
`(P_(s))/(P_(p))= (1.8W)/(7.2W)=(1)/(4)= 1:4`