A resistor ` R_(1)` consumes electrical power `P_(1)` when connected to an `emf epsilon`. When resistor `R_(2)` is connected to the same `emf` , it consumes electrical power `P_(2)` . In terms of `P_(1) and P_(2)`, what is the total electrical power consumed when they are both connected to this emf source
(a) in parallel
(b) in series

Explanation

Two resistance are `R_(1) a R_(2)`.
Both consumed the Power `P_(1) and P_(2)` respectively.
`P=(V^(2))/(R)therefore P_(1)=(V^(2))/(R_(1)) and P_(2)=(V^(2))/(R_(2))`
`"(i) ""Power in series"=(1)/(P_(1))+(1)/(R_(2))=(R_(1))/(V^(2))+(R_(2))/(V^(2))=(R_(1)+R_(2))/(V^(2))`
`"(ii) ""Power in parallel"=P_(1)+P_(2)`
`" "=(V^(2))/(R_(1))+(V^(2))/(R_(2))=(V^(2)(R_(1)+R_(2)))/(R_(1)R_(2))`