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

Answer

(a) `P_(1) + P_(2)`
(b) `(P_(1) P_(2))/( (P_(1) + P_(2))`

Explanation

` P_(1) = epsilon^(2)// R_(1) , so R_(1) = epsilon ^(2)//P_(1)`
` P_(2) = epsilon^(2)// R_(1) , so R_(2) = epsilon ^(2)//P_(1)`
(a) When the resistors are connected in parallel to the emf, the voltage across each resistor and the power dissipated by each resistor are the same as if only one resistor was connected `P("tot") = P_(1) + P_(2)`
when the resistors are connected in series , the equivalent resistance is `R_(eq.) = R_(1) + R_(2)`.
`P_(t o t) = ( epsilon ^(2))/( R_(1) + R_(2)) = ( epsilon^(2))/( epsilon^(2)//P_(1) + epsilon//P_(2)) = ( P_(1)P_(2))/( P_(1) + P_(2))`