A heater coil is rated 100 W, 200 V. It is cut into two identical parts. Both parts are connected together in parallel to the same source of 200 V. Calculate the energy liberated per second in the new combination.

Explanation

Resistance of heater coil,
`R= (V^(2))/(P)= ((200)^(2))/(100)= (200xx200)/(100)= 400Omega`
After cutting, resistance of each part =` (400)/(2)= 200Omega`
When connected in parallel, the net resistance is given by
`(1)/(R_(P))= (1)/(200)+ (1)/(200)`
`implies (1)/(R_(P))= (2)/(200)`
`R_(P)= 100Omega`
Energy liberated `= P xx t`
`= (V^(2))/(R) xx t= ((200)^(2))/(100)xx 1= 400"Joule"`