The coil of a heater connected to a 200 V line , con-sumes a power of 100 W . The coil is divided into two equal parts . The two parts are combined in parallel and connected to a 200 V line . What will be the power consumed by the new combination ?

Explanation

If R be the resistance of the coil , then power con-sumed in the potential difference `V,P =(V^(2))/(R)`
In the first case , `100=(200)^(2)/(R)`
In the second case , resistance case ,resistance of each of the two equals parts `=(R)/(2)`
So , the equaivalent resistance in the parallel combination
`=((R)/(2)xx(R)/(2))/((R)/(2)+(R)/(2))=(R)/(4)`
So , power consumed ,
`p_(2)=(200)^(2)/(R/4)=4xx(200)^(2)/(R/4)=4xx100=400W`