In the circuit shown , the heat produced in `5Omega` resistot is `10"cal"s^(-1)` .The heat produced per sec in `4Omega` resistor will be

Options

A. 1 cal
B. 2 cal
C. 3 cal
D. 4 cal

Answer

b

Explanation

`H=l^(2)Rt`
`therefore(H_(1))/(H_(2))=((l_(1))/(l_(2)^(2)))xx((R_(1))/(R_(2)))`
or `(10)/(H_(2))=[(1)/(((1)/(2)))]xx(5)/(4)`
`H_(2)=2cal`