In the circuit shown in figure the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calorie per second. The heat generated in the 4 ohm resistor is

Explanation

`V_(AB)=I_(2)(4+6)`,

`therefore V_(AB)=10I_(2)`
`"Also "V_(AB)=I_(1)(5)`,
`therefore 5I_(1)=10I_(2) or I_(2)=2I_(2)`
`"Let "H=I^(2)R," "therefore (H_(4))/(H_(5))=(I_(2)^(2)xx4)/(I_(1)^(2)xx5)`
`(H_(4))/(10)=(I_(2)^(2)xx4)/((2I_(2))^(2)xx5) or H_(4)=(10xx4)/(4xx5)=2("cal")/("sec")`