Two wires A and B of same length and same material, have their cross sectional areas in the ratio 1:4. What would be the ratio of heat produced in these wires when the voltage across each is constant?

Explanation

Given `l_(A) = l_(B), rho_(A) = rho_(B), V_(A) = V_(B)`
`A_(A) : A_(B) = 1 : 4`
heat of heat produced in a wire,
`H = i^(2) R = (V^(2))/(R^(2)) R = (V^(2))/(R ) = (V^(2) A)/(rho l)`
`[:' H = i^(2) R = (V^(2))/(R^(2)) R = (V^(2) A)/(rho l)]`
Since, `V, rho, l` are same for both wires A and B,H `prop` A (Area of crossection)
For two wires A and B,
`(H_(A))/(H_(B)) = (A_(A))/(A_(B)) = (1)/(4)`
`:. H_(A) : H_(B) = 1 : 4`.