`100 J` of heat are produced each second in a `4 Omega` resistance. Find the potential difference across the resistor.

Explanation

`H=100J, R=4 Omega, t=1 s, V=?`
From Eq. (12.21) we have the current through the resistor as
`I= sqrt(H//Rt)`
`=sqrt[100J//(4Omegaxx1s)]`
=5A
Thus the potential difference across the resistor, V [ from Eq. [(12.5)] is
V=iR
`=5 Axx4 Omega`
= 20 V.