The total heat produced in resistor `r` in an `RL` circuit when the current in the inductor decreases from `I_(o)` to `0` is

Options

A. `LI_(0)^(2)`
B. `(1)/(2) LI_(0)^(2)`
C. `(3)/(2) LI_(0)^(2)`
D. `(1)/(3) LI_(0)^(2)`

Answer

B

Explanation

the power dissipated in the resistor,
`P = (dW)/(dt) = I^(2)R`
Since the current through resistor varies with time, we must integrate.
The total energy produced as heat in the resistor
`W = int_(0)^(oo) I^(2) R dt`
The current in an `RL` circuit is `I = I_(0)e^(-(R//L)t)`
`W = int_(0)^(oo) I_(0)^(2) e^(-(R//L)t) R dt = (I_(0)^(2) R)/(-2R//L) [e^((-2R)/(L)t)]_(0)^(oo) = (1)/(2)`
We can integrate by substituting
Note thet the total heat produced equals the energy `(1//2)LI_(0)^(2)`
originally stored in the conductor.