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The total heat produced in resistor `r` in an `RL` circuit when the current in the inductor decreases from `I_(o)` to `0` isA. `LI_(0)^(2)`B. `(1)/(2) LI_(0)^(2)`C. `(3)/(2) LI_(0)^(2)`D. `(1)/(3) LI_(0)^(2)` |
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Answer» Correct Answer - B 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. |
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