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An LR circuit consists of a 35-mH inductor, ac resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed

Respuesta :

Answer:

 I = 1.23 A

Explanation:

In an RL circuit current passing is described by

           I = E / R (1 - [tex]e^{-Rt/L}[/tex])

Let's reduce the magnitudes to the SI system

        L = 35 mH = 35 10⁻³ H

        t = 5.0 ms = 5.0 10⁻³ s

let's calculate

         I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))

         I = 1.5 (1- [tex]e^{-1.715}[/tex])

         I = 1.23 A

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