Question

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

Answer 1

I = 1.23 A

Step-by-step explanation:

In an RL circuit current passing is described by

I = E / R (1 -
`e^(-Rt/L)`)

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 -
`e^{-12 .. 5 {10}^(-3)/35 .. {10}^(-3) }`e (- 5 10-3 12/35 10-3))

I = 1.5 (1-
`e^(-1.715)`)

I = 1.23 A