Physics 112

A battery is connected in series with a R = 0.26 Ω resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is τ = 0.10 s, and the maximum current in the circuit is I = 8.4 A. Find the following.
(a) the emf of the battery V?
(b) the inductance of the circuit
mH ?
(c) the current in the circuit after one time constant has elapsed
A ?

(d) the voltage across the resistor and the voltage across the inductor after one time constant has elapsed
resistor V
inductor V

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asked by juan p
  1. a) as t--> oo inductor is a short circuit, no reactance
    so
    V = i at big t * R
    V = (8.4)(.26) = 2.14 Volts

    b) at t = 0, i = 0, so all voltage across L
    2.14 = L di/dt

    let i = 8.4 [ 1 - e^-t/tau ]
    given tau = .1 s
    i = 8.4 [ 1 - e^-10 t ]
    di/dt = 8.4 (10 e^-10 t)
    di/dt = 84 at t = 0
    so
    2.14 = L (84)
    L = .0255 H

    c) when t = tau = .1
    i = 8.4 ( 1 - e^-1) = 8.4(.632) = 5.31 amps

    d) Vresistor = i R = 5.31*.26 = 1.38 volts
    Vinductor = 2.14 - 1.38 = .759 volts

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    posted by Damon

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