An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-Ω resistor, and a switch S, all in series. Initially, the switch is open and has been open for a very long time. At time t = 0 s, the switch is suddenly closed. How long after closing the switch will the potential difference across the inductor be 24 V?
A) 1.6 s
B) 1.4 s
C) 1.8 s
D) 1.9 s
E) 2.1 s

Respuesta :

Answer:

A) 1.6 s

Explanation:

Given that

V = 60 V,

VL = 24 V

R = 24 ohm

L = 42 H

We know that time constant λ given as

λ = L / R = 42 / 24 = 1.75

We know that voltage at time t given as

[tex]V-V_L=V(1-e^{-\frac{t}{\lambda}})[/tex]

Now by putting the values

[tex]60-24=60(1-e^{-\frac{t}{1.75}})[/tex]

[tex](1-e^{-\frac{t}{1.75}})=0.6[/tex]

t/1.75 = ln (2.5)

t/1.75 = 0.916

t = 1.6 s

So answer is A) 1.6 s

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