A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the em across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? A) 1.9 Ω B) 1.50 Ω C) 14 Ω D) 5.0 Ω

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Respuesta :

aristeus aristeus · Answer 1 · 2019-09-12T09:46:59+02:00

Answer:

(a) 1.9 ohm

Explanation:

We have given voltage across LR circuit [tex]V_0=14V[/tex]

Inductance L = 34 Henry

Time t = 4 sec

It is given that emf across the inductor is 80% of maximum value in 4 sec

The voltage across RL circuit is given by

[tex]V=V_0e^\frac{-t}{\tau }[/tex] , here [tex]\tau[/tex] is time constant which is [tex]\frac{L}{R}[/tex] for RL circuit

Now according to question

[tex]14\times 0.8=14e^{\frac{-4}{\tau }}[/tex]

[tex]\tau =17.9sec[/tex]

We know that [tex]\tau =\frac{L}{R}[/tex]

[tex]17.9=\frac{34}{R}[/tex]

R = 1.9 ohm

So option (a) is correct