contestada

What resistance should be added in series with a 3.0-H inductor to complete an LR circuit with a time constant of 4.0 ms? A)0.75 k Ω Β) 12 Ω C) 0.75 Ω D) 2.5 Ω

Respuesta :

ogorwyne

Answer:

O.75KΩ

Explanation:

We measure the time constant τ, using the formula τ = L/R,

t is in seconds, then we have R to be the value of the resistor which is measured in ohms and also L is the value of the inductor which is measured in Henries.

Since t = L/R

We make R subject of the formula

R = L/τ

= 3/4x10-3

= 0.00075

= 0.75 KΩ

So we have it that the first Option (A) is the correct answer to the question

boffeemadrid

The resistance to be added is required.

The resistance added should be A. [tex]0.75\ \text{k}\Omega[/tex]

L = Inductance = 3 H

[tex]\tau[/tex] = Time constant = 4 ms

R = Resistance

Time constant is given by

[tex]\tau=\dfrac{L}{R}\\\Rightarrow R=\dfrac{L}{\tau}\\\Rightarrow R=\dfrac{3}{4\times 10^{-3}}=750\ \Omega=0.75\ \text{k}\Omega[/tex]

The resistance added should be [tex]0.75\ \text{k}\Omega[/tex]

Learn more:

https://brainly.com/question/13201007?referrer=searchResults

Otras preguntas