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2 years ago

A 190 mH inductor is connected to an emf given by

Physics

1 answer:

Bas_tet [7]2 years ago

7 0

(a) The reactance of the inductor is 25.46 ohms.

(b) The expression for the current through the inductor is I(t) = (6.32 A) sin(134t)

<h3>Ractance of the inductor</h3>

The reactance of the inductor is calculated as follows;

XL = ωL

where;

ω is angular frequency
L is 190 mH

v(t) = (161 V) sin(134t)

v(t) = V sin(ωt)

The reactance of the inductor is calculated as follows;

XL = (134) x (190 x 10⁻³)

XL = 25.46 ohms

<h3>Peak current in the circuit</h3>

I₀ = V₀/XL

I₀ = (161) / (25.46)

I₀ = 6.32 A

<h3>Expression for the current through the inductor</h3>

I(t) = (6.32 A) sin(134t)

Learn more about inductance here: brainly.com/question/16765199

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Answer:

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Explanation:

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Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

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A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a) at 4.75 cm

b) at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula $E = \dfrac{kq }{d}$

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}$

E = 1461.95 N/C

c) The electric field E is calculated as:

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