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marin [14]
2 years ago
9

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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M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}

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N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}

The resultant vector is the sum of the components of M and N:

F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}

The magnitude of the resultant vector is:

|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm

And the direction of the vector is:

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hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

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