At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s, how fast is the current changing

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Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

The rate at which the current is changing is;

di/dt = 0.2 A/s

We are given;

Inductance; L = 5 H

Current; I = 3 A

Rate of Increase of energy; dE/dt = 3 J/s

Now, the formula for energy stored in inductor is given as;

E = ¹/₂LI²

Since we are looking for rate at which current is changing, then we differentiate both sides of the energy equation to get;

dE/dt = LI (di/dt)

Plugging in the relevant values gives;

3 = (5 × 3)(di/dt)

di/dt = 3/(5 × 3)

di/dt = 0.2 A/s

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