High School

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Given:

- [tex]v₁(t) = 14e^{-4t} \, V[/tex]
- [tex]r₁ = 8 \, \Omega[/tex]
- [tex]l₁ = 11 \, H[/tex]

Derive the complete solution for [tex]i(t)[/tex] for [tex]t \geq 0[/tex].

Answer :

Final Answer

To find the complete solution for i(t) for t ≥ 0, we first need to identify the given parameters:v₁(t) = 14e(-4t) V, r₁ = 8ω, and l₁ = 11 H. Using these values, the expression for i(t) can be derived. The final answer is:

i(t) = (14 * 8ω) / (11 * r₁) * (e(-4t) - e(4t)) * sin(8ωt) A

Explanation

1. Understanding the Problem:

The problem asks us to find the complete solution for the current i(t) in an RL circuit with given voltage source v₁(t) and known parameters r₁ and l₁. The equation for the current in an RL circuit is:

i(t) = (v(t) / r₁) * (1 - e(-r₁ / (2 * l₁) * t)) * sin(ωt)

where v(t) is the voltage across the inductor, r₁ is the resistance, l₁ is the inductance, and ω is the angular frequency.

2. Analyzing the Given Parameters:

We are provided with the voltage source v₁(t) = 14e(-4t) V, r₁ = 8ω, and l₁ = 11 H. We need to find the angular frequency ω and then substitute the given values into the equation for i(t).

3. Finding the Angular Frequency (ω):

Since r₁ = 8ω, we can express ω as:

ω = r₁ / 8 = (8ω) / 8

Now, we can rewrite the equation for i(t) using the given v₁(t) and the calculated ω:

i(t) = (14 / 11 * r₁) * (1 - e(-r₁ / (2 * l₁) * t)) * sin((r₁ / 8) * t)

Substituting r₁ = 8ω:

i(t) = (14 * 8ω) / (11 * r₁) * (1 - e(-8ω / (2 * 11) * t)) * sin((8ω) / 8 * t)

Now, we can simplify the expression:

i(t) = (14 * 8ω) / (11 * r₁) * (1 - e(-8ω / 22 * t)) * sin(ωt) A

In conclusion, using the given voltage source v₁(t) = 14e^(-4t) V, resistance r₁ = 8ω, and inductance l₁ = 11 H, we derived the complete solution for the current i(t) in the RL circuit. The final expression for i(t) is:

i(t) = (14 * 8ω) / (11 * r₁) * (e(-4t) - e(4t)) * sin(8ωt) A

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