Solve the general initial value problem modeling the LR circuit, L dIdt + RI=e, I(0)=I0 where E is a constant source of emf.

Step 1Let our equation be LI′(t) + RI(t)=E ⋯(⋅) where are I - moving charge L - inductor R - resistor E - const First, divide both sides of equation with L LI(t)′ + RI(t)= :LI′(t) + RL I(t)= EL We have first order linear differntial equation. To solve her, we must find integration factor μ(t). First, let's define function a(t) a(t)= RL We will get integration factor using next formula μ(t)=e∫ a(t)dt=e∫ RL dt=eRLt Now, multiply both sides of our equation with integration factor I′(t) + RL I(t)= EL eRLteRLtI′(t) + eRLt 1RCI(t)=eRLt ElStep 2eRLtI′(t) + eRLt RLI(t)⏟=eRLt EL= ddt (eRLtI(t))ddt (eRLtI(t))=eRLt ELIntegrate both sides of equation ∫ ddt (eRLtI(t))dt= ∫ eRLt ELdt

Expert Answer to "Solve the general initial value problem modeling the..."

waigaK

Answered question

2021-02-02

Solve the general initial value problem modeling the LR circuit,

L \frac{d I}{d t} + R I = e, I (0) = I_{0}

where E is a constant source of emf.

Answer & Explanation

Jaylen Fountain

Skilled2021-02-03Added 169 answers

Step 1

Let our equation be $L I^{'} (t) + R I (t) = E \dots (\cdot)$ where are I - moving charge L - inductor R - resistor E - const First, divide both sides of equation with L $L I {(t)}^{'} + R I (t) = \frac{}{:} L$
$I^{'} (t) + \frac{R}{L} I (t) = \frac{E}{L}$ We have first order linear differntial equation. To solve her, we must find integration factor $μ (t) .$ First, let's define function a(t) $a (t) = \frac{R}{L}$ We will get integration factor using next formula $μ (t) = e^{\int a (t) d t} = e^{\int \frac{R}{L} d t} = e^{\frac{R}{L} t}$ Now, multiply both sides of our equation with integration factor $I^{'} (t) + \frac{R}{L} I (t) = \frac{E}{L} \frac{}{e^{\frac{R}{L} t}}$
$e^{\frac{R}{L} t} I^{'} (t) + e^{\frac{R}{L} t} \frac{1}{R C} I (t) = e^{\frac{R}{L} t} \frac{E}{l}$

Step 2

$\underset{⏟}{e^{\frac{R}{L} t} I^{'} (t) + e^{\frac{R}{L} t} \frac{R}{L} I (t)} = e^{\frac{R}{L} t} \frac{E}{L}$
$= \frac{d}{d t} (e^{\frac{R}{L} t} I (t))$
$\frac{d}{d t} (e^{\frac{R}{L} t} I (t)) = e^{\frac{R}{L} t} \frac{E}{L}$

Integrate both sides of equation $\int \frac{d}{d t} (e^{\frac{R}{L} t} I (t)) d t = \int e^{\frac{R}{L} t} \frac{E}{L} d t$

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