Question

Solve the general initial value problem modeling the LR circuit, frac{dl}{dt}+ RI=E, I(0)= I_{0}, where E is a constant source of emf.

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asked 2020-11-20
Solve the general initial value problem modeling the LR circuit, \(\displaystyle{\frac{{{d}{l}}}{{{\left.{d}{t}\right.}}}}+{R}{I}={E},{I}{\left({0}\right)}={I}_{{{0}}}\), where E is a constant source of emf.

Answers (1)

2020-11-21

Step 1

Let our equation be \(\displaystyle{L}{I}'{\left({t}\right)}+{R}{I}{\left({t}\right)}={E}\ldots{\left(\cdot\right)}\) where are I - moving charge L - inductor R - resistor E - const First, divide both sides of equation with L

\(\displaystyle{L}{I}{\left({t}\right)}'+{R}{I}{\left({t}\right)}=\frac{{E}}{:}{L}\)
\(\displaystyle{I}'{\left({t}\right)}+{\frac{{{R}}}{{{L}}}}{I}{\left({t}\right)}={\frac{{{E}}}{{{L}}}}\)

We have first order linear differential equation. To solve her, we must find integration factor \(\displaystyle\mu{\left({t}\right)}\). First, let's define function a(t) \(\displaystyle{a}{\left({t}\right)}={\frac{{{R}}}{{{L}}}}\)

We will get integration factor using next formula

\(\displaystyle\mu{\left({t}\right)}={e}^{{\int{a}{\left({t}\right)}{\left.{d}{t}\right.}}}={e}^{{\int{\frac{{{R}}}{{{L}}}}{\left.{d}{t}\right.}}}={e}^{{{\frac{{{R}}}{{{L}}}}{t}}}\)

Now, multiply both sides of our equation with integration factor

\(\displaystyle{I}'{\left({t}\right)}+{\frac{{{R}}}{{{L}}}}{I}{\left({t}\right)}=\frac{{\frac{{{E}}}{{{L}}}}}{{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}}\)
\(\displaystyle{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}'{\left({t}\right)}+{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{\frac{{{1}}}{{{R}{C}}}}{I}{\left({t}\right)}={e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{\frac{{{E}}}{{{L}}}}\)

Step 2

\(\displaystyle{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}'{\left({t}\right)}+{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{\frac{{{R}}}{{{L}}}}{I}{\left({t}\right)}={e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{\frac{{{R}}}{{{L}}}}\)
\(\displaystyle={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}{\left({t}\right)}\right)}\)
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}{\left({t}\right)}\right)}={e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{\frac{{{E}}}{{{L}}}}\)

Integrate both sides of equation

\(\displaystyle\int{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}{\left({t}\right)}\right)}{\left.{d}{t}\right.}=\int{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{\frac{{{E}}}{{{L}}}}{\left.{d}{t}\right.}\)
\(\displaystyle{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}{\left({t}\right)}={\frac{{{E}}}{{{L}}}}\int{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}{\left({t}\right)}={\frac{{{E}}}{{{L}}}}{\left({\frac{{{L}}}{{{R}}}}{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}+{C}_{{{1}}}\right)},{C}_{{{1}}}=\text{const}\)
\(\displaystyle{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}{\left({t}\right)}={\frac{{{E}}}{{{R}}}}{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}\)

Divide all equation with

\(\displaystyle{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}\)

\(\displaystyle{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{I}{\left({t}\right)}={\frac{{{E}}}{{{R}}}}{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}+{\frac{{{E}}}{{{L}}}}\frac{{C}_{{{1}}}}{:}{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}\)
\(\displaystyle{I}{\left({t}\right)}={\frac{{{E}}}{{{R}}}}{e}^{{{\frac{{{R}}}{{{L}}}}{t}}}{e}^{{-{\frac{{{R}}}{{{L}}}}{t}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}{e}^{{-{\frac{{{R}}}{{{L}}}}{t}}}\)
\(\displaystyle{I}{\left({t}\right)}={\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}{e}^{{-{\frac{{{R}}}{{{L}}}}{t}}}\ldots\cdot\)

Step 3

Now, let's find solution who satisfying our initial value condition. Let's compute constant \(C_{1}\). Our initial value condition is \(\displaystyle{I}{\left({0}\right)}={I}{\left({0}\right)}\)

From the * we have

\(\displaystyle{I}{\left({t}\right)}={\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{C}_{{1}}{e}^{{-{\frac{{{R}}}{{{L}}}}{t}}}\)
\(\displaystyle\Rightarrow\)
\(\displaystyle{I}{\left({0}\right)}={\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}{e}^{{-{\frac{{{R}}}{{{L}}}}\cdot{0}}}\)
\(\displaystyle{I}_{{{0}}}={\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}{e}^{{{0}}}\)
\(\displaystyle{I}_{{{0}}}={\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}\)
\(\displaystyle{I}_{{{0}}}-{\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}\)

Multiply all equation with \(\displaystyle{\frac{{{E}}}{{{L}}}}\)

\(\displaystyle{I}_{{{0}}}-{\frac{{{E}}}{{{R}}}}={\frac{{{L}}}{{{E}}}}\frac{{C}_{{{1}}}}{\cdot}{\frac{{{L}}}{{{E}}}}\)
\(\displaystyle{\left({I}_{{{0}}}-{\frac{{{E}}}{{{R}}}}\right)}\cdot{\frac{{{L}}}{{{E}}}}={C}_{{{1}}}{\frac{{{E}}}{{{L}}}}{\frac{{{L}}}{{{E}}}}\)
\(\displaystyle{I}_{{{0}}}{\frac{{{L}}}{{{E}}}}-{\frac{{{E}}}{{{R}}}}{\frac{{{L}}}{{{E}}}}={C}_{{{1}}}\)
\(I_{0} \frac{L}{E}-\frac{L}{R}=C_{1} \Rightarrow\)

We can compute our solution:

\(\displaystyle{I}{\left({t}\right)}={\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{C}_{{{1}}}{e}^{{-{\frac{{{R}}}{{{L}}}}{t}}}={\frac{{{E}}}{{{R}}}}+{\frac{{{E}}}{{{L}}}}{\left[{I}_{{{0}}}{\frac{{{L}}}{{{E}}}}-{\frac{{{L}}}{{{R}}}}\right]}{e}^{{-{\frac{{{R}}}{{{L}}}}{t}}}\)

Answer: \(\displaystyle{I}{\left({t}\right)}={\frac{{{E}}}{{{R}}}}+{\left[{I}_{{{0}}}-{\frac{{{E}}}{{{R}}}}\right]}{e}^{{-{\frac{{{R}}}{{{L}}}}{t}}}\)

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