Current in Electric Circuit/L, R in Series/Exponentially Decaying EMF at t = 0

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Theorem

Consider the electric circuit $K$ consisting of:

a resistance $R$
an inductance $L$

in series with a source of electromotive force $E$ which is a function of time $t$.

CircuitRLseries.png


Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.

Let an EMF $E$ be imposed upon $K$ at time $t = 0$ defined by the equation:

$E = E_0 e^{-k t}$

The electric current $I$ in $K$ is given by the equation:

$I = \dfrac {E_0} {R - k L} e^{-k t} + \paren {I_0 - \dfrac {E_0} {R - k L} } e^{-R t / L}$


Proof

From Electric Current in Electric Circuit: L, R in Series:

$L \dfrac {\d I} {\d t} + R I = E_0 e^{-k t}$

defines the behaviour of $I$.

This can be written as:

$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L e^{-k t}$

$(1)$ is a linear first order ODE in the form:

$\dfrac {\d I} {\d t} + \map P t I = \map Q t$

where:

$\map P t = \dfrac R L$
$\map Q t = \dfrac {E_0} L e^{-k t}$


Thus:

\(\ds \int \map P t \rd t\) \(=\) \(\ds \int \dfrac R L \rd t\)
\(\ds \) \(=\) \(\ds \dfrac {R t} L\)
\(\ds \leadsto \ \ \) \(\ds e^{\int P \rd t}\) \(=\) \(\ds e^{R t / L}\)

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\map {\dfrac \d {\d t} } {e^{R t / L} I} = e^{R t / L} \dfrac {E_0} L e^{-k t}$

and the general solution becomes:

$\ds e^{R t / L} I = \int e^{R t / L} \dfrac {E_0} L e^{-k t} \rd t$


and so:

\(\ds e^{R t / L} I\) \(=\) \(\ds \frac {E_0} L \int e^{R t / L} e^{-k t} \rd t\)
\(\ds \) \(=\) \(\ds \frac {E_0} L \int e^{\paren {R / L - k} t} \rd t\)
\(\ds \) \(=\) \(\ds \frac {E_0} {L \paren {\frac R L - k} } e^{\paren {R / L - k} t} + C\)
\(\ds \) \(=\) \(\ds \frac {E_0} {R - k L} e^{\paren {R / L - k} t} + C\)


When $t = 0$, we have $I = I_0$.

So:

\(\ds I_0\) \(=\) \(\ds \frac {E_0} {R - k L} + C\)
\(\ds \leadsto \ \ \) \(\ds C\) \(=\) \(\ds I_0 - \frac {E_0} {R - k L}\)


So:

$I e^{R t/ L} = \dfrac {E_0} {R - k L} e^{\paren {R / L - k} t} + I_0 - \dfrac {E_0} {R - k L}$

Multiplying by $e^{\frac {R t} L}$ and tidying up, we get:

\(\ds I\) \(=\) \(\ds \frac {E_0} {R - k L} e^{- R t / L + R t / L - k t} + I_0 e^{-R t / L} - \dfrac {E_0} {R - k L} e^{-R t / L}\)
\(\ds \) \(=\) \(\ds \frac {E_0} {R - k L} e^{- k t} + I_0 e^{-R t / L} - \frac {E_0} {R - k L} e^{-R t / L}\)
\(\ds \) \(=\) \(\ds \frac {E_0} {R - k L} e^{- k t} + \paren {I_0 - \frac {E_0} {R - k L} } e^{-R t / L}\)

$\blacksquare$


Sources

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