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

Theorem
Consider the electrical 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 $t$.


 * [[File: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:

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:

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

So:

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: