Current in Electric Circuit/L, R in Series/Sinusoidal EMF

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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:

$E = E_0 \sin \omega t$

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

$I = \dfrac {E_0} {\sqrt {R^2 - L^2 \omega^2} } \map \sin {\omega t - \alpha} + \paren {I_0 - \dfrac {E_0 L \omega} {R^2 + L^2 \omega^2} } e^{-R t / L}$

where $\tan \alpha = \dfrac {L \omega} R$.


Proof

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

$L \dfrac {\d I} {\d t} + R I = E_0 \sin \omega t$

defines the behaviour of $I$.

This can be written as:

$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$

$(1)$ 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:

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

and so the general solution becomes:

\(\ds e^{R t / L} I\) \(=\) \(\ds \frac {E_0} L \int e^{R t / L} \sin \omega t \rd t\)
\(\ds \) \(=\) \(\ds \frac {E_0} L \frac {e^{R t / L} \paren {\frac R L \sin \omega t - \omega \cos \omega t} } {\paren {\frac R L}^2 + \omega^2} + C\) Primitive of $e^{a x} \sin b x$
\(\ds \) \(=\) \(\ds \frac {E_0 \paren {R \sin \omega t - L \omega \cos \omega t} } {R^2 + \omega^2 L^2} e^{R t / L} + C\)


When $t = 0, I = I_0$.

\(\ds I_0\) \(=\) \(\ds E_0 \, {\frac {-L \omega} {R^2 + \omega^2 L^2} } + C\)
\(\ds \leadsto \ \ \) \(\ds C\) \(=\) \(\ds I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2}\)
\(\ds \leadsto \ \ \) \(\ds I\) \(=\) \(\ds \frac {E_0 \paren {R \sin \omega t - L \omega \cos \omega t} } {R^2 + \omega^2 L^2} + \paren {I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2} } e^{-R t / L}\)

The result follows by application of Multiple of Sine plus Multiple of Cosine: Sine Form.

$\blacksquare$


Sources