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

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:
 * $E = E_0 \sin \omega t$

The electric current $I$ in $K$ is given by the equation:
 * $I = \dfrac {E_0} {\sqrt {R^2 - L^2 \omega^2} } \sin \left({\omega t - \alpha}\right) + \left({I_0 - \dfrac {E_0 L \omega} {R^2 + L^2 \omega^2} }\right) e^{-R t / L}$

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

Proof
From Electric Current in Electric Circuit: L, R in Series:
 * $L \dfrac {\mathrm d I} {\mathrm d t} + R I = E_0 \sin \omega t$

defines the behaviour of $I$.

This can be written as:
 * $(1): \quad \dfrac {\mathrm d I} {\mathrm d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$

$(1)$ is a linear first order ODE in the form:
 * $\dfrac {\mathrm d I} {\mathrm d t} + P \left({t}\right) I = Q \left({t}\right)$

where:
 * $P \left({t}\right) = \dfrac R L$
 * $Q \left({t}\right) = \dfrac {E_0} L e^{-k t}$

Thus:

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
 * $\dfrac {\mathrm d} {\mathrm d t} \left({e^{R t / L} I}\right) = e^{R t / L} \dfrac {E_0} L \sin \omega t$

and so the general solution becomes:

When $t = 0, I = I_0$.

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