Current in Electric Circuit/L, R in Series

From ProofWiki
Jump to navigation Jump to search

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


The electric current $I$ in $K$ is given by the linear first order ODE:

$L \dfrac {\d I} {\d t} + R I = E$


Constant EMF at $t = 0$

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

Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$.

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

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


Exponentially Decaying EMF at $t = 0$

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:

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


Sinusoidal EMF

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


Condition for Ohm's Law

Ohm's Law is satisfied by $K$ whenever the current $I$ is at a maximum or a minimum.


Minimum Current implies Increasing EMF

Let the current $I$ be at a minimum.

Then the EMF $E$ is increasing.


Maximum Current implies Decreasing EMF

Let the current $I$ be at a maximum.

Then the EMF $E$ is decreasing.


Proof

Let:

$E_L$ be the drop in electromotive force across $L$
$E_R$ be the drop in electromotive force across $R$

From Kirchhoff's Voltage Law:

$E - E_L - E_R = 0$

From Ohm's Law:

$E_R = R I$

From Drop in EMF caused by Inductance is proportional to Rate of Change of Current:

$E_L = L \dfrac {\d I} {\d t}$

Thus:

$E - L \dfrac {\d I} {\d t} - R I = 0$

which can be rewritten:

$L \dfrac {\d I} {\d t} + R I = E$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Current_in_Electric_Circuit/L,_R_in_Series&oldid=610056"