PHYS 208 Lecture 21

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Exam next Tuesday (11/15) over Ch. 27-30

RL-Circuits

Now we add our new inductor elements to circuits

Remember that $V=L{\frac {\mathrm {d} I}{\mathrm {d} t}}$, so using kirchoff's Rule, we can use the voltage drop across a simple circuit with a battery, an inductor, and a resistor:

${\mathcal {E}}{\frac {\mathrm {d} I}{\mathrm {d} t}}-IR=0$

This differential equation looks remarkably similar to the equation for charging a capacitor:

$V-{\frac {Q}{C}}-IR=V-{\frac {Q}{C}}-R{\frac {\mathrm {d} Q}{\mathrm {d} t}}$

So the equation for the current in an RL-circuit is

${\frac {\mathcal {E}}{R}}\left(1-\mathrm {e} ^{-{\frac {R}{L}}t}\right)$

For RL time-constant $\tau =L/R$,
${\frac {\mathcal {E}}{R}}\left(1-\mathrm {e} ^{-{\frac {t}{\tau }}}\right)$

LC-Circuits

When we take a charged capacitor and put it into a circuit with an inductor, replacing $I={\frac {\mathrm {d} Q}{\mathrm {d} t}}$, the equation for the voltage becomes

$-{\frac {Q(t)}{C}}-L{\frac {\mathrm {d} ^{2}Q(t)}{\mathrm {d} t^{2}}}=0$

Recall Simple Harmonic Motion from PHYS 218, the current in this circuit will oscillate. The angular frequency is given by

$\omega ={\sqrt {\frac {1}{LC}}}$

mechanical frequency [Hz] of system given by:

$f={\frac {\omega }{2\pi }}$

Thus the harmonic equation for an LC circuit is

${\begin{aligned}q&=Q_{max}\cos {(\omega t+\phi )}\\i&=-\omega Q_{max}\sin {(\omega t+\phi )}\end{aligned}}$

Note:  $A\cos {(\omega t+\phi )}=\alpha \sin {\omega t}+\beta \cos {\omega t}$ (alternate forms of each other)

Energy in Circuit

$U_{E}={\frac {1}{2}}{\frac {Q_{max}^{2}}{C}}={\frac {1}{2}}LI_{max}^{2}=U_{B}$

Conservation of energy between conversion from electric field to magnetic field.

LRC Damping Circuit

If we put a resistor into the circuit, it dissipates power, producing a damping effect:

$-L{\frac {\mathrm {d} ^{2}Q}{\mathrm {d} t^{2}}}-R{\frac {\mathrm {d} Q}{\mathrm {d} t}}-{\frac {Q}{C}}=0$

The solution for a damped differential equation is

${\begin{aligned}q=&A\mathrm {e} ^{-{\frac {R}{2L}}t}\cos {(\omega 't+\phi )}\\\omega '&={\sqrt {{\frac {1}{LC}}-{\frac {R^{2}}{4L^{2}}}}}\end{aligned}}$