Kirchoff's voltage law says that the voltages of all the elements have to add up to the supplied voltage: the
drop across the resistor, together with the induced voltage in the inductor, must add up to V(t):
V(t) = IR + L\dot{I}.
(4.2)
A representative$^{17}$ example of a time-varying voltage is a sine-wave (hence the symbol), e.g.
V_p k \sin(2\pi ft)
(4.3)
for some frequency $f$. Then the above equation is a differential equation for the current $I(t)$. The solution$^{18}$
is
$I(t) = I_0 e^{-\frac{R}{L}t} + \frac{V_{pk}}{Z} \sin(2\pi ft - \phi),$
(4.4)
where $I_0$ is the initial current (the free parameter from the diff. eq., analogous to the constant of integration
from an integral), $Z = \sqrt{R^2 + (2\pi fL)^2}$ is the impedance$^{19}$, $\phi = \arctan(\frac{2\pi fL}{R})$ is called a phase shift, and,
crucially, $\frac{R}{L}$ is the damping rate.
In the long-run, the first term in the current dies out (hence is called transient) while the second continues,
and we get
$I(t) \approx \frac{V_{pk}}{Z} \sin(2\pi ft - \phi).$
(4.5)
$^{17}$You'll learn in advanced calculus that through Fourier analysis, all signals can be represented as sums of sine waves, so
learning what happens to sine waves actually tells us what happens to all signals.
$^{18}$You'll learn how to solve equations like this in a diff. eq. class. This one is a linear first-order non-homogeneous ordinary
diff. eq.
$^{19}$This differs from some texts' definition, but without using complex numbers, this is the best we can do.
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Exercise 4.1: Non-inducting case
Show that the $L \to 0$ limit of Eq. (4.5) is what you would expect of a circuit with no inductance, but
simply a voltage supplied as Eq. (4.3) across a resistance R.
