Time domain solution

By Kirchhoff's voltage law, the voltage beyond the capacitor, VC, additional the voltage beyond the inductor, VL have to according zero:

V _{C} + V_{L} = 0.\,

Likewise, by Kirchhoff's accepted law, the accepted through the capacitor equals the accepted through the inductor:

i_{C} = i_{L} .\,

From the basal relations for the ambit elements, we aswell apperceive that

V _{L}(t) = L \frac{di_{L}}{dt}\,

and

i_{C}(t) = C \frac{dV_{C}}{dt}.\,

Rearranging and substituting gives the additional adjustment cogwheel equation

\frac{d ^{2}i(t)}{dt^{2}} + \frac{1}{LC} i(t) = 0.\,

The constant ω, the radian frequency, can be authentic as: ω = (LC)−1/2. Using this can abridge the cogwheel equation

\frac{d ^{2}i(t)}{dt^{2}} + \omega^ {2} i(t) = 0.\,

The associated polynomial is s2 +ω2 = 0, thus

s = +j \omega\,

or

s = -j \omega\,

area j is the abstract unit.

Thus, the complete band-aid to the cogwheel blueprint is

i(t) = Ae ^{+j \omega t} + Be ^{-j \omega t}\,

and can be apparent for A and B by because the antecedent conditions.

Since the exponential is complex, the band-aid represents a sinusoidal alternating current.

If the antecedent altitude are such that A = B, again we can use Euler's blueprint to access a absolute sinusoid with amplitude 2A and angular abundance ω = (LC)−1/2.

Thus, the consistent band-aid becomes:

i(t) = 2 A \cos(\omega t).\,

The antecedent altitude that would amuse this aftereffect are:

i(t=0) = 2 A\,

and

\frac{di}{dt}(t=0) = 0.\,