| d²q | + | R | dq | + | 1 | q = 0. | |
| dt² | L | dt | LC |
The above equation has two solutions according as R² / 4L² is greater or less than 1/LC. In the first case the current i in the circuit can be expressed by the equation
| i = Q | α² + β² | e−αt (eβt − e−βt), |
| 2β |
where α = R/2L, β = √(R²/4L² − 1/LC), Q is the value of q when t = 0, and e is the base of Napierian logarithms; and in the second case by the equation
| i = Q | α²+β² | e−αt sin βt |
| β |
where
| α = R/2L, and β = √ | 1 | − | R² | . |
| LC | 4L² |
These expressions show that in the first case the discharge current of the jar is always in the same direction and is a transient unidirectional current. In the second case, however, the current is an oscillatory current gradually decreasing in amplitude, the frequency n of the oscillation being given by the expression
| n = | 1 | √ | 1 | − | R² | . |
| 2π | LC | 4L² |
In those cases in which the resistance of the discharge circuit is very small, the expression for the frequency n and for the time period of oscillation R take the simple forms n = 1, 2π √LC, or T = 1/n = 2π √LC.