Question

In an oscillating series RLCcircuit, with a 6.13 Ω resistor and a 15.2 H inductor, find the time required for the maximum energy present in the capacitor during an oscillation to fall to half of its initial value.

Answer 1

time for maximum energy is 1.718 sec

Step-by-step explanation:

Maximum energy on the capacitor is given as
`=(q_(max)^2)/( 2C)`

Maximum energy
`= ((1/6) Q_2)/(2C)`

The maximum charge is given by

`q_(max) =Qe^(-Rt/2L)`

Or
`(q_(max))/(Q) = e^(-Rt/2L)`

Or
`ln(q_(max))/(Q)  = (-Rt)/(2L)`

solving for t

`t = (2L)/(R)(1)/(2) ln( 2)`

Putting all value to get desired value

`t = ln(2)* (L)/(R)`

`t = 0.693* ((15.2)/(6.13) = 1.71 sec`