Question

In an oscillating series RLC circuit, with a 5.79 ? resistor and a 17.3 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

The max energy on the capacitor =q_max² / 2C

therefore, the max energy = (1/6) Q² /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

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

   t = ln(2) x (L / R) = 0.693 x (17.3 / 5.79) = 2.07 sec