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

Answer:

time for maximum energy is 1.718 sec

Explanation:

Maximum energy on the capacitor is given as  $=\frac{q_{max}^2}{ 2C}$

Maximum energy $= \frac{(1/6) Q_2}{2C}$

The maximum charge is given by

                                    $q_{max} =Qe^{-Rt/2L}$

                       Or$\frac{q_{max}}{Q} = e^{-Rt/2L}$

                    Or$ln\frac{q_{max}}{Q} = \frac{-Rt}{2L}$

solving for t

$t = \frac{2L}{R}\frac{1}{2} ln( 2)$

Putting all value to get desired value

 $t = ln(2)\times \frac{L}{R}$

 $t = 0.693\times \frac{(15.2}{6.13} = 1.71 sec$