Question

What will be the time required to half the amplitude of charge stored by a capacitor and time for the energy stored to be halved in a series LRC circuit with L=41mH, R=0.04 ohms and C=4.7uF?

Answer 1

The underdamped solution for the capacitor charge in an RLC circuit:

$q(t)=q_0e^{Rt/(2L)}cos\omega t$

where angular frequency of the oscillations:

$\omega=\sqrt{\frac{1}{LC}-(\frac{R}{2L})^2}=\sqrt{\frac{1}{41\cdot4.7\cdot10^{-9}}-(\frac{0.04}{2\cdot41\cdot10^{-3}})^2}=\sqrt{0.005\cdot10^9-0.238}=2236$ s^-1

then:

$q_0/2=q_0e^{Rt/(2L)}cos\omega t$

$1/2=e^{0.5t}cos2236t$

using Bisection mehtod in online calculator atozmath.com, we get:

$t=1.6$ s

energy stored in a series LRC circuit:

$W=U_L+U_C=LI^2/2+q^2/(2C)\approx e^{-Rt/L}$

for energy stored to be halved:

$1/2=\frac{1}{e^{-Rt/L}}$

$t=-Lln0.5/R=-41\cdot10^{-3}\cdot ln0.5/0.04=0.71$ s