The amplitude of the charge oscillating in an LCR circuit decreases exponentially as $Q=Q_0 e^{-R t

Question

The amplitude of the charge oscillating in an LCR circuit decreases exponentially as $Q=Q_0 e^{-R t / 2 L}$, where $Q_0$ is the initial charge. The time at which the charge amplitude decreases to $0.25 Q_0$ is nearly [Given that $R=1.5 \Omega, L=15 \mathrm{mH}, \ln (2)=0.693$ ]

Professor PC Thomas Classes · Answer

(b) : Given, $R=1.5 \Omega$, $L=15 \mathrm{mH}$, $\ln 2=0.693$. At $t=t_1$, $Q = 0.25 Q_0 = \frac{1}{4} Q_0$. Therefore, $\frac{1}{4} = e^{-R t_1 / 2 L}$, which gives $4 = e^{R t_1 / 2 L}$. Taking natural log on both sides, $\ln 4 = \frac{R t_1}{2 L}$. Since $\ln 4 = 2 \ln 2 = 2 	imes 0.693 = 1.386$. Now, $t_1 = \frac{2 L \ln 4}{R} = \frac{2 	imes 15 	imes 10^{-3} 	imes 1.386}{1.5} = 20 	imes 10^{-3} 	imes 1.386 = 27.72 	imes 10^{-3} \mathrm{~s} = 27.72 \mathrm{~ms}$.