The amplitude of the charge oscillating in a circuit decreases exponentially as $Q=Q_0 e^{-R t / 2 L

Question

The amplitude of the charge oscillating in a circuit decreases exponentially as $Q=Q_0 e^{-R t / 2 L}$, where $Q_0$ is the charge at $t=0 \mathrm{~s}$. The time at which charge amplitude decreases to $0.50 Q_0$ is nearly [Given that $R=1.5 \Omega, L=12 \mathrm{mH}, \ln (2)=0.693$ ]

Professor PC Thomas Classes · Answer

(b) : Given, $R=1.5 \Omega$ \ $ L=12 \mathrm{mH} ; \ln 2=0.693 $ \ At $t=0, Q=Q_0 e^{-R t / 2 L}$ At $t=t_1, Q=0.50 Q_0$ $0.50 Q=Q_0 e^{-R 11 / 2 L}$ $\frac{1}{2}=e^{-R t_1 / 2 L} \quad$ or $\quad 2=e^{R t_1 / 2 L}$ \ Taking $\log$ both sides, $\ln 2=\frac{R t_1}{2 L}$ or $0.693=\frac{1.5 t}{24 \mathrm{mH}}$ $	herefore \quad t_1=11.09 \mathrm{~ms}$