The amplitude of the charge oscillating in a circuit decreases exponentially as $Q=Q_0 e^{-R t 2 L}$, where…
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 $R=1.5 \Omega, L=12 \mathrm{mH}, \ln (2)=0.693$ ]
19.01 ms
11.09 ms
19.01 s
11.09 s
Solution
Given $Q=Q_0 e^{-R t / 2 L}$
$R=1.5 \Omega, L=12 \mathrm{mH}, \operatorname{In}(2)=0.693$
$Q=0.5 Q_0, t=$ ?
$0.5 Q_0=Q_0 e^{-R t / 2 L}$
$\Rightarrow \quad \frac{1}{2}=e^{-R t / 2 L}$
Taking log on both sides
$\ln \left(\frac{1}{2}\right)=\ln e^{-R t / 2 L}$
$\Rightarrow \ln 2=\frac{R t}{2 L}$
$t=\frac{2 L \operatorname{In} 2}{R}=\frac{2 \times 12 \times 10^{-3} \times 0.693}{1.5}$
$t=11.09 \mathrm{~ms}$