Question

The amplitude of the charge oscillating in a circuit decreases exponentially as $Q = Q_0 e^{-Rt/2L}$, where $Q_0$ is the charge at $t=0s$. The time at which charge amplitude decreases to 0.50 $Q_0$ is nearly:
[Given that $R = 15 \Omega$, $L = 12 mH$, $ln(2) = 0.693$]

• A. 19.01 ms
• B. 11.09 ms
• C. 19.01 s
• D. 11.09 s

Answer 1

The Correct Option is B

Given: $Q = Q_0 e^{-Rt/2L}$

When $Q = 0.50 Q_0$:

$0.50 Q_0 = Q_0 e^{-Rt/2L}$

$0.50 = e^{-Rt/2L}$

$\ln(0.50) = -\frac{Rt}{2L}$

$\ln\left(\frac{1}{2}\right) = -\frac{Rt}{2L}$

$-\ln 2 = -\frac{Rt}{2L}$

\[ t = \frac{2L \ln 2}{R} \]
Given: $R = 1.5 \, \Omega$, $L = 12 \, \text{mH} = 12 \times 10^{-3} \, H$, $\ln 2 = 0.693$
\[ t = \frac{2(12 \times 10^{-3} \, H)(0.693)}{1.5 \, \Omega} \approx 11.088 \times 10^{-3} \, s \approx 11.09 \, ms \]