Question

The amplitude of the charge oscillating in a circuit decreases exponentially as \(Q = Q_0e^{ -Rt/2L}\) , where \(Q_0\) is the charge at t = 0 s. The time at which charge amplitude decreases to \(0.50Q_0\) is nearly:

(1) 19.01 ms

(2) 11.09 ms

(3) 19.01 s

(4) 11.09 s

Answer 1

Correct option is (2) 11.09 ms

Given \(Q=Q_0 e^{-R t / 2 L}\)

\(R=1.5 \Omega, L=12 \mathrm{mH}, \ln (2)=0.693\)

\(Q=0.5, Q_0, t=?\)

\(0.5 Q_0=Q_0 e^{-R t / 2 L}\)

\(\Rightarrow \frac{1}{2}=e^{-R t / 2 L}\)

Taking log on both sides

\(\ln \left(\frac{1}{2}\right)=\ln e^{-R t / 2 L} \)

\(\Rightarrow \ln 2=\frac{R t}{2 L}\)

\(t=\frac{2 L \ln 2}{R}=\frac{2 \times 12 \times 10^{-3} \times 0.693}{1.5}\)

\(t=11.09 \mathrm{~ms}\)