The amplitude of the charge oscillating in a circuit decreases exponentially as where is the charge at The time at which charge amplitude decreases to is nearly [2024]

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Solution

Q.
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 $t = 0$ $s.$ The time at which charge amplitude decreases to $0.50$ $Q_{0}$ is nearly

[Given that $R = 1.5 Ω, L = 12$ $mH, \ln (2) = 0.693$ ] [2024]

1 19.01 ms

2 11.09 ms

3 19.01 s

4 11.09 s

Ans.
(2)

Given, $R = 1.5 Ω$

$L = 12$ $mH; \ln$ $2 = 0.693$

At $t = 0, Q = Q_{0} e^{- R t / 2 L}$ ...(i)

At $t = t_{1}, Q = 0.50$ $Q_{0}$ ...(ii)

$0.50$ $Q = Q_{0} e^{- R t_{1} / 2 L}$

$\frac{1}{2} = e^{- R t_{1} / 2 L} or 2 = e^{R t_{1} / 2 L}$

Taking log of both sides, $\ln 2 = \frac{R t_{1}}{2 L} or 0.693 = \frac{1.5 t}{24 mH}$

$∴$ $t_{1} = 11.09$ $ms$

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