The amplitude of the charge oscillating in a circuit decreases exponentially as Q = Q₀ e-Rt/2L, where Q₀ is the charge at t = 0 s. The time at which charge amplitude decreases to 0.50Q₀ is nearly: [Given that R = 1.5 Ω, L = 12 mH, ln (2) = 0.693 ] (a) 19.01 ms (b) 11.09 ms (c) 19.01 s (d) 11.09 s

Answer: (b). 0.5Q₀ = Q₀ e-Rt/2L ⇒ Rt/2L = ln 2 t = (2Lln 2)/(R) = 2 × 12×10⁻³ × 0.6931.5 t = 1.109×10⁻² s So the answer is (b) 11.09 ms.

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LC Oscillations

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 $t = 0\text{ s}$ . The time at which charge amplitude decreases to $0.50Q_0$ is nearly: [Given that $R = 1.5\,\Omega,\ L = 12\text{ mH},\ \ln(2) = 0.693$ ]

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Answer: (b)

$0.5Q_0 = Q_0 e^{-Rt/2L} \Rightarrow \frac{Rt}{2L} = \ln 2$
$t = \frac{2L\ln 2}{R} = \frac{2 \times 12\times10^{-3} \times 0.693}{1.5}$
$t = 1.109\times10^{-2}\text{ s}$
So the answer is (b) $11.09\text{ ms}$ .

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