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In an LC circuit shown in figure, C = 1 F, L= 4H. At time t = 0, charge in the capacitor is 4 C and it is decreasing at the rate of . $\sqrt{5} C s^{- 1}$ . Choose the correct statement.

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Maximum charge in the capacitor can be 6 C

Charge in the capacitor will be maximum after time $3 \sin^{- 1} (2 / 3) s$

Maximum charge in the capacitor can be 8 C

None of these

answer is A.

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Detailed Solution

$i = \sqrt{5} A$

$\frac{q_{m}^{2}}{2 C} = \frac{q^{2}}{2 C} + \frac{1}{2} L i^{2} \Rightarrow q_{m a x} = 6 C$

$q = Q \cos (ω t + δ), a t t = 0, q = 4$

$\Rightarrow 4 = 6 \cos δ \Rightarrow \cos δ = \frac{2}{3} \Rightarrow δ = \cos^{- 1} (\frac{2}{3})$

Charge on the capacitor is maximum at t = t₁
for this $ω t_{1} + δ = π$

$\Rightarrow t_{1} = \frac{1}{ω} (π - δ) = \sqrt{L C} (π - δ) = 2 [π - \cos^{- 1} (\frac{2}{3})]$

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