In an LC circuit shown in figure, C = 2 F and L = 2H. At time t = 0, charge on the capacitor is 3 coulomb and it is decreasing with rate of $\sqrt{4}$ coulomb $\sec^{- 1}$ . Then choose the correct statement.

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maximum charge on the capacitor can be 6C.

maximum charge on the capacitor can be 4C.

charge on the capacitor will be maximum after time $\frac{127 π}{90} \sec$

charge on the capacitor will be maximum after time $\frac{153 π}{90} \sec$

answer is C.

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Detailed Solution

$\begin{array}{l} i = \sqrt{4} amp \\ \frac{q^{2}}{2 C} - \frac{q^{2}}{2 C} + \frac{1}{2} {LC}^{2} \\ q_{m}^{2} = 9 + 4 (\sqrt{4})^{2} \\ q_{m} = 5 C \\ q = Qcos (ωt + δ) \\ at t = 0 \\ 3 = 5 \cos δ \Rightarrow δ = 53^{\circ} \end{array}$

Let at t = t₁, charge on capacitor is maximum

$\begin{array}{l} {ωt}_{1} + δ = π \\ t_{1} = (\frac{π - δ}{ω}) = (\sqrt{LC}) (π - δ) \\ = \frac{2 (180 - 53^{\circ}) π}{180} = (\frac{127 π}{90}) \sec \end{array}$