If the total charge stored in the LC circuit is Q₀, then for t ≥ 0, [JEE Adv 2006 P1]

Correct Answer is : the charge on the capacitor is Q = - LC d2Qdt2d² Q/dt²dt2d2Q

the charge on the capacitor is Q = Q₀ ≤ft( π/2 + {1}{{LC}} ).

the charge on the capacitor is Q₀ ≤ft( π/2 - {1}{{LC}} ).

the charge on the capacitor is Q = - LC d² Q/dt²

the charge on the capacitor is Q = - {1}{{LC}} d² Q/dt²

If the total charge stored in the LC circuit is Q₀, then for t ≥ 0, [JEE Adv 2006 P1]

Correct Answer is : the charge on the capacitor is Q = - LC d2Qdt2d² Q/dt²dt2d2Q

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IIT JEE Advanced 2006 Paper 1

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This section contains 4 paragraphs. Based on each paragraph, there are 3 questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 21–23:

The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch

S_1

while keeping switch

S_2

open. The capacitor can be connected in series with an inductor L by closing switch

S_2

and opening

S_1

Section: Physics

Question : 23 of 120

Marks: +1, -0

If the total charge stored in the LC circuit is

Q_0

, then for t ≥ 0,

[JEE Adv 2006 P1]

the charge on the capacitor is Q = $Q_0 \cos \left( \frac{\pi}{2} + \frac{1}{\sqrt{LC}} \right)$ .
the charge on the capacitor is $Q_0 \cos \left( \frac{\pi}{2} - \frac{1}{\sqrt{LC}} \right)$ .
the charge on the capacitor is Q = - LC $\frac{d^2 Q}{dt^2}$
the charge on the capacitor is Q = - $\frac{1}{\sqrt{LC}} \frac{d^2 Q}{dt^2}$

Validate

Solution:

We have the change Q:

Q =

Q_0 \cos \omega t

\frac{d^2 Q}{dt^2}

-Q_0 \omega^2 \cos \omega t

-\omega^2 Q

-\frac{Q}{LC}

Therefore, the total change on the capacitor is

Q =

- LC \; \frac{d^2 Q}{dt^2}

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