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A capacitor of capacitance 2 \mu \mathrm{F} is charged to a potential difference of 12V. It is then connected across an inductor of inductance 0.6 mH. What is the current in the circuit at a time when the potential difference across the capacitor is 6.0V?

Option: 1

0.5A


Option: 2

0.6A


Option: 3

0.7A


Option: 4

0.8A


Answers (1)

best_answer

As the capacitor is charged to a p.d. of 12V, the initial charge on the capacitor is

\mathrm{\mathrm{q}_0=\mathrm{C} \mathrm{V}_0=2 \times 10^{-6} \times 12 \text { Coul. }}          ....[1]

At any instant as the capacitor discharges through the inductor (LC circuit), the instantaneous charge on the capacitor is given by

\mathrm{q=q_0 \cos \omega t}              ....[2]                    \mathrm{\text { [because at } t=0, q=q_0 \text { ] }}

But q = CV . . . . (3)

where V is the p.d. at the instant ‘t’.

From (1) and (3) we obtain \mathrm{\frac{q}{q_0}=\frac{V}{V_0}}

Putting the value of V and Vo we obtain

\mathrm{\begin{aligned} & \frac{\mathrm{q}}{\mathrm{q}_0}=\frac{1}{2} \quad \Rightarrow \cos \omega \mathrm{t}=\frac{1}{2} \\ & \Rightarrow \omega \mathrm{t}=\cos ^{-1}\left(\frac{1}{2}\right) \\ & \Rightarrow \omega \mathrm{t}=\pi / 3 \text { rad. } \end{aligned}}           ......[4]

\mathrm{\begin{aligned} & \text { Here } \omega=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\left[0.6 \times 10^{-3} \times 2 \times 10^{-6}\right]^{\frac{1}{2}}} \\ & \Rightarrow \quad \omega=\frac{10^5}{2 \sqrt{3}} \mathrm{rad} / \mathrm{sec} . \end{aligned}}...[5]

The current through the circuit at that instant is given by,

\mathrm{\begin{aligned} & \mathbf{i}=\frac{\mathrm{dq}}{\mathrm{dt}} \\ & \mathrm{i}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{q}_0 \cos \omega \mathrm{t}\right] \\ & \Rightarrow \mathrm{i}=-\mathrm{q}_0 \omega \sin \omega \mathrm{t} \\ & \Rightarrow|\mathrm{i}|=\mathrm{q}_0 \omega \sin \omega \mathrm{t} \end{aligned}}

Putting the value of q_0 from (1), \omega from (5) and  \omega t from (4) we obtain.

\mathrm{|\mathrm{i}|=2 \times 10^{-7} \times 12 \times \frac{10^5}{2 \sqrt{3}} \sin (\pi / 3) \Rightarrow|\mathrm{i}|=0.6 \mathrm{~A} .}

Posted by

shivangi.shekhar

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