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A condenser of capacity ' $\mathrm{C}$ ' is charged to a potential difference of ' $\mathrm{V}_1$ '. The plates of the condenser are then connected to an ideal inductor of inductance ' $L$ '. The current through an inductor |when the potential difference across the condenser reduces to ' $\mathrm{V}$ ' is
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Verified Answer
The correct answer is:
$\left[\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)^{\frac{1}{2}}}{\mathrm{~L}}\right]$
The correct option is (C).
Concept: For the LC circuit the potential drop across the circuit can be written as: $\frac{\mathrm{q}}{\mathrm{C}}+\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=0$
On rewriting, $\frac{\mathrm{q}}{\mathrm{C}}+\mathrm{L}\left(\frac{\mathrm{dq}}{\mathrm{dt}}\right) \frac{\mathrm{di}}{\mathrm{dt}}=0$
Therefore, $\frac{\mathrm{q}}{\mathrm{C}}+\mathrm{Li} \frac{\mathrm{di}}{\mathrm{dq}}=0$
On integrating, $\int_{\mathrm{cv}_1}^{\mathrm{cv}_2} \frac{\mathrm{q}}{\mathrm{c}} \mathrm{dq}=-\int_0^{\mathrm{i}}$ Lidi
On solving for $I=\left[\frac{\mathrm{c}\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right)^{1 / 2}}{\mathrm{~L}}\right]$
Concept: For the LC circuit the potential drop across the circuit can be written as: $\frac{\mathrm{q}}{\mathrm{C}}+\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=0$
On rewriting, $\frac{\mathrm{q}}{\mathrm{C}}+\mathrm{L}\left(\frac{\mathrm{dq}}{\mathrm{dt}}\right) \frac{\mathrm{di}}{\mathrm{dt}}=0$
Therefore, $\frac{\mathrm{q}}{\mathrm{C}}+\mathrm{Li} \frac{\mathrm{di}}{\mathrm{dq}}=0$
On integrating, $\int_{\mathrm{cv}_1}^{\mathrm{cv}_2} \frac{\mathrm{q}}{\mathrm{c}} \mathrm{dq}=-\int_0^{\mathrm{i}}$ Lidi
On solving for $I=\left[\frac{\mathrm{c}\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right)^{1 / 2}}{\mathrm{~L}}\right]$
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