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MHT CET · Physics · Alternating Current

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

  1. A \(\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}\)
  2. B \(\frac{\mathrm{C}\left(\mathrm{V}_1^2+\mathrm{V}_2^2\right)}{\mathrm{L}}\)
  3. C \(\left[\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)^{\frac{1}{2}}}{\mathrm{~L}}\right]\)
  4. D \(\left[\frac{\mathrm{C}\left(\mathrm{V}_1-\mathrm{V}_2\right)^{\frac{1}{2}}}{\mathrm{~L}}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)^{\frac{1}{2}}}{\mathrm{~L}}\right]\)

Step-by-step Solution

Detailed explanation

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 \(I=\left[\frac{\mathrm{c}\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right)^{1 / 2}}{\mathrm{~L}}\right]\)