Question

A condenser of capacity C is charged to a potential difference \[{{V}_{1}}\]. The plate of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential across the condenser reduces to \[{{V}_{2}}\] is:
A. \[{{\left( \dfrac{C{{\left( V_{1}^{{}}-V_{2}^{{}} \right)}^{2}}}{L} \right)}^{\dfrac{1}{2}}}\]
B. \[\dfrac{C\left( V_{1}^{2}-V_{2}^{2} \right)}{L}\]
C. \[\dfrac{C\left( V_{1}^{2}+V_{2}^{2} \right)}{L}\]
D. \[{{\left( \dfrac{C\left( V_{1}^{2}-V_{2}^{2} \right)}{L} \right)}^{\dfrac{1}{2}}}\]

Answer 1

Hint: In the question we have been given that, at first only the capacitor is connected to the condenser. Later the inductor and condenser both are connected together to the condenser. Therefore, in both instances, the energy will be varied. So, we will apply the law of conservation of energy to equate the energy at these separate instances and find the current in the inductor in relation to capacitance.
Formula Used: Law of conservation of energy,
\[{{E}_{1}}={{E}_{2}}\]

Complete step-by-step solution:
At first, only the capacitor is connected to the condenser. Therefore the energy is stored only in the capacitor due to the charging of the capacitor. This energy say \[{{E}_{1}}\] is given by,
\[{{E}_{1}}=\dfrac{C{{V}_{1}}^{2}}{2}\] ……………………. (1)
Later, when the inductor is connected, then the net energy will be distributed between both
Therefore,
The energy in inductor say \[{{E}_{i}}\], is given by
\[{{E}_{i}}=\dfrac{L{{I}^{2}}}{2}\]
Similarly, the energy in capacitor say \[{{E}_{c}}\] when the potential across the condenser is reduced \[{{V}_{2}}\], is given by
\[{{E}_{c}}=\dfrac{C{{V}_{2}}^{2}}{2}\]
Therefore, the net energy becomes
\[{{E}_{2}}\]= \[{{E}_{i}}\] + \[{{E}_{c}}\]
\[{{E}_{2}}\]= \[\dfrac{L{{I}^{2}}}{2}\]+ \[\dfrac{C{{V}_{2}}^{2}}{2}\] ……………………. (2)
Now applying energy conservation,
\[{{E}_{1}}={{E}_{2}}\]
From (1) and (2)
 We get,
\[\dfrac{CV_{1}^{2}}{2}=\dfrac{L{{I}^{2}}}{2}+\dfrac{C{{V}_{2}}^{2}}{2}\]
\[\therefore L{{I}^{2}}=C(V_{1}^{2}-V_{2}^{2})\]
On solving,
\[I=\sqrt{\dfrac{C(V_{1}^{2}-V_{2}^{2})}{L}}\]
This can also be written as,
\[I={{\left( \dfrac{C(V_{1}^{2}-V_{2}^{2})}{L} \right)}^{\dfrac{1}{2}}}\]
Therefore, the correct answer is option D.

Note: This question can be solved by using another method as given below.
The given circuit is LC circuit where voltage across capacitor is given by,
\[V={{V}_{0}}\cos \omega t\]
Where,
\[\omega =\sqrt{\dfrac{1}{LC}}\]
Also,
\[V={{V}_{1}}\cos \omega t\] ……………………. (1)
The charge on capacitor is given by,
\[q=CV\]
Therefore, substituting values
 We get,
\[q=C{{V}_{1}}\cos \omega t\]
The current is given by,
\[i=\dfrac{dq}{dt}\]
\[\therefore i=-\omega C{{V}_{1}}\sin \omega t\]
Substitute,
\[t={{t}_{1}},V={{V}_{2}}\] in equation (1)
We get,
\[{{V}_{2}}={{V}_{1}}\cos \omega {{t}_{1}}\]
\[\therefore \cos \omega {{t}_{1}}=\dfrac{{{V}_{2}}}{{{V}_{1}}}\]
Now considering only the magnitude of current,
\[i=\omega C{{V}_{1}}\sin \omega t\]
\[i=\omega C{{V}_{1}}\sqrt{1-{{\cos }^{2}}\omega {{t}_{1}}}\]
\[\therefore i=\sqrt{\dfrac{1}{LC}}\times C{{V}_{1}}\sqrt{1-{{\left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)}^{2}}}\]
Therefore,
\[i={{\left( \dfrac{C(V_{1}^{2}-V_{2}^{2})}{L} \right)}^{\dfrac{1}{2}}}\]
We get the same results using both the methods. However, the second method is lengthy and involves a lot of formulae which can be confusing.