"Live as if you were to die tomorrow.
Learn as if you were to live forever."

– Mahatma Gandhi

Questions appearing in AIPMT Main (Physics) question paper
are generally not as simple as those appearing in AIPMT Preliminary question paper.
Those who get qualified (by passing the preliminary examination) should bear
this in mind and prepare accordingly. I give below two questions involving the
discharge of a charged capacitor through an inductor:

(1) A condenser of capacity

*C*is charged to a potential difference of*V*_{1}. The plates of the condenser are then connected to an ideal inductor of inductance*L*. The current through the inductor when the potential difference across the condenser reduces to*V*_{2 }is
(1) [

*C*(*V*_{1 }–*V*_{2})^{2}/*L*]^{1/2}^{}
(2)

*C*(*V*_{1}^{2}_{ }–*V*_{2}^{2})/*L*
(3)

*C*(*V*_{1}^{2}_{ }+*V*_{2}^{2})/*L*
(4) [

*C*(*V*_{1}^{2}_{ }–*V*_{2}^{2})/*L*]^{1/2}^{}
The above question appeared in AIPMT Main 2010
question paper.

The
initial energy of the charged capacitor is ½

*CV*_{1}^{2}. When the potential difference across the capacitor reduces to*V*_{2}_{ }the energy is ½*CV*_{2}^{2}. Therefore, the energy lost by the capacitor is ½*C*(*V*_{1}^{2}_{ }–*V*_{2}^{2}). This amount of energy is gained by the inductor.
If

*I*represents the discharge current at the instant when the potential difference across the capacitor reduces to*V*_{2}we have
½

*C*(*V*_{1}^{2}_{ }–*V*_{2}^{2}) = ½*LI*^{2}^{}
[Remember that the energy of an inductor carrying
a current

*I*is ½*LI*^{2}and the energy is store in the magnetic field]
From the above equation we obtain

*I =*[

*C*(

*V*

_{1}

^{2}

_{ }–

*V*

_{2}

^{2})/

*L*]

^{1/2}

^{}

(2) Capacitance of 6 μF is charged by 6 V battery. Now it is
connected with inductor of 5 mH. Find the current in the inductor when 1/3rd of
total energy is magnetic.

This question was asked as half the part of a
free response (and not a multiple choice) question in the AIPMT Main 2007
question paper.

[Current practice is to ask objective type
(multiple choice) questions].

The
initial energy

*E*_{1}*of the charged capacitor is given by**E*

_{1}=

*½*

*CV*

^{2}= ½ ×(6×10

^{–6})×6

^{2}

Since one-third of total energy is mentioned as

*magnetic*in the question, we understand that the energy of the inductor carrying the discharge current*I*is*E*_{1}/3.
Therefore we have

*E*

_{1}/3 = ½

*LI*

^{2}

Or, (1/3) ×[½
×(6×10

^{–6})×6^{2}] = ½ ×(5×10^{–3})*I*^{2}
This gives

*I =*0.12 A
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