## Wednesday, May 09, 2012

### AIPMT Main Exam – Two Questions Involving Discharge of a Capacitor Through an Inductor

"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 V1. 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 V2 is
(1) [C(V1 V2)2/L]1/2
(2) C(V12 V22)/L
(3) C(V12 + V22)/L
(4) [C(V12 V22)/L]1/2
The above question appeared in AIPMT Main 2010 question paper.
The initial energy of the charged capacitor is ½ CV12. When the potential difference across the capacitor reduces to V2  the energy is ½ CV22. Therefore, the energy lost by the capacitor is ½ C(V12 V22). 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 V2 we have
½ C(V12 V22) = ½ LI2
[Remember that the energy of an inductor carrying a current I is ½ LI2 and the energy is store in the magnetic field]
From the above equation we obtain
I = [C(V12 V22)/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 E1 of the charged capacitor is given by
E1 = ½ CV2 = ½ ×(6×10–6)×62
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 E1/3.
Therefore we have
E1/3 = ½ LI2
Or, (1/3) ×[½ ×(6×10–6)×62] = ½ ×(5×10–3)I2
This gives I = 0.12 A