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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 to V₂ is
Options
(a) [ C (V₁² – V₂²) / L ]¹/²
(b) [ C (V₁ – V₂)² / L ]¹/²
(c) C (V₁² – V₂²) / L
(d) C (V₁ – V₂) / L
Correct Answer:
[ C (V₁² – V₂²) / L ]¹/²
Explanation:
q = CV₁ cos wt
i = dq/ dt = – ?Cv₁ sin ?t
Also, ?ᅡᄇ = 1 / LC and V = V₁ cos ?t
At t = t₁, V = V₂ and i = – ?CV₁ sin ?t₁
cos ?t₁ = V₂ / V₁ (-ve sign gives direction)
Hence, i = V √(C/L) [ 1 – V₂² / V₁²) ]¹/²
= [ C (V₁² – V₂²) / L ]¹/²
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Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If force(F),velocity(v) and time(T) are taken as units, then the dimensions of mass are
- The instantaneous angular position of a point on a rotating wheel is given by
- A thin wire of resistance 4 ohm is bent to form a circle. The resistance across
- The ionisation potential of hydrogen-atom is -13.6 eV. An electron in the groundstate
- Three cells, each of e.m.f 1.5 V and internal resistance 1 ohm are connected in parallel
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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