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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
Related Questions: - For high frequency a capacity offers
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Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- For high frequency a capacity offers
- The p-type semiconductor is
- Which one of the following electrical meter has the smallest resistance?
- A small block of super dense material has mass 3 x 10²⁴ kg.It is situated at a height
- The weight of a body on surface of earth is 12.6 N. When it is raised to a height
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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