JEE Advanced2006PhysicsAlternating CurrentActual
Paragraph: The capacitor of capacitance C can be charged (with the help of a resistance R ) by a voltage source V , by closing switch S₁ while keeping switch S₂ open. The capacitor can be connected in series with an inductor L by closing switch S₂ and opening S₁ . Question: Initially, the capacitor was uncharged. Now, switch S₁ is closed and S₂ is kept open. If time constant of this circuit is , then
Options
- Aafter time interval , charge on the capacitor is C V / 2
- Bafter time interval 2 , charge on the capacitor is C V / (1-e⁻² )
- Cthe work done by the voltage source will be half on the heat dissipated when the capacitor is fully charged
- Dafter time interval 2 , charge on the capacitor is C V (1-e⁻¹ )
Correct answer
B. after time interval 2 , charge on the capacitor is C V / (1-e⁻² )
Step-by-step solution
Charge on capacitor at time t is q=q₀ (1-e^ -t / ) array ll Here, & q₀=C V and t=2 & q=C V (1-e^ -2 / )=C V (1-e⁻² ) array