JEE Main 20137 April
- Exam
- JEE Main
- Year
- 2013
- Date
- 7 April
- Subject
- Physics
- Chapter
- Capacitance
JEE Main 2013 · 7 April · Physics · Capacitance
In an LCR circuit as shown below both switches are open initially....
In an circuit as shown below both switches are open initially. Now switch is closed, kept open. ( is charge on the capacitor and is capacitive time constant). Which of the following statement is correct?
Options
- AAt $\text{t} = 2 τ \text{, q} = \text{CV} \left(1 - {\text{e}}^{- 2}\right)
- BAt $\text{t} = \dfrac{τ}{2} \text{, q} = \text{CV} \left(1 - {\text{e}}^{-1}\right)
- CWork done by the battery is half of the energy dissipated in the resistor
- D\text{At t} = τ \text{, q} = \text{CV} / 2
Correct answer
A. At $\text{t} = 2 τ \text{, q} = \text{CV} \left(1 - {\text{e}}^{- 2}\right)
Step-by-step solution
\text{q} = \text{CV} \left(1 - {\text{e}}^{\text{t} / τ}\right)\text{t} = 2 τ\text{q} = \text{CV} \left(1 - {\text{e}}^{- 2}\right)