Question

In an LCR circuit as shown in figure, both switches are open initially. Now switch S₁ is closed, S₂ kept open. (q is charge on the capacitor and $\tau$ = RC is capacitive time constant). Which of the following statement is correct?

• A. $Att=\frac{\tau }{2},q=CV\left(1-e^{-1}\right)$
• B. Work done by the battery is half of the energy dissipated in the resistor
• C. $Att=\tau ,q=CV/2$
• D. $Att=2\tau ,q=CV\left(1-e^{-2}\right)$

Answer 1

Answer: (D)

As switch S₁ is closed and switch S₂ is kept open. Now, capacitor is charging through a resistor R. Charge on a capacitor at any time t is
$q=q_0\left(1-e^{t/\tau }\right)=CV\left(1-e^{-t/\tau }\right)$
$\left[As{q}_0=CV\right]$
At t $=\frac{\tau }{2},q=CV\left(1-e^{-\tau 2/\tau }\right)=CV\left(1-e^{-1/2}\right)$
At t $=\tau ,q=CV\left(1-e^{-\tau /\tau }\right)=CV=\left(1-e^{-1}\right)$
At t $=2\tau ,q=CV\left(1-e^{-2\tau /\tau }\right)=CV\left(1-e^{-2}\right)$