Question
In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S_1}$$ is closed, $${S_2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau = RC$$ is Capacitive time constant). Which of the following statement is correct ?
In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S_1}$$ is closed, $${S_2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau = RC$$ is Capacitive time constant). Which of the following statement is correct ?
A.
Work done by the battery is half of the energy dissipated in the resistor
B.
At $$t = \tau ,\,q = \frac{{CV}}{2}$$
C.
At $$t = 2\tau ,\,q = CV\left( {1 - {e^{ - 2}}} \right)$$
D.
At $$t = 2\tau ,\,q = CV\left( {1 - {e^{ - 1}}} \right)$$
Answer :
At $$t = 2\tau ,\,q = CV\left( {1 - {e^{ - 2}}} \right)$$
Solution :
Charge on he capacitor at any time t is given by
$$\eqalign{ & q = CV\left( {1 - {e^{\frac{t}{\tau }}}} \right) \cr & {\text{at}}\,t = 2\tau \cr & q = CV\left( {1 - {e^{ - 2}}} \right) \cr} $$
Charge on he capacitor at any time t is given by
$$\eqalign{ & q = CV\left( {1 - {e^{\frac{t}{\tau }}}} \right) \cr & {\text{at}}\,t = 2\tau \cr & q = CV\left( {1 - {e^{ - 2}}} \right) \cr} $$
