JEE MAIN - Physics (2013 (Offline) - No. 9)
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 Capacitance time constant). Which of the following statement is correct ?
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Work done by the battery is half of the energy dissipated in the resistor
$$t = \,\tau ,\,q = CV/2$$
At $$t = \,2\tau ,\,q = CV\left( {1 - {e^{ - 2}}} \right)$$
At $$t = \,2\tau ,\,q = CV\left( {1 - {e^{ - 1}}} \right)$$
Giải thích
Charge on he capacitor at any time $$t$$ is given by
$$q = CV\left( {1 - {e^{t/\tau }}} \right)$$
at $$t = 2\tau $$
$$q = CV\left( {1 - {e^{ - 2}}} \right)$$
$$q = CV\left( {1 - {e^{t/\tau }}} \right)$$
at $$t = 2\tau $$
$$q = CV\left( {1 - {e^{ - 2}}} \right)$$