A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is then discharg

Question

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is then discharged by connecting it directly across a 100-k\Omega100−kΩ resistor. At what point in time after the resistor is connected will the capacitor have 13.5~\mu C13.5 μC of charge remaining on it?

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Thiên Ân 5 months 2021-09-04T10:43:44+00:00 1 Answers 17 views 0

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    0
    2021-09-04T10:45:10+00:00

    Explanation:

    The given data is as follows.

           C = 20 \times 10^{-6} F

            R = 100 \times 10^{3} ohm

            Q_{o} = 100 \times 10^{-6} C

              Q = 13.5 \times 10^{-6} C

    Formula to calculate the time is as follows.

              Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}]

           13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]

                   0.135 = e^{\frac{-t}{2}}

             e^{\frac{t}{2}} = \frac{1}{0.135}

                             = 7.407

               \frac{t}{2} = ln (7.407)

                          t = 4.00 s

    Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

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