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Oliga [24]
3 years ago
10

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 t

hen 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?
Physics
1 answer:
sertanlavr [38]3 years ago
4 0

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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E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

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E = \frac{\lambda}{2\pi \epsilon_0 r}

4 0
3 years ago
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VladimirAG [237]

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How many possible outcomes are there in rolling a spinner 1 to 8?; What is the probability of spin an even number on a 1 to 8 sp
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<h3>What is probability?</h3>

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Total number of outcomes = 8.

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brainly.com/question/13604758

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I agree the best choice is A............

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