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?

Answer 1

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.

Answer 2

Answer:

Capacitor is a device that can store electricity for a short period of time. After 4 seconds, the charge of the capacitor remains $\bold{13.5 \mu C}$.

Capacitor:

It is a device that can store electricity for a short period of time. It is made up of two metallic plates , separated by insulator.

Given here,

Resistance R =  100 k$\Omega$

Capacitance  C =  $\BOLD {20\times 10^3 F}$

Initial Charge $\bold { Q_o = 100 \times 10^-^6 C}$

Charge $\bold{Q_t }$ = $\bold{13.5 \times 10^-^6 C}$

The time can be calculated using the formula,

$\bold{Q_t = Q_o ( e ^\frac{-t}{2} })$

Put the values in the formula,

$\bold{13.5 \times 10^-^6 = 100 \times 10^-^6 ( e ^\frac{-t}{2} )}$

$\bold{ ( e ^\frac{-t}{2} ) = 0.135}\\\\\bold{ ( e ^\frac{t}{2} ) =\frac{1}{0.135} }\\\\\bold{ ( \frac{t}{2} ) = log ( 7.407)}\\\\\bold {t = 4.0}$

Therefore, After 4 seconds the charge of the capacitor remains $\bold{13.5 \mu C}$.

To know more capacitor, refer to the link:

https://brainly.com/question/16918033