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A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.

Capacitance C = 5.90 mF

詯 = 28.0V

Emf has negligible resistance.

Calculate the charge on the capacitor for different positions of switches.

(a)

for the charge on the capacitor, a long time after S is moved to position 2.

The capacitor is charged by the battery in a series resistor when the switch is in position 2.

Charge on the capacitor, a long time after S moves,

$$\begin{gathered} Q_f=\left(\text{28.0}V\right)\left(\text{5.90}脳{\text{10}}^{\text{- 6}}F\right) \\ {Q}_f=\text{165.2}脳{\text{10}}^{\text{- 6}}C\left(\frac{{\text{10}}^{\text{- 6}}C}{\text{1}渭C}\right) \\ {Q}_f=\text{165.2}渭C \end{gathered}$$

(b)

forthe value of the resistance R.

(1)

Substituting the given data in the above equation we get,

Thus, the value of the resistance R is

(c)

for the charge on the capacitor to be equal to 99.0% of the final value found in part (a).

Time is given as,

$t=-RC脳ln\left(1-\frac{q}{Q_f}\right)$ (2)

for the capacitor value = 99.0% ,the value of part (a) is,

$\frac{q}{Q_f}=\text{0.99}$

Substitute these values in equation (2), and we get

$$\begin{gathered} t=-\left(\text{463}惟\right)\left(\text{5.90}脳{\text{10}}^{\text{- 6}}F\right)脳ln\left(1-0.99\right) \\ t=\text{0.0126}s\left(\frac{1ms}{{10}^{-3}s}\right) \\ t=\text{12.6}ms \end{gathered}$$

thus, the charge on the capacitor is equal to 99.0% of the final value found in part (a) is 12.6 ms