91影视

A capacitor is a device that storeselectrical energyin anelectric field. It is apassiveelectronic componentwith twoterminals.

The effect of a capacitor is known ascapacitance.

From equation 26.16, in a circuit,

R - C the charge on a discharging capacitor as a function of time is given by:
$\mathrm{q}={\mathrm{Q}}_{\mathrm{f}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}$
Where Q f is the initial charge on the capasitor and T = RC is the time constant of the circuitFrom equation 24.], the capacitance C of a capacitor is the ratio of the charge to the potential difference betWeen its plates.
$\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$

Given
The initial voltage difference across the capacitor is ${\mathrm{V}}_{\mathrm{O}}=12\mathrm{V}$,

$\mathrm{R}=3.40脳{10}^6\mathrm{惟}$ the internal resistance of the voltameter is and after time the voltameter reads a voltage of : V = 12 V
Calculations
Solving equation (2) for V, we get:
$\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$

So,bydividingequation(1)byC,Wegetthevoltageacrossadischargingcapacitorasafunctionoftime:

$\mathrm{V}={\mathrm{V}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}$
$\mathrm{V}={\mathrm{V}}_{\mathrm{O}}={\mathrm{Q}}_{\mathrm{f}}/\mathrm{C}$'is the initialvoltage difference-

When we connect the capacitor to the voltameter, the capacitor (?scharges its charge in the internal resistance of the

voltameter.

And from Kirchhoff's loop rule, the voltameter is reading the voltage difference across the capacitor.:

(a)Inordertogetthecapacitanceofthecapacitor,Wesolveequation(3)forCasfollow

Now, we enter our given values, so we get:

$\frac{\mathrm{V}}{{\mathrm{V}}_{\mathrm{O}}}={\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}$

$$\begin{gathered} \frac{\mathrm{t}}{\mathrm{RC}}=\mathrm{In}\left(\frac{{\mathrm{V}}_{\mathrm{o}}}{\mathrm{V}}\right) \\ \mathrm{C}=\frac{\mathrm{t}}{\mathrm{R}\mathrm{In}\left({\mathrm{V}}_{\mathrm{o}}/\mathrm{V}\right)} \\ \mathrm{C}=\frac{4}{3.40脳{10}^6\mathrm{In}\left(12/3\right)} \\ =0.849\mathrm{uF} \\ \mathrm{T}=\mathrm{RC} \\ =\left(3.40脳{10}^6\right)\left(0.849脳{10}^{-6}\right) \\ =2.89\sec \end{gathered}$$

Therefore the capacitance is = 0.849uF and the time required is 2.89sec