91Ó°ÊÓ

• a)The capacitance of the capacitor,$C=0.220\mathrm{μ}F=0,220×{10}^{-6}F.$.
• b)The potential difference of capacitor,$V_0=5.00V$.
• c)The decrease in the potential difference of the plates,$V=0.800V.$ .
• d)Discharging time value range,$t=10×{10}^{-6}sto6.0×{10}^{-2}s$.

The simple concept of charging and discharging a given capacitor also gives us the idea that the potential drop for the same capacitor is directly dependent on the charge separation value at a given time. Again, a higher resistance implies a longer discharging time and vice-versa.

Formula:

The potential difference across a RC connection, $V=V_0\left(e^{-t/RC}\right)$ (i)

Now, using the given data in equation (i), the lower value of the resistance R range of the resistor can be given for the shorter discharging time as follows:

$R=\frac{t}{CIn(V_0/V)}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} R=\frac{10×{10}^{-6}s}{\left(0.0220×{10}^{-6}F\right)In\left(5.0V/0.800V\right)} \\ =24.8\mathrm{Î\copyright } \end{gathered}$$

Hence, the value of lower resistance is $24.8\mathrm{Î\copyright }$.

Similarly, using the given data in equation (i), the higher value of the resistance range of the resistor can be given for the shorter discharging time as follows:

$R=\frac{t}{CIn\left(V_0/V\right)}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} R=\frac{6×{10}^{-2}s}{\left(0.0220×{10}^{-6}F\right)In\left(5.0V/0.800V\right)} \\ =1.49×{10}^4\mathrm{Î\copyright } \end{gathered}$$

Hence, the value of lower resistance is$1.49×{10}^4\mathrm{Î\copyright }$ .