91Ó°ÊÓ

It is known that the capacitors are connected in parallel, therefore the equivalent capacitance is given by:

$\frac{\mathbf{1}}{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{3}}}$

Given data,

C_1 =15.0pF, C_2=20.0pF, C_3=10.0pF and resistor R=25ohm. Each of the capacitor has a charge of magnitude Q=3.50nC.

Since the capacitors are connected in parallel therefore the charge over the plates also has the same magnitude, given by:

$\mathrm{Q}={\mathrm{Q}}_0{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}}$

And the stored energy is given by:

$\mathrm{U}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$

Now, from both the equation we get:

$$\begin{gathered} \mathrm{U}=\frac{{\left({\mathrm{Q}}_0{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}}\right)}^2}{2\mathrm{C}} \\ =\left(\frac{{\mathrm{Q}}_0^2}{2\mathrm{C}}\right){\mathrm{e}}^{\frac{-21}{\mathrm{RC}}} \end{gathered}$$

But,${\mathrm{U}}_0=\frac{{\mathrm{Q}}_0^2}{2\mathrm{C}}$

Thus:$\mathrm{U}={\mathrm{U}}_0{\mathrm{e}}^{\frac{-21}{\mathrm{RC}}}$

Now, if the capacitor has lost 80% of its stored energy, then the stored energy is 20% of the total energy. Or it means that

Or it can also be written as:

$\frac{1}{5}={\mathrm{e}}^{\frac{-21}{\mathrm{RC}}}$

Solving for t we get:

Since ${\mathrm{C}}_1=\frac{3}{2}{\mathrm{C}}_3$and${\mathrm{C}}_2=2{\mathrm{C}}_3$ therefore we can also write:

$$\begin{gathered} \frac{1}{\mathrm{C}}=\frac{1}{3{\mathrm{C}}_3}+\frac{1}{2{\mathrm{C}}_3}+\frac{1}{{\mathrm{C}}_3} \\ =\frac{13}{6{\mathrm{C}}_3} \end{gathered}$$

Substituting the values for t we get:

$$\begin{gathered} \mathrm{t}=-\frac{6{\mathrm{RC}}_3}{26}\mathrm{In}\left(\frac{1}{5}\right) \\ =-\frac{3{\mathrm{RC}}_3}{13}\mathrm{In}\left(\frac{1}{5}\right) \end{gathered}$$

Putting the values, we get:

Current in a circuit is given by:

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{dQ}}{\mathrm{dt}} \\ =\left(\frac{{\mathrm{Q}}_0}{\mathrm{RC}}\right){\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}} \end{gathered}$$

Putting the values, we get:

$$\begin{gathered} \mathrm{I}=\left(\frac{3.50×{10}^{-9}\mathrm{C}}{\left(25.0\mathrm{Î\copyright }\right)\left(10.0×{10}^{-12}\mathrm{F}\right)}\right){\mathrm{e}}^{-\left(\frac{9.29×{10}^{-12}\mathrm{S}}{\left(25.0\mathrm{Î\copyright }\right)\left(10.0×{10}^{-12}\mathrm{F}\right)}\right)} \\ =13.5\mathrm{A} \end{gathered}$$

Therefore, the current is:13.5 A