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:

$Q=Q_0{e}^{\frac{-t}{RC}}$

And the stored energy is given by:

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

Now, from both the equation we get:

$$\begin{gathered} U=\frac{{\left(Q_0{e}^{\frac{-t}{RC}}\right)}^2}{2C} \\ =\left(\frac{{Q_0}^2}{2C}\right)e^{\frac{-2t}{RC}} \end{gathered}$$

But,$U_0=\frac{{Q_0}^2}{2C}$

Thus:

$U=U_0{e}^{\frac{-2t}{RC}}$

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

$$\begin{gathered} U=20\%U_0 \\ =\frac{U_0}{5} \end{gathered}$$

Or it can also be written as:

$\frac{1}{5}=e^{-\frac{2t}{RC}}$

Solving for t we get:

$t=-\frac{RC}{2}\ln \left(\frac{1}{5}\right)$

Since$C_1=\frac{3}{2}{C}_3$ and$C_2=2C_3$ therefore we can also write:

$$\begin{gathered} \frac{1}{C}=\frac{2}{3C_3}+\frac{1}{2C_3}+\frac{1}{C_3} \\ =\frac{13}{6C_3} \end{gathered}$$

Substituting the values for t we get:

$$\begin{gathered} t=-\frac{6R{C}_3}{26}\ln \left(\frac{1}{5}\right) \\ =-\frac{3R{C}_3}{13}\ln \left(\frac{1}{5}\right) \end{gathered}$$

Putting the values, we get:

$$\begin{gathered} t=-\frac{3\left(25.0\mathrm{Î\copyright }\right)\left(10.0×{10}^{-12}F\right)}{13}\ln \left(\frac{1}{5}\right) \\ =9.29×{10}^{-12}s \end{gathered}$$

Current in a circuit is given by:

$$\begin{gathered} I=\frac{dQ}{dt} \\ =\left(\frac{Q_0}{RC}\right)e^{\frac{-t}{RC}} \end{gathered}$$

Putting the values, we get:

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

Therefore, the current is: 13.5 A