91Ó°ÊÓ

Electromotive force (Emf) is the total amount of electrical potential energy a battery can supply before it is connected to an external circuit

Given,

Firstly, we need to find the emf of the battery when the current has a magnitude of$\mathrm{l}=3.00\mathrm{A}$ and the charge on the capacitor is$\mathrm{Q}=40.0×{10}^{-6}\mathrm{C}$

Therefore, the charge over the capacitor is given by the equation:

$\mathrm{Q}=\mathrm{°äÎ\mu }(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{Ï„}}})$

Now, the current in the circuit is given by:

Dividing the above two equations we get:

$\frac{\mathrm{Q}}{\mathrm{A}}=\mathrm{CR}\left(\frac{1-{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{Ï„}}}}{{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{Ï„}}}}\right)$

Multiply by $\frac{{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{Ï„}}}}{{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{Ï„}}}}$ we get:

$\frac{\mathrm{Q}}{\mathrm{I}}=\mathrm{CR}\left({\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{Ï„}}}-1\right)$

Solving for t we get:

$\mathrm{t}=\mathrm{Ï„\pm õ²Ô}\left(\frac{\mathrm{Q}}{\mathrm{ICR}}+1\right)$

Substituting the values, we get,

$$\begin{gathered} \mathrm{t}=(6.00×{10}^{-5}\mathrm{S})\mathrm{In}\left(\frac{40.0×{10}^{-6}\mathrm{C}}{(3.00\mathrm{A})(12.0\mathrm{Î\copyright })(5.00×{10}^{-6}\mathrm{F})}+1\right) \\ =1.20×{10}^{-5}\mathrm{S} \end{gathered}$$

Now, substitute the values in,

$\mathrm{Î\mu }={\mathrm{IRe}}^{\frac{-\mathrm{t}}{\mathrm{Ï„}}}$

We get,

Therefore, the emf is 44.0 V

From the above parts, we see that the current in the circuit has a magnitude of I = 3.00 A and the charge on the capacitor is$\mathrm{Q}=40.0×{10}^{-6}\mathrm{C}$ , now the time at which the circuit has the above magnitude and charge is given by:

$\mathrm{t}=1.20×{10}^{-5}\mathrm{S}$

Now the energy stored is given by:

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

Therefore, the rate of energy stored in the capacitor is:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{C}}=\frac{\mathrm{QI}}{\mathrm{C}} \\ {\mathrm{P}}_{\mathrm{C}}=\frac{(40.0×{10}^{-6}\mathrm{C})(3.00\mathrm{A})}{5.00×{10}^{-6}\mathrm{F}} \\ =24.0\mathrm{W} \end{gathered}$$

Therefore, the rate of energy stored in the capacitor is 24 W

Again, the rate of energy supplied by the battery is:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{Î\mu }}=\mathrm{\pm õÎ\mu } \\ =(3.00\mathrm{A})(44.0\mathrm{V}) \\ =132\mathrm{W} \end{gathered}$$

Therefore, the energy supplied by the battery is 132 W