Question

You connect a battery, resistor, and capacitor as in Fig. 26.20a, where R = 12.0 Ω and C = 5.00 x 10^-6 F. The switch S is closed at t = 0. When the current in the circuit has a magnitude of 3.00 A, the charge on the capacitor is 40.0 x 10^-6 C. (a) What is the emf of the battery? (b) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10^-6 C? (c) When the current has magnitude 3.00 A, at what rate is energy being (i) stored in the capacitor, (ii) supplied by the battery

Short answer

(a) The emf on the battery is 44.0 V

(b) After$1.20\times {10}^{-5}\mathrm{S}$ the switch is closed the charge on the capacitor is equal to$40.0\times {10}^{-6}\mathrm{C}$

(c) When the current has magnitude$3.00\mathrm{A}$ then the energy being stored in the capacitor is 24 W and supplied by the battery is 32 W

Step-by-step solution

EMF of the battery

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

Current through the circuit

Given,

$\mathrm{C}=5.00\times {10}^{-6}\mathrm{F},\mathrm{R}=12.0\mathrm{\Omega }\mathrm{and}\mathrm{time}\mathrm{t}=0$

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$\mathrm{Q}=40.0\times {10}^{-6}\mathrm{C}$

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

$\mathrm{Q}=\mathrm{C\epsilon }(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}})$

Now, the current in the circuit is given by:

$$\begin{gathered} \mathrm{l}=\frac{\mathrm{dQ}}{\mathrm{dt}} \\ =\frac{\mathrm{\epsilon }}{\mathrm{R}}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}} \end{gathered}$$

Calculating the emf

Dividing the above two equations we get:

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

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

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

Solving for t we get:

$\mathrm{t}=\mathrm{\tau In}\left(\frac{\mathrm{Q}}{\mathrm{ICR}}+1\right)$

Substituting the values, we get,

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

Now, substitute the values in,

$\mathrm{\epsilon }={\mathrm{IRe}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}}$

We get,

role="math" localid="1655792248682" $$\begin{gathered} \mathrm{\epsilon }=(3.00\mathrm{A})\times (12.0\mathrm{\Omega })\times {\mathrm{e}}^{\left(\frac{1.20\times {10}^{-5}\mathrm{S}}{6.00\times {10}^{-5}\mathrm{S}}\right)} \\ =44.0\mathrm{V} \end{gathered}$$

Therefore, the emf is 44.0 V

Energy stored in both capacitors

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$\mathrm{Q}=40.0\times {10}^{-6}\mathrm{C}$ , now the time at which the circuit has the above magnitude and charge is given by:

$\mathrm{t}=1.20\times {10}^{-5}\mathrm{S}$

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:

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

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

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

$$\begin{gathered} {\mathrm{P}}_{\mathrm{\epsilon }}=\mathrm{I\epsilon } \\ =(3.00\mathrm{A})(44.0\mathrm{V}) \\ =132\mathrm{W} \end{gathered}$$

Therefore, the energy supplied by the battery is 132 W