Question

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

Answer:

(a) The emf on the battery is 44.0 V

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

(c) When the current has magnitude then the energy being stored in the capacitor is 24W and supplied by the battery is 132 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,

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

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

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

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

Now, the current in the circuit is given by:

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

Calculating the emf

Dividing the above two equations we get:

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

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

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

Solving for t we get:

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

Substituting the values, we get,

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

Now, substitute the values in,

$\epsilon =I{Re}^{\frac{t}{\tau }}$

We get,

$$\begin{gathered} \epsilon =(3.00A)\times (12.0\Omega )\times e^{\left(\frac{1.20\times {10}^{-5}s}{6.00\times {10}^{-5}s}\right)} \\ =44.0V \end{gathered}$$

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

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

Now the energy stored is given by:

role="math" localid="1655816376742" ${U}_C=\frac{Q^2}{2C}$

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

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

Therefore, the rate of energy stored in the capacitor is

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

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

Therefore, the energy supplied by the battery is 132 W