Question

In the circuit shown in Fig. E26.41, both capacitors are initially charged to 45.0 V. (a) How long after closing the switch S will the potential across each capacitor be reduced to 10.0 V, and (b) what will be the current at that time?

Short answer

Answer:

(a) After$4.21\times {10}^{-3}s$ closing the switch the potential across each capacitor is reduced to .

(b) The at that time will be$0.125\text{A}$

Step-by-step solution

Concept

A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.

Adding capacitance and resistance

From the above circuit we can see that$C_1=15.0\mu F$,$C_2=20.\mu F$, and $R_1=30.0\Omega$,$R_2=50.0\Omega$the switch is closed at the time .

Now, since both the capacitors are initially charged to we are required to find the equivalent capacitance and resistance to reduce the circuit to a simple circuit. Now we know that the capacitance adds in parallel and the resistance adds in series, hence we can write the thing in:

$$\begin{gathered} R_{eq}=R_1+R_2 \\ {C}_{eq}=C_1+C_2 \end{gathered}$$

Now let us substitute the values to get:

$$\begin{gathered} R_{eq}=(30.0+50.0)\Omega \\ =80.0\Omega \\ {C}_{eq}=(15.0+20.0)\mu F \\ =35.0\mu F \end{gathered}$$

Calculating the time taken

While discharging the capacitor, the charge on the capacitor is given by:

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

Where $\tau =R_{eq}{C}_{eq}$is the time constant, $Q=C{V}_C$but and $Q=C{V}_{C0}$, we can write:

$V_C=V_{C0}{e}^{\frac{-t}{\tau }}$

Now, we need to find how long after the closing of the switch S will the potential defense across each of the capacitors reduced to , therefore we need to solve for :

$t=-\tau \ln \left(\frac{V_C}{V_{C0}}\right)$

Now, putting the values we get:

$$\begin{gathered} t=-(2.80\times {10}^{-3}s)\ln \left(\frac{10.0V}{45.0V}\right) \\ =4.21\times {10}^{-3}s \end{gathered}$$

Therefore, the time taken is:

Calculating current

The current at a time is defined by

$$\begin{gathered} I=\frac{V_{C0}}{R_{eq}} \\ =\frac{V_{C0}}{R_{eq}}{e}^{\frac{-t}{\tau }} \end{gathered}$$

Plugging in the values we get:

$$\begin{gathered} I=\frac{V_{C0}}{R_{eq}}{e}^{\frac{-t}{\tau }} \\ =\frac{45.0}{80.0}{e}^{-\left(\frac{4.21\times {10}^{-3}s}{2.80\times {10}^{-3}s}\right)} \\ =0.125A \end{gathered}$$

Therefore, the current is 0.125 A