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

(a) After$4.21\times {10}^{-3}$ closing the switch S the potential across each capacitor is reduced to 10.0 V.

(b) The at that time will be 0.125 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,${\mathrm{C}}_1=15.0\mathrm{\mu F},{\mathrm{C}}_2=20.0\mathrm{\mu F},{\mathrm{R}}_1=30.0\mathrm{\Omega }\mathrm{and}\mathrm{R}=50.0\mathrm{\Omega }$, , and the switch is closed at the time$\mathrm{t}=0$ .

Now, since both the capacitors are initially charged to${\mathrm{V}}_{\mathrm{co}}=45.0\mathrm{V}$ 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:

$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_1+{\mathrm{R}}_2 \\ {\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2 \end{gathered}$$

Now let us substitute the values to get:

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

Calculating the time taken

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

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

Where $\mathrm{\tau }={\mathrm{R}}_{\mathrm{eq}}{\mathrm{C}}_{\mathrm{eq}}$ is the time constant, but $\mathrm{Q}={\mathrm{CV}}_{\mathrm{C}}\mathrm{and}\mathrm{Q}={\mathrm{CV}}_{\mathrm{CO}}$we can write:

${\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_{\mathrm{CO}}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}}$

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 10.0 V , therefore we need to solve for t :

$\mathrm{t}=-\mathrm{\tau In}\left(\frac{{\mathrm{V}}_{\mathrm{C}}}{{\mathrm{V}}_{\mathrm{CO}}}\right)$

Now, putting the values we get:

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

Therefore, the time taken is:$4.21\times {10}^{-3}$S

Calculating current

The current at a time is defined by

$$\begin{gathered} \mathrm{I}=\frac{{\mathrm{V}}_{\mathrm{CO}}}{{\mathrm{R}}_{\mathrm{eq}}} \\ =\frac{{\mathrm{V}}_{\mathrm{CO}}}{{\mathrm{R}}_{\mathrm{eq}}}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}} \end{gathered}$$

Plugging in the values we get:

$$\begin{gathered} \mathrm{I}=\frac{{\mathrm{V}}_{\mathrm{CO}}}{{\mathrm{R}}_{\mathrm{eq}}}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}} \\ =\frac{45.0}{80.0}{\mathrm{e}}^{-\left(\frac{4.21\times {10}^{-3}\mathrm{S}}{2.80\times {10}^{-3}\mathrm{S}}\right)} \\ =0.125\mathrm{A} \end{gathered}$$

Therefore, the current is 0.125 A