Question

Capacitor 3 in Figure 25-41ais a variable capacitor(its capacitance can be varied). Figure 25-41bgives the electric potential ${\mathit{V}}_{\mathbf{1}}$across capacitor 1 versus ${\mathit{C}}_{\mathbf{3}}$. The horizontal scale is set by ${\mathit{C}}_{\mathbf{3}\mathit{s}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{\mu F}$. Electric potential ${\mathbf{V}}_{\mathbf{1}}$approaches an asymptote of 10V as ${\mathit{C}}_{\mathbf{3}}\mathbf{\to }\mathbf{\infty }$. (a) What are the electric potential V across the battery? (b) ${\mathit{C}}_{\mathbf{1}}$, and (c) ${\mathit{C}}_{\mathbf{2}}$?

Short answer

1. The electric potential V across the battery is 10 V .
2. The value of capacitance ${\mathrm{C}}_1$is 8.0$\mathrm{\mu F}$
3. The value of capacitance ${\mathrm{C}}_2$ is 2.0$\mathrm{\mu F}$ .

Step-by-step solution

The given data

1. Capacitance value set by the horizontal scale, ${\mathrm{C}}_{3\mathrm{s}}=12\mathrm{\mu F}$
2. Electric potential approaches an asymptote,${\mathrm{V}}_1=10\mathrm{V}$

Understanding the concept of the equivalent capacitance

We need to find the equivalent capacitance for the series and parallel capacitor first. Now, using this relation and the information from the graph, we will find the voltage across the battery and the capacitance.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{{\mathrm{C}}_{\mathrm{equivalent}}}=\sum _{}\frac{1}{{\mathrm{C}}_{\mathrm{i}}}$ ...(i).

The equivalent capacitance of a parallel connection of capacitors,

${\mathrm{C}}_{\mathrm{equivalent}}\underset{}{=\sum }{\mathrm{C}}_{\mathrm{i}}$ ...(ii)

The charge stored between the plates of the capacitor, q = CV ...(iii)

(a) Calculation of the electric potential of a battery

In the given circuit diagram, capacitor and are parallel. So, the equivalent capacitance of the two capacitors is given using equation (ii) by:

$C_{23}=C_2+C_3$

Now, we have the equivalent capacitorwhich is in series with the capacitor.

So, the equivalent capacitance ofandis given by using equation (i) as:

$$\begin{gathered} \frac{1}{C_{123}}=\frac{1}{C_1}+\frac{1}{C_2+C_3} \\ =\frac{C_1+C_2+C_3}{C_1\left(C_2+C_3\right)} \end{gathered}$$ ...(iv)

Now using equation (iii), we get that$q=C_{123}V$and

$$\begin{gathered} q=q_1 \\ =C_1{V}_1 \end{gathered}$$

Thus, using the above values, we get the equation of potential as:

$$\begin{gathered} V_1=\frac{q_1}{C_1} \\ =\frac{q}{C_1} \\ =\frac{C_{123}}{C_1}V \end{gathered}$$

Substituting equation (a) in the above value of potential, we get that

$$\begin{gathered} V_1=\frac{C_1+\left(C_2+C_3\right)}{C_1\left(C_1+C_2+C_3\right)}V \\ =\frac{C_2+C_3}{C_1+C_2+C_3}V \end{gathered}$$ ...(v)

Substituting the value of $C_3\to \infty$in equation (v), we get that ( $V_1$approach 10 V in this limit as$C_3\to \infty$)

$$\begin{gathered} V_1=V \\ V=10V \end{gathered}$$

Hence, the value of the potential is 10 V.

(b) Calculation of the capacitance C1

From the above graph at $C_3=0$, the graph shows

${\mathrm{V}}_1=2.0\mathrm{V}$

Substituting these values in equation (v), we get the following equation as:

$$\begin{gathered} 2.0V=\frac{C_2+0}{C_1+C_2+0}10V \\ 2C_1+2C_2=10C_2 \\ 2C_1=8C_2 \\ {C}_1=4C_2 \end{gathered}$$ ...(vi)

From the graph we see that when $C_3=6\mu F$the voltage across $C_1$ is exactly half the battery voltage, that is

$$\begin{gathered} {\mathrm{V}}_1=\frac{10}{2} \\ =5\mathrm{V} \end{gathered}$$

Substituting these values in equation (v), we get the following equation as:

$$\begin{gathered} 5\mathrm{V}=\frac{{\mathrm{C}}_2+6\mathrm{\mu F}}{{\mathrm{C}}_1+{\mathrm{C}}_2+6\mathrm{\mu F}}10\mathrm{V} \\ {\mathrm{C}}_1+{\mathrm{C}}_2+6\mathrm{\mu F}=\left({\mathrm{C}}_2+6\mathrm{\mu F}\right)\times 2 \\ {\mathrm{C}}_1+\frac{{\mathrm{C}}_1}{4}+6\mathrm{\mu F}=2\frac{{\mathrm{C}}_1}{4}+12\mathrm{\mu F}(\mathrm{from}\mathrm{equation}(\mathrm{vi}\left)\right) \end{gathered}$$

Hence, the value of the required capacitance is 8.0 $\mathrm{\mu F}$.

(c) Calculation of the capacitance C2

Using the value of capacitance ${\mathrm{C}}_1$in equation (vi), we can get the value of capacitance as follows:

$$\begin{gathered} {\mathrm{C}}_2=\frac{8\mathrm{\mu F}}{4} \\ =2.0\mathrm{\mu F} \end{gathered}$$

Hence, the value of the capacitance is 2.0 $\mathrm{\mu F}$.