Question

In Fig. 25-50, the battery potential difference Vis 10.0 Vand each of the seven capacitors has capacitance $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu F}$.What is the charge on (a) capacitor 1 and (b) capacitor 2?

Short answer

1. The charge on capacitor 1 is $100\mathrm{\mu C}$.
2. The charge on capacitor 2 is$20\mathrm{\mu C}$ .

Step-by-step solution

The given data

1. The potential difference of the battery, V = 10 V
2. Capacitance of each capacitor from the seven capacitors, C = 10$\mathrm{\mu F}$

Understanding the concept of the equivalent capacitance and charge

The potential difference on each capacitor is the same in the parallel combination but the charge on the capacitors may be different. In a series combination, the charges on the capacitor are the same but the potential difference may be different. By using the relation between charge and capacitance, and the concept of equivalent capacitance, we can find the charge on both the capacitors.

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,

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

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

(a) Calculation of charge on capacitor 1

The potential difference on each capacitor is the same in parallel combination. If the potential across the capacitor 1 is V , the charge can be calculated as using equation (iii) as:

$$\begin{gathered} {\mathrm{q}}_1=\left(10\mathrm{\mu F}\right)\times \left(10\mathrm{v}\right)\begin{array}{c}\\ \end{array} \\ =100\mathrm{\mu C} \end{gathered}$$

Hence, the value of the charge is$100\mathrm{\mu C}$ .

(b) Calculation of the charge on capacitor 2

Since${\mathrm{C}}_2$is in series with another capacitor of the same capacitance, we find the equivalent capacitance using equation (i) as follows:

$$\begin{gathered} \mathrm{C}\text{'}=\frac{{\mathrm{C}}_2\mathrm{C}}{{\mathrm{C}}_2+\mathrm{C}} \\ =\frac{100{\left(\mathrm{\mu F}\right)}^2}{20\mathrm{\mu F}} \\ =5\mathrm{\mu F} \end{gathered}$$

This combination is in parallel with another$10\mathrm{\mu F}$capacitor. So, we can calculate the equivalent capacitancefor the parallel combination by using equation (ii) as follows:

$$\begin{gathered} \mathrm{C}"=\mathrm{C}\text{'}+\mathrm{C} \\ =5\mathrm{\mu F}+10\mathrm{\mu F} \\ =15\mathrm{\mu F} \end{gathered}$$

Again this combination is in series with another$10\mathrm{\mu F}$capacitor, thus the equivalent capacitance is given using equation (i) as follows:

Thus,

$$\begin{gathered} \mathrm{C}\text{'}\text{'}\text{'}=\frac{\mathrm{C}"\mathrm{C}}{\mathrm{C}"+\mathrm{C}} \\ =\frac{15\mathrm{\mu F}\times 10\mathrm{\mu F}}{25\mathrm{\mu F}} \\ =6\mathrm{\mu F} \end{gathered}$$

So for the bottom part, the equivalent capacitance is $\mathrm{C}"=6\mathrm{\mu F}$

The charge on the bottom loop can be calculated using equation (iii) as follows:

$$\begin{gathered} \mathrm{q}=6\mathrm{\mu F}\times 10\mathrm{V} \\ =60\mathrm{\mu C} \end{gathered}$$

In the bottom loop,$C_2$ is in series with the other capacitor, so they both must have the same charge.
So, the potential across the bottom right capacitor is given using equation (iii) as:

$$\begin{gathered} \mathrm{V}=\frac{60\mathrm{\mu C}}{10\mathrm{\mu F}} \\ =6\mathrm{V} \end{gathered}$$

So, the potential across the group of capacitors in the upper right part is given by,

$$\begin{gathered} \mathrm{V}=10\mathrm{V}-6\mathrm{V} \\ =4\mathrm{V} \end{gathered}$$

This potential must be equally divided by the capacitorand the other capacitor below it.

So, the potential cross capacitor 2 is${\mathrm{V}}_2=2\mathrm{V}$

Therefore, the charge on capacitoris given using equation (iii) as follows:

$$\begin{gathered} {\mathrm{q}}_2=10\mathrm{\mu F}\times 2\mathrm{V} \\ =20\mathrm{\mu C} \end{gathered}$$

Hence, the value of the charge is$20\mathrm{\mu C}$ .