91Ó°ÊÓ

1. Capacitance,$C_1=1.00\mathrm{μ¹ó}$
2. Capacitance, role="math" localid="1661746858346" $C_2=2.00\mathrm{μ¹ó}$
3. Capacitance$C_3=3.00\mathrm{μ¹ó}$
4. Capacitance $C_4=4.00\mathrm{μ¹ó}$
5. Voltage on battery, V = 12.0 V

We find the equivalent capacitance for the series and parallel combination of capacitors by using the given formula. Using the relation of charge and capacitance we find the charge on each capacitor.

Formulae:

The equivalent capacitance of a series connection of capacitors,

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

The equivalent capacitance of a parallel connection of capacitors,

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

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

When switch$S_1$is closed

In this situation capacitor andare in series, and therefore, the charges on both capacitors are same. Thus,

$$\begin{gathered} q_1=q_3 \\ =C_{13}V \end{gathered}$$

Equivalent capacitance is then given using equation (i) as follows:

$$\begin{gathered} \frac{1}{C_{13}}=\frac{1}{C_1}+\frac{1}{C_3} \\ {C}_{13}=\frac{C_1{C}_3}{C_1+C_3} \end{gathered}$$

Using the given values in equation (iii), the charge on the capacitances 1 and 3 is given as:

$$\begin{gathered} q_1=q_3 \\ =\frac{\left(1\mathrm{μ¹ó}\right)\left(3\mathrm{μ¹ó}\right)}{\left(1\mathrm{μ¹ó}\right)+\left(3\mathrm{μ¹ó}\right)}\left(12V\right) \\ =9.00\mathrm{μ}C \end{gathered}$$

Hence, the value of the charge is $9.00\mathrm{μ°ä}$.

When switch $S_1$is closed

In this situation capacitor and are in series, and therefore, the charges on both capacitors are same. Thus, using given values and equations (i) and (iii), we get the charge value as:

$$\begin{gathered} q_2=q_4 \\ =\frac{C_2{C}_4}{C_2+C_4}V \\ =\frac{\left(2\mathrm{μ¹ó}\right)\left(4\mathrm{μ¹ó}\right)}{\left(2\mathrm{μ¹ó}\right)+\left(4\mathrm{μ¹ó}\right)}\left(12V\right) \\ =16.0\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the charge is $16.0\mathrm{μ°ä}$.

From the calculations of part (a), we can get the value of the charge on the capacitance $C_3$islocalid="1661747960660" $16.00\mathrm{μ°ä}$.

When switch $S_1$and $S_2$is closed,

With switch 2 closed, the potential difference ${\mathrm{V}}_1$across ${\mathrm{C}}_1$must be equal to the potential difference across ${\mathrm{C}}_2$thus, the value of the potential is given using equation (ii) as follows:

$$\begin{gathered} {\mathrm{V}}_1=\frac{\left({\mathrm{C}}_3+C_4\right)}{{\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3+C_4}\mathrm{V} \\ =\frac{\left(3\mathrm{μ¹ó}+4\mathrm{μ¹ó}\right)12\mathrm{V}}{1\mathrm{μ¹ó}+2\mathrm{μ¹ó}+3\mathrm{μ¹ó}+4\mathrm{μ¹ó}} \\ =8.40\mathrm{V} \end{gathered}$$

Now, the value of the charge on the capacitance is given using equation (iii) as:

$$\begin{gathered} {\mathrm{q}}_1=\left(1\mathrm{μ¹ó}\right)\left(8.40\mathrm{V}\right) \\ =8.40\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the charge is $8.40\mathrm{μ°ä}$.

Similarly, from the calculated potential in part (e), we can get the charge on the capacitance using equation (iii) as follows:

$$\begin{gathered} {\mathrm{q}}_2=\left(2\mathrm{μ¹ó}\right)\left(8.40\mathrm{V}\right) \\ =16.8\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the charge is $16.8\mathrm{μ°ä}$.

Now, we can get the charge on the capacitance ${\mathrm{C}}_3$using equation (iii) as follows:

$$\begin{gathered} {\mathrm{q}}_3={\mathrm{C}}_3(\mathrm{V}-{\mathrm{V}}_1) \\ =\left(3\mathrm{μ¹ó}\right)\left(12\mathrm{V}-8.40\mathrm{V}\right) \\ =10.8\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the charge is $10.8\mathrm{μ°ä}$.

Now, we can get the charge on the capacitance $C_4$ using equation (iii) as follows:

$$\begin{gathered} {\mathrm{q}}_4={\mathrm{C}}_4(\mathrm{V}-{\mathrm{V}}_1) \\ =\left(4\mathrm{μ¹ó}\right)\left(12\mathrm{V}-8.40\mathrm{V}\right) \\ =14.4\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the charge is $14.4\mathrm{μ°ä}$.