91Ó°ÊÓ

$$\begin{gathered} \mathrm{V}=100\mathrm{V} \\ {\mathrm{C}}_1=10.0\mathrm{μ¹ó} \\ {\mathrm{C}}_2=5.00\mathrm{μ¹ó} \\ {\mathrm{C}}_3=4.00\mathrm{μ¹ó} \end{gathered}$$

First, find the equivalent capacitance by using the formula for series and parallel combinations. Using the equivalent capacitance , find the potential difference${\mathit{V}}_{\mathbf{1}}$across${\mathit{C}}_{\mathbf{1}}$and also the increase in potential difference. Using the concept of capacitance, find the increase in the charge on the capacitor .

Formulae are as follows:

$C_{eq}=\underset{j=1}{\overset{n}{∑}}{C}_J$

For parallel combination,

$\frac{1}{C_{eq}}=\underset{j=1}{\overset{n}{∑}}\frac{1}{C_j}$

For series combination,

q=CV

Where C is capacitance, V is the potential difference, and q is the charge on the particle.

First, find the equivalent capacitance as$C_1$ and$C_2$ are in parallel combination.

$$\begin{gathered} C\text{'}=C_1+C_2 \\ =10\mathrm{μ¹ó}+5\mathrm{μ¹ó} \\ =15\mathrm{μ¹ó} \end{gathered}$$

This$C\text{'}$ is a series with$C_3$,

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C\text{'}}+\frac{1}{C_3} \\ \frac{1}{C_{eq}}=\frac{C_3+C\text{'}}{C_{3}C\text{'}} \end{gathered}$$

By substituting the values,

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{4\mathrm{μ}F+15\mathrm{μ}F}{15\mathrm{μ}F×4\mathrm{μ}F} \\ =\frac{19}{60\mathrm{μ}F} \\ =\frac{1}{3.16\mathrm{μ}F} \\ {C}_{eq}=3.16\mathrm{μ¹ó} \end{gathered}$$

The potential difference$V_1$ across$C_1$ is given by,

$V_1=\frac{C_{eq}V}{C_1+C_2}$

By substituting the value,

$$\begin{gathered} {\mathrm{V}}_1=\frac{\left(3.16×{10}^{-6}\mathrm{F}\right)×100\mathrm{V}}{\left(10×{10}^{-6}\mathrm{F}\right)+\left(5×{10}^{-6}\mathrm{F}\right)} \\ =21.1\mathrm{V} \end{gathered}$$

Thus, the increase in potential difference across capacitors 1 is,

$$\begin{gathered} ∆V_1=V-V_1 \\ =100V-21.1V \\ =78.9V \end{gathered}$$

Now, we can find the increase in charge on the capacitor .

Since,

q=CV

We can write,

$$\begin{gathered} ∆\mathrm{q}={\mathrm{C}}_1∆{\mathrm{V}}_1 \\ ∆\mathrm{q}=\left(10×{10}^{-6}\mathrm{F}\right)×78.9\mathrm{V} \\ =7.89×{10}^{-4}\mathrm{C} \end{gathered}$$

Hence, the increase in the charge on the capacitor 1 is$7.89×{10}^{-4}C$

From part a), it can be concluded that the increase in potential difference across capacitors 1 is 78.9 V.

Hence, the increase in the potential difference across capacitors 1 is$\mathrm{Δ}{V}_1=78.9V$

Therefore, find the increase in charge and potential difference across the capacitor 1 can be found by using the concept of capacitance and the formula for equivalent capacitance for a series and parallel combination.