Question

In Fig. 25-29, a potential difference of V= 100.0 V is applied across a capacitor arrangement with capacitances,${\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu F}$,${\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}$, and${\mathbf{C}}_3\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}$.If capacitor 3 undergoes electrical breakdown so that it becomes equivalent to conducting wire, (a) What is the increase in the charge on capacitor 1? (b) What is the increase in the potential difference across capacitor 1?

Short answer

a) The increase in the charge on the capacitor 1 is$\Delta q=7.89\times {10}^{-4}C$

b) The increase in the potential difference across capacitors 1 is$\Delta V_1=78.9V$

Step-by-step solution

Given

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

Determining the concept

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:

$C_{eq}=\sum _{j=1}^n{C}_J$

For parallel combination,

$\frac{1}{C_{eq}}=\sum _{j=1}^n\frac{1}{C_j}$

For series combination,

q=CV

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

(a) Determining the increase in the charge on the capacitor 1

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

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

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,

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

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,

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

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,

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

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

(b) Determining the increase in the potential difference across the capacitor

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$\Delta V_1=78.9V$

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.