Question

(a) IfC = 50 $\mathit{\mu }\mathit{F}$in Fig. 25-52, what is the equivalent capacitance between points Aand B? (Hint:First imagine that a battery is connected between those two points.) (b) Repeat for points Aand D.

Short answer

1. The equivalent capacitance between points A and B is $41\mathrm{\mu F}$.
2. The equivalent capacitance between points A and D is $42\mathrm{\mu F}$.

Step-by-step solution

The given data

Value of the capacitance, C = 50$\mathrm{\mu F}$

Understanding the concept of the equivalent capacitance

If the capacitors are connected in parallel, the equivalent capacitance can be calculated and if capacitors are connected in series, the equivalent capacitance can be calculated using the given formula. So, we have to use the concept of equivalent capacitance in series and parallel to find the equivalent capacitance between different points.

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)

(a) Calculation of the capacitance between A and B

In this case, D is not attached to anything. So, thecapacitors, 6C and 4C are in series. Thus, the equivalent capacitance of this combination can be given using equation (i) as follows:

$\begin{array}{l}{\mathrm{c}}_{45}=\frac{6\mathrm{C}\times 4\mathrm{C}}{6\mathrm{C}+4\mathrm{C}}\\ =\frac{24\mathrm{C}}{10}\\ =2.4\mathrm{C}\end{array}$

This combination is then in parallel with the 2C capacitor.So, the equivalent capacitance of this combination can be given using equation (ii) as follows:

$$\begin{gathered} {\mathrm{C}}_{246}=2.4\mathrm{C}+2\mathrm{C} \\ =4.4\mathrm{C} \end{gathered}$$

Again, this combination is in series withC , so, the equivalent capacitance between A and B can be given as follows:

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\frac{\mathrm{C}\times 4.4\mathrm{C}}{\mathrm{C}+4.4\mathrm{C}} \\ =0.82\mathrm{C} \\ =0.82\times 50\mathrm{\mu F} \\ =41\mathrm{\mu F} \end{gathered}$$

Hence, the value of the capacitance is $41\mathrm{\mu F}$.

 Step 4: (b) Calculation of the capacitance between A and D

In this case, B is not attached to anything.

So, the capacitors, 6C and 2C are in series. Thus, the equivalent capacitance can be given using equation (i) as follows:

$$\begin{gathered} {\mathrm{C}}_{26}=\frac{6\mathrm{C}\times 2\mathrm{C}}{6\mathrm{C}+2\mathrm{C}} \\ =\frac{12\mathrm{C}}{8} \\ =1.5\mathrm{C} \end{gathered}$$

This combination is then in parallel with the 4C capacitor, thus, the equivalent capacitance can be given using equation (ii) as follows:

$$\begin{gathered} {\mathrm{C}}_{26}=1.5\mathrm{C}+4\mathrm{C} \\ =5.5\mathrm{C} \end{gathered}$$

Again this combination is in series with C .

So, the equivalent capacitance between A and D is given using equation (i) as:

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\frac{\mathrm{C}\times 5.5\mathrm{C}}{\mathrm{C}+5.5\mathrm{C}} \\ =0.85\mathrm{C} \\ =0.85\times 50\mathrm{\mu F} \\ =42\mathrm{\mu F} \end{gathered}$$

Hence, the value of the charge is $42\mathrm{\mu F}$.