91Ó°ÊÓ

The capacitance${\mathrm{C}}_1=10.0\mathrm{μ¹ó}$

The capacitance ${\mathrm{C}}_2=5.00\mathrm{μ¹ó}$

The capacitance ${\mathrm{C}}_3=4.00\mathrm{μ¹ó}$

Using the equation 25-19, find the equivalent capacitance of C₁ and C₂. Then using 25-20, find the equivalent capacitance of the given combination.

Formulae are as follows:

Capacitors in series combination,

$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}$

Capacitors in parallel combination,

${\mathbf{C}}_{\mathbf{eq}}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}$

From Figure, it can be seen that, ${\mathrm{C}}_1$and ${\mathrm{C}}_2$are connected in parallel. Therefore,

From the equation 25-19, the equivalent capacitance is given by,

${\mathrm{C}}_{12}={\mathrm{C}}_1+{\mathrm{C}}_2$

From the above figure, we can see that ${\mathrm{C}}_{12}$and ${\mathrm{C}}_3$are connected in a series.

From the equation 25-20, the equivalent capacitance is given by,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{C}}_{12}}+\frac{1}{{\mathrm{C}}_3}$

Therefore,

${\mathrm{C}}_{\mathrm{eq}}=\frac{{\mathrm{C}}_{12}{\mathrm{C}}_3}{{\mathrm{C}}_{12}+{\mathrm{C}}_3}$

Therefore,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\frac{\left(10.0\mathrm{μ¹ó}+5.00\mathrm{μ¹ó}\right)×4.00\mathrm{μ¹ó}}{\left(10.0\mathrm{μ¹ó}+5.00\mathrm{μ¹ó}\right)+4.00\mathrm{μ¹ó}} \\ {\mathrm{C}}_{\mathrm{eq}}=3.16\mathrm{μ¹ó} \end{gathered}$$

Hence, the equivalent capacitance of the combination is 3.16$\mathrm{μ¹ó}$

Therefore, by using the formula of capacitors in series and parallel combinations, equivalent capacitance can be determined.