91Ó°ÊÓ

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

Using the equation 25-20, find the equivalent capacitance of c₁and c₂. Then using 25-19, find the equivalent capacitance of this combination.

Formulae are as follows:

Capacitors in series combination,

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

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 the capacitors ${\mathrm{C}}_1\mathrm{and}{\mathrm{C}}_2$are connected in a series. Therefore,

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

$\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}$

Therefore,

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

Fig 25-28 implies that ${\mathrm{C}}_{12}$is in parallel with ${\mathrm{C}}_3$.

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

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

Therefore,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\left(\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}\right)+{\mathrm{C}}_3 \\ {\mathrm{C}}_{\mathrm{eq}}=\left(\frac{10.0\mathrm{μ¹ó}×5.00\mathrm{μ¹ó}}{10.0\mathrm{μ¹ó}+5.00\mathrm{μ¹ó}}\right)+4.00\mathrm{μ¹ó} \\ {\mathrm{C}}_{\mathrm{eq}}=7.33\mathrm{μ¹ó} \end{gathered}$$

Hence, theequivalent capacitance of the combination is$C_{\mathrm{eq}}=7.33\mathrm{μ¹ó}$.

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