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Q11P

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Found in: Page 740

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# In Fig. 25-29, find the equivalent capacitance of the combination. Assume that ${{\mathbf{C}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\mu F}}$, ${{\mathbf{C}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\mu F}}$ and ${{\mathbf{C}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\mu F}}$

The equivalent capacitance of the combination is 3.16 $\mathrm{\mu F}$

See the step by step solution

## Step 1: Given

The capacitance ${\mathrm{C}}_{1}=10.0\mathrm{\mu F}$

The capacitance ${\mathrm{C}}_{2}=5.00\mathrm{\mu F}$

The capacitance ${\mathrm{C}}_{3}=4.00\mathrm{\mu F}$

## Step 2: Determining the concept

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,

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

## Step 3: Determining the equivalent capacitance of the combination

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,

${\mathrm{C}}_{\mathrm{eq}}=\frac{\left(10.0\mathrm{\mu F}+5.00\mathrm{\mu F}\right)×4.00\mathrm{\mu F}}{\left(10.0\mathrm{\mu F}+5.00\mathrm{\mu F}\right)+4.00\mathrm{\mu F}}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{\mathrm{eq}}=3.16\mathrm{\mu F}$

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

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

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