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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Suppose you have a $$9.00\;V$$ battery, a $$2.00{\rm{ }}\mu F$$ capacitor, and a $$7.40{\rm{ }}\mu F$$capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series. (b) Do the same for a parallel connection.

The charge and energy stored if the capacitors are connected to the battery in series is $$Q = 14.1{\rm{ }}\mu C$$ and $${U_E} = 6.35 \times {10^{ - 5}}\;J$$.

b) The charge and energy stored, if the capacitors are connected to the battery in parallel is $$Q = 84.6{\rm{ }}\mu C$$ and $${U_E} = 3.80 \times {10^{ - 4}}\;J$$.

See the step by step solution

## Step 1: Definition of Capacitor

A capacitor is a device used to store electrical energy and works in an electric field. It is a passive electrical component with two terminals. The term "capacitance" is used to characterize a capacitor's effect. The ratio of $$C = \frac{Q}{{\Delta V}}...(1)$$ determines the capacitance $$C$$.

The following is the amount of energy held in a capacitor with capacitance $$C$$ that is charged to a potential difference $$\Delta V$$: $${U_E} = \frac{1}{2}C{(\Delta V)^2}...(2)$$.

## Step 2: Explaining about Capacitors in Series and Parallel

Capacitors in Series: The reciprocal of the equivalent capacitance $${C_{eq{\rm{ }}}}$$equals the total of the reciprocals of the individual capacitances when capacitors with capacitances $${C_1},{C_2}, \ldots$$ are connected in series:

$$\frac{1}{{{C_{eq{\rm{ }}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots (3)$$

Capacitors in Parallel: The equivalent capacitance $${C_{eq{\rm{ }}}}$$ is the sum of the individual capacitances when capacitors with capacitances $${C_1},{C_2}, \ldots$$ are connected in parallel:

$${C_{eq{\rm{ }}}} = {C_1} + {C_2} + \ldots (4)$$

## Step 3: Given information

• The potential difference across the battery is: $$\Delta V = 9.00\;V$$.
• The capacitance of the first capacitor is: $${C_1} = 2.00{\rm{ }}\mu F$$.
• The capacitance of the second capacitor is: $${C_2} = 7.40{\rm{ }}\mu F$$.

## Step 4: Find the charge and energy stored if the capacitors are connected to the battery in series.

a)

Equation $$(3)$$ yields the total capacitance of the two capacitors when linked in series:

\begin{aligned}{c}\frac{1}{{{C_{eq{\rm{ }},s}}}} &= \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}\\{C_{eq,s}} &= \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}\end{aligned}

Substitute the following numerical values:

\begin{aligned}{}{C_{eq{\rm{ }},s}} &= \frac{{(2.00{\rm{ }}\mu F)(7.40{\rm{ }}\mu F)}}{{2.00{\rm{ }}\mu F + 7.40{\rm{ }}\mu F}}\\ &= 1.57{\rm{ }}\mu F\\ = (1.57{\rm{ }}\mu F)\left( {\frac{{1\;F}}{{{{10}^6}{\rm{ }}\mu F}}} \right)\\ &= 1.57 \times {10^{ - 6}}\;F\end{aligned}

Then, Equation $$(1)$$ is used to calculate the charge stored in the two capacitors when they are linked in series:

$$Q = {C_{eq,{\rm{ }}s}}\Delta V$$

Now, replacing the values of $${C_{eq,{\rm{ }}s}}$$and $$\Delta V$$:

We get,

\begin{aligned}{}Q &= (1.57{\rm{ }}\mu F)(9.00\;V)\\ &= 14.1{\rm{ }}\mu C\end{aligned}

Then, Equation $$(2)$$ calculates the energy stored in the two capacitors when they are linked in series:

$${U_E} = \frac{1}{2}{C_{eq,s}}{(\Delta V)^2}$$

Substituting, the values of $${C_{eq,{\rm{ }}s}}$$ and $$\Delta V$$:

\begin{aligned}{}{U_E} &= \frac{1}{2}\left( {1.57 \times {{10}^{ - 6}}\;F} \right){(9.00\;V)^2}\\ &= 6.35 \times {10^{ - 5}}\;J\end{aligned}

Hence, the charge and energy values are obtained as $$Q = 14.1{\rm{ }}\mu C$$ and $${U_E} = 6.35 \times {10^{ - 5}}\;J$$.

## Step 5: Finding the charge and energy stored if the capacitors are connected to the battery in parallel.

b) Equation $$(4)$$ calculates the total capacitance of the capacitors when they are linked in parallel:

$${C_{eq,p{\rm{ }}}} = {C_1} + {C_2}$$

Substitute the following numerical values:

\begin{aligned}{}{C_{eq,p{\rm{ }}}}2.00{\rm{ }}\mu F + 7.40{\rm{ }}\mu F &= 9.40{\rm{ }}\mu F\\ &= (9.40{\rm{ }}\mu F)\left( {\frac{{1\;F}}{{{{10}^6}{\rm{ }}\mu F}}} \right)\\ &= 9.40 \times {10^{ - 6}}\;F\end{aligned}

Now, Equation $$(1)$$ is used to calculate the charge stored in the two capacitors when they are linked in parallel:

$$Q = {C_{eq,{\rm{ p}}}}\Delta V$$

We get the following results by substituting in the values for $${C_{eq,{\rm{ p}}}}$$and $$\Delta V$$

\begin{aligned}{}Q &= (9.40{\rm{ }}\mu F)(9.00\;V)\\Q &= 84.6{\rm{ }}\mu C\end{aligned}

Then, Equation $$(2)$$ calculates the energy stored in the two capacitors when they are linked in parallel:

$${U_E} = \frac{1}{2}{C_{eq,p}}{(\Delta V)^2}$$

By giving the values for $${C_{eq,{\rm{ p}}}}$$and $$\Delta V$$, we get:

\begin{aligned}{}{U_E} &= \frac{1}{2}\left( {9.40 \times {{10}^{ - 6}}\;F} \right){(9.00\;V)^2}\\ &= 3.80 \times {10^{ - 4}}\;J\end{aligned}

Therefore, the charge and energy values are obtained as $$Q = 84.6{\rm{ }}\mu C$$ and $${U_E} = 3.80 \times {10^{ - 4}}\;J$$.