Q66PE - College Physics (Urone)

Question

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.

Step-by-Step Solution

Answer

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 \({U_E} = 3.80 \times {10^{ - 4}}\;J\).

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:

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

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, 

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

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 \(\Delta V\):

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

 

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:

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

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 \(\Delta V\)

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

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:

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

 

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