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Q66PE

Expert-verifiedFound in: Page 699

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

**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 **\(\Delta V\)**: **\({U_E} = \frac{1}{2}C{(\Delta V)^2}...(2)\)**.**

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:

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

- 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: \({C_2} = 7.40{\rm{ }}\mu F\).

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 \({U_E} = 6.35 \times {10^{ - 5}}\;J\).

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

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