Suppose you have a 9.00-V battery, a 2.00 - \mu F capacitor, and a 7.40

vousetmoiec 2021-11-11 Answered
Suppose you have a 9.00-V battery, a 2.00 - \(\displaystyle\mu{F}\) capacitor, and a \(\displaystyle{7.40}-\mu{F}\) capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series

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Expert Answer

Coldst
Answered 2021-11-12 Author has 884 answers
Identify the unknown:
The charge and energy stored if the capacitors are connected in series
Set Up the Problem:
Capacitance of a serie combination:
\(\displaystyle{\frac{{{1}}}{{{C}_{{S}}}}}={\frac{{{1}}}{{{C}_{{1}}}}}+{\frac{{{1}}}{{{C}_{{2}}}}}+{\frac{{{1}}}{{{C}_{{3}}}}}+\ldots\)
\(\displaystyle{\frac{{{1}}}{{{C}_{{S}}}}}={\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{7.4}}}}\)
\(\displaystyle{C}_{{S}}={\frac{{{2}\times{7.4}}}{{{2}+{7.4}}}}={1.575}\mu{F}={1.575}\times{10}^{{-{6}}}{F}\)
Capacitance of a series combitation is given by:
\(\displaystyle{C}_{{S}}={\frac{{{Q}}}{{{V}}}}\)
Then the charges stored in the series combination is:
\(\displaystyle{Q}={C}_{{S}}{V}\)
Energy stored in the series combination is:
\(\displaystyle{U}_{{C}}={\frac{{{1}}}{{{2}}}}{V}^{{2}}{C}_{{S}}\)
Solve the problem:
\(\displaystyle{Q}={1.575}\times{10}^{{-{6}}}\times{9}={1.42}\times{10}^{{-{5}}}{C}\)
\(\displaystyle{U}_{{C}}={\frac{{{1}}}{{{2}}}}\times{\left({9}\right)}^{{2}}\times{1.575}\times{1}{)}^{{-{6}}}={6.38}\times{10}^{{-{5}}}\) J
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