Net Capacitance of three identical in series is 1 μF. What will be their net capacitance, if connected in parallel? Find the ratio of energy stored in the two configurations, if they are both connected to the same source.
Answer
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Hint: The ability of a component or circuit to collect and store energy in the form of an electrical charge is known as capacitance. Capacitors are energy-storage devices that come in a variety of shapes and sizes. They are made up of two conductor plates (usually thin metal) sandwiched between an insulator made of ceramic, film, glass, or other materials, including air.
Complete step by step answer:
When capacitors are connected in series, the total capacitance is lower than the individual capacitances of the series capacitors. When two or more capacitors are connected in series, the result is a single (equivalent) capacitor with the sum total of the individual capacitors' plate spacings. As we've seen, an increase in plate spacing, when combined with all other factors, can result in a significant increase in plate spacing.Formula for when capacitors are connected in series is given as,
\[\dfrac{1}{{{C}_{n}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}......\dfrac{1}{{{C}_{n}}}\]
When capacitors are connected in series, the total capacitance equals the sum of the capacitances of the individual capacitors. When two or more capacitors are connected in parallel, the result is a single equivalent capacitor with the sum total of the individual capacitors' plate areas. As we've seen, increasing plate area while keeping all other factors constant results in increased capacitance.
Formula for when capacitors are connected in parallel is given as,
\[{{C}_{total=}}{{C}_{1}}+{{C}_{2}}+{{C}_{3}}.....{{C}_{n}}\]
In above example, in series combination,
\[\dfrac{1}{{{C}_{s}}}=\dfrac{1}{C}+\dfrac{1}{{{C}_{{}}}}+\dfrac{1}{C} \\
\Rightarrow C=3{{C}_{s}} \\
\Rightarrow C_s=3\mu F \\ \]
In parallel combination,
\[{{C}_{p}}=3C \\
\Rightarrow {{C}_{p}}=9F \\ \]
\[\Rightarrow \dfrac{{{U}_{s}}}{{{U}_{p}}}=\dfrac{\dfrac{1}{2}{{C}_{s}}{{V}^{2}}}{\dfrac{1}{2}{{C}_{p}}{{V}^{2}}}=\dfrac{{{C}_{s}}}{{{C}_{p}}} \\
\therefore U_s:U_p=1:9 \]
Hence, the ratio of energy stored in the two configurations is 1:9.
Note: The most obvious reason for connecting two or more capacitors in series is to increase the available capacitance. It may be more practical to use two or more smaller capacitors rather than a single larger one. Again, availability may be an issue, necessitating the use of two parallel capacitors. In many circuits, capacitors are connected in series. It's very useful to know how to calculate the overall value, even if it's just a rough calculation in your head. The online series capacitor calculator can be very useful if a more precise value is required.
Complete step by step answer:
When capacitors are connected in series, the total capacitance is lower than the individual capacitances of the series capacitors. When two or more capacitors are connected in series, the result is a single (equivalent) capacitor with the sum total of the individual capacitors' plate spacings. As we've seen, an increase in plate spacing, when combined with all other factors, can result in a significant increase in plate spacing.Formula for when capacitors are connected in series is given as,
\[\dfrac{1}{{{C}_{n}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}......\dfrac{1}{{{C}_{n}}}\]
When capacitors are connected in series, the total capacitance equals the sum of the capacitances of the individual capacitors. When two or more capacitors are connected in parallel, the result is a single equivalent capacitor with the sum total of the individual capacitors' plate areas. As we've seen, increasing plate area while keeping all other factors constant results in increased capacitance.
Formula for when capacitors are connected in parallel is given as,
\[{{C}_{total=}}{{C}_{1}}+{{C}_{2}}+{{C}_{3}}.....{{C}_{n}}\]
In above example, in series combination,
\[\dfrac{1}{{{C}_{s}}}=\dfrac{1}{C}+\dfrac{1}{{{C}_{{}}}}+\dfrac{1}{C} \\
\Rightarrow C=3{{C}_{s}} \\
\Rightarrow C_s=3\mu F \\ \]
In parallel combination,
\[{{C}_{p}}=3C \\
\Rightarrow {{C}_{p}}=9F \\ \]
\[\Rightarrow \dfrac{{{U}_{s}}}{{{U}_{p}}}=\dfrac{\dfrac{1}{2}{{C}_{s}}{{V}^{2}}}{\dfrac{1}{2}{{C}_{p}}{{V}^{2}}}=\dfrac{{{C}_{s}}}{{{C}_{p}}} \\
\therefore U_s:U_p=1:9 \]
Hence, the ratio of energy stored in the two configurations is 1:9.
Note: The most obvious reason for connecting two or more capacitors in series is to increase the available capacitance. It may be more practical to use two or more smaller capacitors rather than a single larger one. Again, availability may be an issue, necessitating the use of two parallel capacitors. In many circuits, capacitors are connected in series. It's very useful to know how to calculate the overall value, even if it's just a rough calculation in your head. The online series capacitor calculator can be very useful if a more precise value is required.
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