Question

Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors \[{C_1}\] and \[{C_2}\;\]​ with their capacitances in the ratio \[1:2\] so that the energy stored in the two cases becomes the same.

Answer 1

Hint: Capacitors are the devices that store electrical energy on their plates which is in the form of an electrical charge. These capacitors consist of two parallel plates and they do not touch each other. They are prevented from touching each other by an insulating material called a dielectric. When these capacitors are charged fully there will be potential differences developed between the plates. Capacitors are connected through series or parallel connections. We can find the voltage between them using this formula given below.

Complete step by step solution:
When the capacitors are connected in series connection, then the total capacitance is given by,
\[\dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} + .........\]
Here \[{C_s}\]is the total capacitance of the series connection
\[{C_1}\],\[{C_2}\],\[{C_3}\] are the capacitance of the individual capacitors connected in series.
When the capacitors are connected in parallel connection, then the total capacitance is given by,
\[{C_p}\; = \;{C_1}\; + \;{C_2}\; + \;{C_3}\; + {\text{ }} \ldots \]
Similarly \[{C_p}\] is the total capacitance of the parallel combination
\[{C_1}\],\[{C_2}\],\[{C_3}\] are the capacitance of the individual capacitors connected parallel.
Given that two capacitors are connected in both parallel and series. And it is also given the ratio of their capacitance is \[1:2\]. Therefore we can write it as,
\[\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{1}{2}\]
\[{C_2} = 2{C_1}\]
For parallel combination,\[{C_p}\]=\[{C_1}\; + \;{C_2}\;\]
Substituting\[{C_1}and{C_2}\]in the above equation.
\[{C_p} = {C_1} + 2{C_1} = 3{C_1}\]
For series combination, \[\dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}\]
\[{C_s} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}\]
\[{C_s} = \dfrac{{{C_1} \times 2{C_1}}}{{{C_1} + 2C}} = \dfrac{{2{C_1}^2}}{{3{C_1}}}\]
\[{C_s} = \dfrac{2}{3}{C_1}\]
The potential energy stored in a capacitor is given by,
\[u = \dfrac{1}{2}c{v^2}\]
Here\[v\]is the potential applied
Let \[{u_1}\] be the potential energy stored when the capacitors are connected in parallel and \[{u_2}\]be the potential energy stored in the capacitors when they are connected in a series combination.
Let us try and find \[{u_1}\]and \[{u_2}\].
\[{u_1} = \dfrac{1}{2}{c_p}{v_1}^2\]
Substituting all the known values.
\[{u_1} = \dfrac{3}{2}{C_1}{v_1}^2\]
Similarly\[{u_2}\], potential energy stored in the capacitors when they are connected in series combination is given as,
\[{u_2} = \dfrac{1}{3}{C_1}{v_1}^2\]
Given that the potential energy stored in the two cases becomes the same. Therefore,
\[{u_1} = {u_2}\]
\[\dfrac{3}{2}{C_1}{v_1}^2 = \dfrac{1}{3}{C_1}{v_1}^2\]
\[{(\dfrac{{{v_1}}}{{{v_2}}})^2} = \dfrac{2}{9}\]
\[\dfrac{{{v_1}}}{{{v_2}}}\]=\[\dfrac{{\sqrt 2 }}{3}\]
\[\dfrac{{{v_1}}}{{{v_2}}} = 0.471\]
This gives us the required ratio of potential differences.

Note:
Capacitors are used in various devices such as calculators, flash lamps, etc. We have already used the formula for potential energy stored in capacitors. But there are also other forms of potential energy used in capacitors. The other forms are
\[{E_{cap}} = \dfrac{{QV}}{2}\]where \[Q\] is the amount of charge stored in the capacitor and
\[{E_{cap}} = \dfrac{{{Q^2}}}{{2C}}\]here \[C\] is the capacitance of the capacitor.