Question

Net capacitance of three identical capacitors in series is $1\mu F$. What will be their net capacitance, if connected in parallel? Find the ratio of energy stored in these two configurations, if they are both connected to the same source.

Answer 1

Hint:Use the formula for net capacitance in series connection to get the capacitance value for each capacitor. Then use the formula for net capacitance in parallel connection and calculate the net capacitance in a parallel connection. To get the energy ratio, first calculate the energy in each configuration and then divide them accordingly.

Formula Used:
Net capacitance for 3 capacitors in series connection, $\dfrac{1}{{{C_S}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}$
Where, ${C_1},\,{C_2},\,{C_3}$ are the capacitance of the three capacitors respectively.
Net capacitance for 3 capacitors in parallel connection, ${C_P} = {C_1} + {C_2} + {C_3}$
Where, ${C_1},\,{C_2},\,{C_3}$ are the capacitance of the three capacitors respectively.
Energy in a series connection, ${U_S} = \dfrac{1}{2}{C_S}{V^2}$
Where, ${C_S}$ is the net capacitance for 3 capacitors in series connection, and $V$ is the potential difference applied across them.
Energy in a parallel circuit, ${U_P} = \dfrac{1}{2}{C_P}{V^2}$
Where, ${C_P}$ is the net capacitance for 3 capacitors in parallel connection, and $V$ is the potential difference applied across them.

Complete step by step solution:
We are given that the net capacitance for 3 capacitors in series connection, ${C_S} = 1\mu F$
Using the formula for net capacitance in series connection, we get
$\dfrac{1}{1} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}$
The capacitors are identical. Therefore, ${C_1} = {C_2} = {C_3}$
$ \Rightarrow \dfrac{1}{1} = \dfrac{3}{C}$ which gives, $C = 3\mu F$
Therefore, ${C_1} = {C_2} = {C_3} = 3\mu F$
When connected in parallel, net capacitance is calculated using the formula ${C_P} = {C_1} + {C_2} + {C_3}$
Therefore, ${C_P} = 3 + 3 + 3 = 9\mu F$
Now, we are given that these two configurations, series and parallel, are connected to the same source. Let the potential for both of them be equal to $V$ .
Now we will calculate the energy in series connection first.
We have, ${U_S} = \dfrac{1}{2}{C_S}{V^2}$
$ \Rightarrow {U_S} = \dfrac{1}{2} \times 1 \times {V^2}$
This gives, ${U_S} = \dfrac{{{V^2}}}{2}$
Now calculate the energy in parallel connection.
We have, ${U_P} = \dfrac{1}{2}{C_P}{V^2}$
$ \Rightarrow {U_P} = \dfrac{1}{2} \times 9 \times {V^2}$
This gives, ${U_P} = \dfrac{{9{V^2}}}{2}$
Therefore, the ratio ${U_S}:{U_P} = \dfrac{{{V^2}}}{2}:\dfrac{{9{V^2}}}{2} = 1:9$ (rest all gets cancelled)

This is the required answer.

Note:Do not get confused with the formulas for net capacitance in series connection and net capacitance in parallel connection with each other. Make sure the values are in SI units before using them in the question. If not, convert them accordingly.