Question

Two capacitors of unknown capacitances ${C_1}$ and ${C_2}$ are connected first in series and then in parallel across a battery of $100V$ . If the energy stored in the two combinations is $0.045J$ and $0.25J$ respectively, determine the value of ${C_1}$ and ${C_2}$ . Also calculate the charge on each capacitor in parallel combination.

Answer 1

Resultant capacitance for two capacitors connected in series is the sum of their inverse capacitances. And the resultant capacitance for two capacitors connected in parallel is the sum of their individual capacitances. Recall the formula for energy stored in a capacitor or capacitors.

Formula used:
$E = \dfrac{1}{2}C{V^2}$
Where $E$ is the energy stored,
$C$ is the capacitance and
$V$ is the potential difference.
$Q = CV$ Where Q is the charge on capacitor

Complete step by step solution:
We need to find the capacitances of both the capacitors and the charge stored in each of them, we are given with energy.
By using the formula of Energy stored, we can find the capacitance and then it is easy to find the charge stored.
$E = \dfrac{1}{2}C{V^2}$
Now $0.045J$ energy is stored when capacitors are connected in series
For series combination, the resultant capacitance is given as
$\dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$ where ${C_s}$ is resultant capacitance
$\Rightarrow {C_s} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$ ---- equation (1)
For parallel combination, the resultant capacitance is given as
$\Rightarrow {C_p} = {C_1} + {C_2}$ -----equation (2)
where ${C_p}$ is the resultant capacitance
The energy stored in series combination is given as $0.045J$ and
the energy stored in parallel combination is $0.25J$, therefore we can have
${E_1} = \dfrac{1}{2}{C_s}{V^2}$
${E_2} = \dfrac{1}{2}{C_p}{V^2}$
Substituting the values from equation $1,2$ and given values of potential difference, we can have
$0.045J = \dfrac{1}{2}(\dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}){(100)^2}$
$0.25J = \dfrac{1}{2}({C_1} + {C_2}){(100)^2}$
Solving this, we get
$(\dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}) = 0.09 \times 1{0^{ - 4}}F$
$\Rightarrow (\dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}) = 9 \times 1{0^{ - 6}}F$
$\Rightarrow ({C_1} + {C_2}) = 50 \times {10^{ - 6}}F$
$\Rightarrow (\dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}) = 9\mu F$
$\Rightarrow ({C_1} + {C_2}) = 50\mu F$
Where $F$ is the unit of capacitance.
Solving this we have.
${C_1} = 11.8\mu F$
${C_2} = 38.2\mu F$
Now we can easily calculate the charge stored in each capacitor:
${Q_1} = {C_1}V$
${Q_2} = {C_2}V$
$\Rightarrow {Q_1} = 11.8 \times {10^{ - 6}} \times 100$
$\Rightarrow {Q_2} = 38.2 \times {10^{ - 6}} \times 100$
$\Rightarrow {Q_1} = 0.12 \times 1{0^{ - 2}}C$
$\Rightarrow {Q_2} = 0.38 \times 1{0^{ - 2}}C$

The values of capacitances are ${C_1} = 11.8\mu F$ and ${C_2} = 38.2\mu F$ and the charges on them are ${Q_1} = 0.12 \times 1{0^{ - 2}}C$ and ${Q_2} = 0.38 \times 1{0^{ - 2}}C$ respectively.

Note: Remember that capacitor is a device which stores electrical energy in the electrical field.
The capacitor is a passive device which means it does not produce energy. While solving the problem be careful with the units of capacitances and convert them accordingly in farad and microfarads.