#### Question

Two capacitors of unknown capacitances C_{1} and C_{2} are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C_{1} and C_{2}. Also calculate the charge on each capacitor in parallel combination.

#### Solution

When the capacitors are connected in parallel,

Equivalent capacitance, C_{P}=C_{1}+C_{2}

The energy stored in the combination of the capacitors, `E_P=1/2C_pV^2`

`=>E_P=1/2(C_1+C_2)(100^2)=0.25J`

⇒(C_{1}+C_{2})=5×10^{−5} .....(i)

When the capacitors are connected in series,

Equivalent capacitance, `C_S=(C_1C_2)/(C_1+C_2)`

The energy stored in the combination of the capacitors,

`E_S=1/2C_SV^2`

`=>E_S=1/2(C_1C_2)/(C_1+C_2)(100)^2=0.045J`

`1/2(C_1C_2)/(5xx10^(-5))(100)^2=0.045J`

⇒C_{1}C_{2}=0.045×10^{−4}×5×10^{−5}×2=4.5×10^{−10}

(C_{1}−C_{2})^{2}=(C_{1}+C_{2})^{2}−4C_{1}C_{2}

⇒(C_{1}−C_{2})^{2}=25×10^{−10}−4×4.5×10^{−10}=7×10^{−10}

`=>(C_1-C_2)=sqrt(7xx10^(-10))=2.64xx10^(-5)`

C_{1}−C_{2}=2.64×10^{−5} .....(ii)

Solving (i) and (ii), we get

*C*_{1} = 35 μF and *C*_{2} = 15 μF

When the capacitors are connected in parallel, the charge on each of them can be obtained as follows:

Q_{1}=C_{1}V=35×10^{−6}×100=35×10^{−4} C

Q_{2}=C_{2}V=15×10^{−6}×100=15×10^{−4} C