Question

Two capacitors of unknown capacitances C₁ and C₂ 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 C₁ and C₂. Also calculate the charge on each capacitor in parallel combination.

Answer 1

When the capacitors are connected in parallel, 

Equivalent capacitance, C_P=C₁+C₂

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`

⇒(C₁+C₂)=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`

⇒C₁C₂=0.045×10⁻⁴×5×10⁻⁵×2=4.5×10⁻¹⁰

(C₁−C₂)²=(C₁+C₂)²−4C₁C₂

⇒(C₁−C₂)²=25×10⁻¹⁰−4×4.5×10⁻¹⁰=7×10⁻¹⁰

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

C₁−C₂=2.64×10⁻⁵  .....(ii)

Solving (i) and (ii), we get

C₁ = 35 μF and C₂ = 15 μF

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

Q₁=C₁V=35×10⁻⁶×100=35×10⁻⁴ C

Q₂=C₂V=15×10⁻⁶×100=15×10⁻⁴ C