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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.

Solution:

When the two capacitors are in series, the equivalent capacitance, Cs = C1C2/(C1+C2)
When the two capacitors are in parallel, the equivalent capacitance, Cp = C1 + C2
Here, Us = (1/2)CsV^2 or 0.045 = 0.5 × (C1C2/(C1+C2)) × (100)^2
or 9(C1+C2) = 10^6C1C2.. (1)
Now, Up = (1/2)CpV^2 or 0.25 = 0.5 × (C1+C2) × (100)^2
or 0.02(C1+C2) × 10^6 = 1.. (2)
(1)/(2) ⇒ 450 = 10^12C1C2.. (3)
using, (2) ,(3), 0.02(C1 + 450 × 10^-12/C1) × 10^6 = 1
Solving, C1 = 25 + 5√7 μF
C2 = 25 - 5√7 μF
In parallel, voltage across the capacitors is equal to the supply voltage (100 V).
Q1 = CV = (2.5 + 0.5√7) mC
Q2 = CV = (2.5 - 0.5√7) mC