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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.
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Solution
When the capacitors are connected in parallel,
Equivalent capacitance, CP=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
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