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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 100V. If the energy stored in 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 capacitor are connected in parallel,Equivalent Capacitance,CP=C1+C2The energy stored in the combinationEP=12CPV2⇒0.25=12(C1+C2)(100)2⇒C1+C2=5×10−5⟶(i)Similarly when the capacitor are connected in seriesCs=C1C2C1+C2Energy Stored Es=12CsV2=12C1C2C1+C2(100)2=0.045⇒C1C2=4.5×10−10Now, (C1−C2)2=(C1+C2)2−4C1C2=(5×10−5)2−4(4.5×10−10)=7×10−10C1−C2=√7×10−10=2.64×10−5⟶(ii)on Solving (i) and (ii) we get,C1=38μF, C2=1.2μFcharges Q1=C1V=38×10−6×100=38×10−4CQ2=C2V=12×10−6×100=12×10−4C

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