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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 two capacitors are in series, the equivalent capacitance, Cs=C1C2C1+C2When the two capacitors are in parallel, the equivalent capacitance, Cp=C1+C2Here, Us=12CsV2 or 0.045=0.5×C1C2C1+C2(100)2 or 9(C1+C2)=106C1C2...(1)Now, Up=12CpV2 or 0.25=0.5×(C1+C2)(100)2 or 0.02(C1+C2)×106=1...(2)(1)/(2)⇒450=1012C1C2...(3)using, (2) ,(3), 0.02(C1+450×10−12C1)×106=1 Solving, C1=25+5√7μF C2=25−5√7μFIn parallel, voltage across the capacitors is equal to the supply voltage (100 V).Q1=CV=(2.5+.5√7)mCQ2=CV=(2.5−.5√7)mC

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