Electrostatics Question 605

Question: Two capacitors C1=2μF and C2=6μF in series, are connected in parallel to a third capacitor C3=4μF . This arrangement is then connected to a battery of e.m.f. = 2V, . How much energy is lost by the battery in charging the capacitors [MP PET 2001]

Options:

A) 22×106J

B) 11×106J

C) (323)×106J

D) (163)×106J

Show Answer

Answer:

Correct Answer: B

Solution:

Ceq=C1C2C1+C2+C3=2×62+6+4=5.5μF

Energy supplied (E)=QV=CV2=22×106J

P.E. stored(U)=12CeqV2=12×5.5×(2)2=11×106J

therefore Energy lost=EU=11×106J



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