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1. Two capacitors (2.0 μF and 4.0 μF) are connected in parallel across a 300-V potential difference. Calculate the total stored energy in the system.
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Answer #1

As the capacitors connected in parallel so equivalent capacitance wil be (2+4)×10^-6=6×10^-6C

Total energy of system =1/2×CV^2=1/2×6×10^-6×300×300=0.27J

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