Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 25 - Capacitance - Problems - Page 742: 31



Work Step by Step

Total energy stored $U$ can be determined as: $U=\frac{1}{2}CV^2$....................eq(1) Since the capacitors are connected in parallel, the equivalent capacitance $C$ is $C=C_1+C_2=2+4=6\mu F$ We then put the values in eq(1) and solve: $U=\frac{1}{2}({6\times10^{-6}}{)(300)}^2$ $U=0.27J$
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