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C1 = C2 = 1 Farad. C1 is charged to 2 volts and C2 is discharged.
At t = 0, the switch closes. The charge on C1 divides between C1
and C2, so both capacitors carry 1 volt. The charge on a capacitor
is described by:
Q = CV
In this case, C1 started at 2 volts, so it held 2 Coulombs; after
the switch is flipped, each capacitor holds 1 Coulomb.
The energy stored in a capacitor is:
1/2 (CV2)
Before and after analysis of the energy stored (Joules):
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C1
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C2
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TOTAL
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BEFORE
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2
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0
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2
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AFTER
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1/2
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1/2
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1
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Question:
What happened to the missing Joule?
Answer
See all EE-Quiz Questions
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