Question
Two capacitors, 2 µF and 3 µF, are charged separately
to 100 V and 50 V respectively, and then connected positive to positive, negative to negative. Find the loss in energy due to redistribution.Solution
We are given:
- Capacitor 1: C₁ = 2 µF, charged to V₁ = 100 V
- Capacitor 2: C₂ = 3 µF, charged to V₂ = 50 V
- They are connected positive to positive, negative to negative ⇒ charges redistribute, no reversal of polarity
- We are to find the loss in energy after redistribution
Two capacitors of capacitance 6 µF and 3 µF are connected in series across a battery of constant voltage.
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