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A Capacitor of capacitance `C=5muF` is connected to source of emf `varepsilon_(2)=200V` with the switch S in the position 1 (Figure 6.2). Subsequently the switch is pushed to the position2. Find the amount of heat generated in `R_(1)=500 Omega if R_(2)=300Omega`.
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Correct Answer - `H=C epsilon^(2)R_(1)//2(R_(1)+R_(2)) =62.5mJ`

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