A capacitor of capacitance $150 μ F$ is charged to a potential difference of 200V and then connected across the discharged tube which conducts until the potential difference across it has fallen to 80 volt. The energy dissipated in the tube is $P \times 10^{- 2} J$ . Find P.

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answer is 252.

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Detailed Solution

$C = 150 \times 10^{- 6} = 15 \times 10^{- 5} F, V_{1} = 200 V$

Energy $U_{1} = \frac{1}{2} C V_{1}^{2} = \frac{1}{2} 15 \times 10^{- 5} \times 4 \times 10^{4} = \frac{6}{2} = 3 J$

Energy left on capacitor at 80V $(V_{2} = 80 V)$

$U_{2} = \frac{1}{2} C V_{2}^{2} = \frac{1}{2} 15 \times 10^{- 5} \times 6400 = 32 \times 15 \times 10^{- 3} = 480 \times 10^{- 3} = 0.48 J$

Energy dissipated $Δ U = 3 - 0.48 = 2.52 = 252 \times 10^{- 2} J$

P=252

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