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PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

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A $40 μ F$ capacitor in a defibrillator is charged to $3000 V$ . The energy stored in the capacitor is sent through the patient during a pulse of duration $2 m s$ . The power delivered to the patient is

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Answer:

Solution:

Power, $P = \frac{E n e r g y}{t i m e} = \frac{\frac{1}{2} C V^{2}}{t}$

$P = \frac{C V^{2}}{2 t}$

$P = \frac{40 \times 1 6^{- 6} \times 9 \times 1 0^{6}}{2 \times 2 \times 1 0^{- 3}}$

$P = 90 \times 1000$

$P = 90 k W$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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