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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorForce & Energy analysisHard2 minPYQ_2004

PhysicsHardsingle choice

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 $2 ms$ . The power delivered to the patient is

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Solution:

The energy stored in capacitor $= \frac{1}{2} C V^{2}$ $= \frac{1}{2} \times 40 \times 1 0^{- 6} \times (3000)^{2} = 20 \times 9 \times 1 0^{6} \times 1 0^{- 6} = 180 J .$ The power delivered in $2 ms$ is $= \frac{180}{2 \times 1 0 ^{- 3}} = 90 \times 1 0^{3} W = 90 kW .$

Stream:AIIMSSubject:PhysicsTopic:CapacitorSubtopic:Force & Energy analysis

⏱ 2mℹ️ Source: PYQ_2004

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