A 40 μ F capacitor in a defibrillator is charged to 3000,V . energy stored in capacitor is sent thr

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2. Electric Potential and Capacitance

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A $40$ $\mu F$ 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......$kW$

A

$45$

B

$90$

C

$180$

D

$360$

(AIIMS-2004)

Solution

Power $ = \frac{{\frac{1}{2}C{V^2}}}{t} = \frac{{1 \times 40 \times {{10}^{ – 6}} \times {{(3000)}^2}}}{{2 \times 2 \times {{10}^{ – 3}}}} = 90\,kW$

Standard 12

Physics

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