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

45 k W

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90 k W

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180 k W

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360 k W

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Solution

The correct option is B $90 k W$
The work done by the capacitor $, W =$ energy stored in it $= \frac{1}{2} C V^{2} = \frac{1}{2} \times (40 \times 10^{- 6}) \times (3000)^{2} = 180 J$
Power $P = \frac{W}{t} = \frac{180}{2 \times 10^{- 3}} = 90 k W$

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A 40μF capacitor in a defibrillator is charged to 3000V.The energy stored in the capacitor is set through the patient during a pulse of duration 2ms, The power delivered to the patient is :

The correct option is B 90kWThe work done by the capacitor ,W= energy stored in it =12CV2=12×(40×10−6)×(3000)2=180JPower P=Wt=1802×10−3=90kW

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

The correct option is B $90 k W$
The work done by the capacitor $, W =$ energy stored in it $= \frac{1}{2} C V^{2} = \frac{1}{2} \times (40 \times 10^{- 6}) \times (3000)^{2} = 180 J$
Power $P = \frac{W}{t} = \frac{180}{2 \times 10^{- 3}} = 90 k W$