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A 50μf capacitor in a defibrillator is charged to 3000V. The energy stored in the capacitor is sent through the victim during a pulse of duration 2 ms; the power of the pulse is close to: 
1. 50 Kw
2. 75 kW
3. 112.5 kW
4. 200 kW

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Correct Answer - Option 3 : 112.5 kW

Concept:

Energy stored in condenser:

A capacitor is a device which stores energy in the form of charge.

The process of charging up a capacitor involves the transferring of electric charges from one plate to another.

The energy stored in the capacitor is;

\(U = \;\frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\) ----(1)
Where, 

Q = charge stored on the capacitor 

U = energy stored in the capacitor 

C = capacitance of the capacitor  

V = Electric potential difference

Energy: The ability to do work is defined as Energy.

Power: It is defined as the ratio of work done to that of the time taken

Power = work / time

So, Energy = Power × Time  ----(2)

Calculation: 

Given:

C = 50μf, V = 3000V, t = 2 ms

From equation (1);

\(U=\frac{1}{2}\times50\times 10^{-6}\times(3000)^2\)

U = 225 J

From equation (2);

\(Power = \frac{225}{2\times 10^{-3}}\)

Power = 112.5 KilloWatt

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