What is the energy (in J) stored in the 31.0 µF capacitor of a heart defibrillator charged to 6.70 ✕ 103 V? J

Medicine

2 answers:

yuradex [85]4 years ago

6 0

Answer:

Energy stored = 695.8joules

Explanation:

Energy stored in a capacitor = C * v²/2

Where c = capacitance = 31ụf

V = voltage = 6.70 *10^3

So therefore:

Energy = (31 * 10^-6) * ( 6.70 * 10^3)²/2

= 695.8 joules

dezoksy [38]4 years ago

6 0

Answer: Note that voltage of the defibrillator is given as 6.70*103V instead of 6.70*10^3volts

The answer is:

Ecap=6957950joules

Explanation:

Ecap=CV^2/2

Where Ecap( Energy in joules) =?

V(Voltage in volt)=6.70*10^3volts

C(capacitance )=31.0µF =31*10^-6F

Ecap=(31*10^-6F)*(6.70*10^3volts)^2/2=13915900/2

Ecap=6957950joules

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