Defibrillator: charge, capacitance, power and current problem

[bmandrade]

Homework Statement

A heart dysfunction that can cause death is ventricular fibrillation. This is an
uncoordinated quivering of the heart as opposed to regular beating. An electric shock to
the chest can cause momentary paralysis of the heart muscle, after which the heart will
sometimes start organized beating again. A defibrillator is a device that applies a strong
electric shock to the chest over a time of a few milliseconds. The device contains a
capacitor of several microfarads, charged to several thousand volts. Electrodes called
paddles, about 8 cm across and coated with conducting paste, are held against the chest
on both sides of the heart. Their handles are insulated to prevent injury to the operator,
who calls “Clear!” and pushes a button on one paddle to discharge the capacitor through
the patient's chest.


Assume that the capacitor in the defibrillator is 20.0 μF and is charged to 4,000 Volts.

a. How much charge is stored in the capacitor before it is discharged?
b. How much energy is released when the capacitor is discharged?
c. If the capacitor completely discharges in 2.0 ms, what is the average current
delivered by the defibrillator?
d. What is the average power delivered?

Homework Equations

C= Q/V
Energy stored= 1/2 C*V^2
I = Q\t
P=I*V

The Attempt at a Solution

a. How much charge is stored in the capacitor before it is discharged?
C= Q/V
Q = C*V = 20E-6 * 4000 = .08 C

b. How much energy is released when the capacitor is discharged?
Energy stored= 1/2 C*V^2 = 1/2 (20E-6)(4000^2 = 160 J

c. If the capacitor completely discharges in 2.0 ms, what is the average current
delivered by the defibrillator?
I = Q/t = .08/.002 = 40 A

d. What is the average power delivered?
P = I*V = 40 (4000)= 1.6X10^5W


I don't know if i did this problem right can someone help me

 

[Redbelly98]

a, b, and c look good.

A problem with d:
V is 4000v only at the beginning of the discharge, and drops to 0V by the end of the discharge. So P=I*V=1.6X10^5W is only true at the beginning; the average power will be less.

Do you know any other equations that deal with power?

 

[nlightner]

?

Your calculations for parts a-d appear to be correct. The only thing to note is that in part c, the units for average current should be in Amperes (A) rather than Amps (which is just a shortened form of Amperes). Overall, your response is a good explanation of the concepts of charge, capacitance, power, and current in relation to a defibrillator. However, as a scientist, it would also be important to mention any potential safety concerns or limitations of using a defibrillator, as well as any current research or advancements in the technology. Additionally, it would be helpful to provide context for why this information is important and how it relates to the overall goal of using a defibrillator to save lives.