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A defibrillator is a device used to shock the heart back to normal beat patterns. To do this, it discharges a 15 μF capacitor through paddles placed on the skin, causing charge to flow through the heart. Assume that the capacitor is originally charged with 5.0 kV .Part AWhat is the charge initially stored on the capacitor?3×10−9 C7.5×104 C7.5×10−2 C7.5×10−5 CPart BWhat is the energy stored on the capacitor?What is the energy stored on the capacitor?1.9×108 J380 J190 J1.9×10−4 JPart CIf the resistance between the two paddles is 100 Ω when the paddles are placed on the skin of the patient, how much current ideally flows through the patient when the capacitor starts to discharge?5×105 A50 A2×10−2 A5×10−2 APart DIf a defibrillator passes 17 A of current through a person in 90 μs . During this time, how much charge moves through the patient?If a defibrillator passes 17 {\rm A} of current through a person in 90 {\rm \mu s} . During this time, how much charge moves through the patient?190 mC1.5 C1.5 mC17 C

User Jj Tuibeo
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1 Answer

3 votes

Answer:

a) q = 7.5 10⁻² C , b) 190 J , c) I₀ = 50 A , d) 1.5 mC

Step-by-step explanation:

The expression for capacitance is

C = q / DV

q = C DV

let's reduce the magnitudes to the SI system

ΔV = 5 kV = 5000 V

C = 15 μF = 15 10⁻⁶ F

t = 90 μs = 90 10⁻⁶ s

q = 15 10⁻⁶ 5000

q = 7.5 10⁻² C

b) the energy in a capacitor is

U = ½ C ΔV²

U = ½ 15 10⁻⁶ 5000²

U = 1,875 10² J

answer 190 J

c) At the moment the discharge begins, all the current is available and it decreases with time,

whereby

V = I R

in the first instant I = Io

I₀ = V / R

I₀ = 5000/100

I₀ = 50 A

but this is for a very short time

answer 50 A

d) The definition of current is

i = dq / dt

in this case they give us the total current and the total time, so we can find the total charge

i = q / t

q = i t

q = 17 90 10⁻⁶

q = 1.53 10⁻³ C

answer is 1.5 mC

User Behram Mistree
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