Question

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 A

What is the charge initially stored on the capacitor?

a. 3×10−9 C
b. 7.5×104 C
c. 7.5×10−2 C
d. 7.5×10−5 C

Part B

What is the energy stored on the capacitor?

a. 1.9×108 J
b. 380 J
c. 190 J
d. 1.9×10−4 J

Part C

If 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?

a. 5×105 A
b. 50 A
c. 2×10−2 A
d. 5×10−2 A

Part D

If 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?

a. 190 mC
b. 1.5 C
c. 1.5 mC
d. 17 C

Answer 1

Answer:

Part A: Q = 0.075C

Part B: E = 187.5J

Part C: I = 50A

Part D: ΔQ 1.53C

Explanation:

Part A

Q = C×V

Given

C = 15μF = 15×10‐⁶ F, V = 5kV = 5000V

Q = 15×10‐⁶× 5000 = 0.075C

Part B

Energy = E = 1/2 ×CV² = 1/2 × 15×10‐⁶ × 5000² = 187.5J

Part C

Given R = 100Ω

V = IR, I = V/R = 5000/100 = 50A

Part D

I = ΔQ/Δt

Given Δt= 90ms = 90×10-³s, I = 17A

ΔQ = I × Δt = 17×90×10-³ = 1.53 C