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

Answer 2

Part A. The charge be "0.075 C".

Part B. The energy stored be "187.5 J".

Part C. The current be "50 A".

Part D. The charge be "1.53 C".

Resistance, Charge and Current:

According to the question,

Capacitor, C = 15 μF or,

                     = 15 × 10⁻⁶ F

Capacity, V = 5.0 kV or,

                   = 5000 V

Resistance, R = 100 Ω

Time, Δt = 90 ms or,

              = 90 × 10⁻³

Defibrillator passes through current, I = 17 A

Part A:

We know the formula,

The charge will be:

→ Q = C × V

By substituting the values,

      = 15 × 10⁻⁶ × 5000

      = 0.075 C

Part B:

We know the formula,

The energy stored will be:

→ E = $\frac{1}{2}$ × CV²

By substituting the values,

     =  $\frac{1}{2}$ × 15 × 10⁻⁶ (5000)²  

     = 187.5 J

Part C:

We know the formula,

→ Voltage, V = IR

or,

→ Current, I = $\frac{V}{R}$

By substituting the values,

                   = $\frac{5000}{100}$

                   = 50 A  

Part D:

We know the formula,

→ I = $\frac{\Delta Q}{\Delta t}$

or,

Charge, ΔQ = I × Δt

By substituting the values,

                    = 17 × 90 × 10⁻³

                    = 1.53 C

Thus the above responses are correct.

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