In a portable defibrillator, energy is stored in a capacitor charged to a high potential difference. The capacitor

is discharged through the patient’s body in a short, high-powered pulse.
a) The capacitor in a portable defibrillator is charged to 5000 V and stores 800 J of electrical energy. What is its
capacitance?
b) During a pulse, 200 J of energy is delivered to the patient’s body. What is the potential difference across the plates
of the capacitor after a single pulse?
c) We can model the discharging of the capacitor through the patient’s body as an RC circuit, with the heart acting
as the resistor. If a pulse lasts 5 ms, What is the resistance of the path through the heart?

(a) Use E = (1/2) C V^2 to solve for C.

(b) Since only 1/4 of the stored energy is delivered, you still have sqrt(3/4) = 86.6% of the original voltage on the plates.
(c) "Time constant" = 5*10^-3 s = R*C
Solve for R

a) To find the capacitance (C) of the capacitor, we can use the formula:

Energy (E) = 1/2 * C * V^2

Where E is the stored energy, C is the capacitance, and V is the potential difference across the capacitor.

We are given:
Energy (E) = 800 J
Potential difference (V) = 5000 V

Plugging in these values into the formula, we can solve for the capacitance (C):

800 J = 1/2 * C * (5000 V)^2

Simplifying the equation:
1600 J = C * (5000 V)^2

Dividing both sides by (5000 V)^2:
C = 1600 J / (5000 V)^2

Calculating the value:
C = 1600 J / (25,000,000 V^2)

Thus, the capacitance of the capacitor is C = 6.4 x 10^(-6) F.

b) To find the potential difference across the plates of the capacitor after a single pulse, we need to subtract the energy delivered to the patient (200 J) from the initial stored energy (800 J). The remaining energy will be the energy stored in the capacitor.

Remaining energy = Stored energy - Energy delivered
Remaining energy = 800 J - 200 J
Remaining energy = 600 J

Using the previous formula, we can find the potential difference (V) with the remaining energy (600 J) and the capacitance we found in part a (C = 6.4 x 10^(-6) F):

600 J = 1/2 * (6.4 x 10^(-6) F) * V^2

Simplifying the equation:
1200 J = (6.4 x 10^(-6) F) * V^2

Dividing both sides by (6.4 x 10^(-6) F):
V^2 = 1200 J / (6.4 x 10^(-6) F)
V^2 = 187,500,000 V

Taking the square root of both sides:
V = √(187,500,000 V)

Thus, the potential difference across the plates of the capacitor after a single pulse is V ≈ 13,674 V.

c) To determine the resistance (R) of the path through the heart, we can use the time constant (τ) of the RC circuit and the pulse duration (t).

τ = R * C

We are given the pulse duration (t) = 5 ms (which can also be written as 0.005 s) and the capacitance (C) which we found in part a (C = 6.4 x 10^(-6) F).

Plugging in these values:
τ = R * (6.4 x 10^(-6) F)

Since we have the pulse duration (t), we know that the time constant (τ) is equal to t:

t = R * (6.4 x 10^(-6) F)

Solving for resistance (R):
R = t / (6.4 x 10^(-6) F)

Substituting the given values:
R = 0.005 s / (6.4 x 10^(-6) F)

R ≈ 781.25 ohms

Therefore, the resistance of the path through the heart is approximately 781.25 ohms.