Question

The 88 F capacitor in a defibrillator unit supplies an average of 6500 W of power to the chest of the patient during a discharge lasting 5.0 ms. Part A To what voltage is the capacitor charged? Express your answer with the appropriate units. MA ?

Answer 1

To find the voltage to which the capacitor in a defibrillator is charged, the energy delivered during a discharge is calculated using power and time, and then combined with the capacitance value to solve for voltage. The calculated voltage is approximately 27.16 kV.

Step-by-step explanation:

To calculate the voltage to which the 88 F (farad) capacitor in the defibrillator unit is charged, we would first need to determine the amount of energy stored in the capacitor using the power and time of the discharge. The formula for power is power = energy/time, so we can rearrange the formula to solve for energy: energy = power * time.

Using the provided values, the energy (E) stored in the capacitor can be calculated as:
E = 6500 W * 5.0 ms
E = 6500 W * 5.0 * 10-3 s
E = 32.5 J

Next, we use the formula for the energy stored in a capacitor: E = 1/2 * C * V2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts. We can solve for V:
V = sqrt(2 * E / C)

We then substitute the values for E and C:
V = sqrt(2 * 32.5 J / 88 F)
V = sqrt(2 * 32.5 J / 0.088 F)
V ≈ sqrt(737.5)
V ≈ 27.16 kV

Therefore, the capacitor is charged to approximately 27.16 kV (kiloVolts).