A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 47.0 μF capacitor is charged to 5.91 kV. Paddles are used to make an electrical connection to the patient's chest. A pulse of current lasting 1.00 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is 220 Ω .
a) What is the initial energy stored in the capacitor?
b) What is the initial current through the patient?
c) How much energy is dissipated in the patient during the 1.00 ms?
d) If it takes 2.14 s to recharge the capacitor, compare the average power supplied by the power source with the average power delivered to the patient.
The average power supplied by the power source is ?????? times that delivered to the patient.
I have figured out:
a) 8.21×102 J
b) 2.69×101 A
c) 1.44×102 J
I know these answers are correct. But I am having difficulties figuring out d).
I've been going over that problems so many different ways and my answers are always wrong. Somewhere I am making a mistake. Can someone help me please.
average power suppied=answerA/time
in the mean time (nine) minus (one) (eight)
They're gonna want his milk money next poor Pluto
Bring back Pluto
This is Russian dolls that getting smaller and smaller still, jesus pleebus man now tell wtf am i supposed to do
we were busy putting bars on a big iron gate
To calculate the average power supplied by the power source and compare it with the average power delivered to the patient, we need to calculate the energy dissipated in the patient per unit time.
Here's how you can solve this:
Step 1: Calculate the charge on the capacitor.
The charge stored on the capacitor can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.
Q = (47.0 μF) × (5.91 kV) = 2.775 C
Step 2: Calculate the initial energy stored in the capacitor.
The energy stored in the capacitor can be calculated using the formula E = 1/2 CV^2, where E is the energy and C is the capacitance, and V is the voltage.
E = 1/2 × (47.0 μF) × (5.91 kV)^2 = 434.674 J
Step 3: Calculate the energy dissipated in the patient during the 1.00 ms.
The energy dissipated can be calculated using the formula E = 1/2 Q^2 / C, where E is the energy, Q is the charge, and C is the capacitance.
E = 1/2 × (2.775 C)^2 / (47.0 μF) = 102.340 J
Step 4: Calculate the average power supplied by the power source.
The average power supplied by the power source can be calculated using the formula P = E/t, where P is the power, E is the energy, and t is the time.
P = 434.674 J / 2.14 s = 203.148 W
Step 5: Calculate the average power delivered to the patient.
The average power delivered to the patient can be calculated using the formula P = E/t, where P is the power, E is the energy, and t is the time.
P = 102.340 J / 0.001 s = 102340 W
Step 6: Compare the average power supplied by the power source with the average power delivered to the patient.
The average power supplied by the power source is 203.148 W, and the average power delivered to the patient is 102,340 W.
Therefore, the average power supplied by the power source is approximately 1/504 times that delivered to the patient.
I hope this helps you solve the problem and understand the process! Let me know if you have any further questions.