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Chapter 17: Problem 85

A defibrillator consists of a 15 - \(\mu\) F capacitor that is charged to $9.0 \mathrm{kV} .\( (a) If the capacitor is discharged in \)2.0 \mathrm{ms}$, how much charge passes through the body tissues? (b) What is the average power delivered to the tissues?

Short Answer

Expert verified
Answer: The total charge passing through the body tissues during the capacitor discharge is 135 mC, and the average power delivered to the body tissues during this process is 303.75 MW.

Step by step solution

01

Calculate the charge on the capacitor

To calculate the charge on the capacitor, we can simply use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor. In this case, C = 15渭F and V = 9.0kV, so the formula becomes: Q = (15 * 10^{-6} F) * (9 * 10^3 V)
02

Evaluate the charge on the capacitor

Using the formula from Step 1, we can calculate the charge Q: Q = (15 * 10^{-6} F) * (9 * 10^3 V) = 135 * 10^{-3} C = 135 mC So the total charge passing through the body tissues during the capacitor discharge is 135 mC.
03

Calculate the energy stored in the capacitor

The energy stored in the capacitor can be found using the formula E = (1/2)CV^2, where E is the energy, C is the capacitance, and V is the voltage across the capacitor. Plugging the given values into this equation, we get: E = (1/2) * (15 * 10^{-6} F) * (9 * 10^3 V)^2
04

Evaluate the energy stored in the capacitor

Using the formula from Step 3, we can calculate the energy E: E = (1/2) * (15 * 10^{-6} F) * (9 * 10^3 V)^2 = 607.5 * 10^{3} J = 607.5 kJ The energy stored in the capacitor is 607.5 kJ.
05

Find the average power delivered to the tissues

The average power delivered to the tissues can be found by dividing the total energy by the discharge time. Therefore, the average power P is given by: P = E/t = (607.5 * 10^3 J) / (2.0 * 10^{-3} s)
06

Evaluate the average power delivered to the tissues

Using the formula from Step 5, we can calculate the average power P: P = (607.5 * 10^3 J) / (2.0 * 10^{-3} s) = 303.75 * 10^{6} W = 303.75 MW The average power delivered to the body tissues during the capacitor discharge is 303.75 MW.

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