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Chapter 17: Problem 84

A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is $9 \mu \mathrm{F}\( and the energy delivered is to be \)300 \mathrm{J},$ to what potential difference must the capacitor be charged?

Short Answer

Expert verified
Based on the given capacitance of a defibrillator and the energy it should deliver, calculate the potential difference that the capacitor must be charged to. Given: - Capacitance, C = 9 μF - Energy, E = 300 J Formula for energy stored in a capacitor: E = (1/2)CV^2 Solution: 1. Rearrange the formula to solve for potential difference, V: V = √(2E/C) 2. Substitute the given values for E and C: V = √(2 × 300J / (9 × 10^-6 F)) 3. Calculate the potential difference: V ≈ 2294.7 V The capacitor must be charged to a potential difference of approximately 2294.7 V to deliver the required energy.

Step by step solution

01

Write down the given information

We are given the following information: - Capacitance, C = \(9\mu F = 9 \times 10^{-6} F\) - Energy, E = \(300J\) The task is to find the potential difference, V.
02

Write down the formula for energy of a capacitor

The formula for the energy stored in a capacitor is: \(E = \frac{1}{2}CV^2\)
03

Solve for the potential difference

We can rearrange the formula to solve for the potential difference, V: \(V = \sqrt{\frac{2E}{C}}\)
04

Substitute the given values and calculate V

Now, we can plug the given values for E and C into the formula: \(V = \sqrt{\frac{2 \times 300J}{9 \times 10^{-6}F}}\)
05

Calculate the potential difference

After calculating the square root, we get: \(V \approx 2294.7 V\) So, the capacitor must be charged to a potential difference of approximately \(2294.7 V\) to deliver the required energy.

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Most popular questions from this chapter

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