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A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in this device is charged to \(7.5 \mathrm{kV}\) and stores \(1200 \mathrm{~J}\) of energy. What is its capacitance?

Short Answer

Expert verified
The capacitance is \( 42.7 \mathrm{~\mu F} \).

Step by step solution

01

Understand the Formula

The energy stored in a capacitor is given by the formula: \( E = \frac{1}{2} C V^2 \), where \( E \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage.
02

Rearrange the Formula for Capacitance

We need to find the capacitance \( C \). Rearrange the formula to make \( C \) the subject: \( C = \frac{2E}{V^2} \).
03

Convert Voltage to Volts

The given voltage is \( 7.5 \mathrm{kV} \). Convert this to volts: \( 7.5 \mathrm{kV} = 7500 \mathrm{V} \).
04

Substitute Values into Formula

Substitute the given values into the formula: \( C = \frac{2 \times 1200 \mathrm{~J}}{(7500 \mathrm{~V})^2} \).
05

Perform the Calculations

Calculate the numerator: \( 2 \times 1200 = 2400 \). Calculate the denominator: \( (7500)^2 = 56250000 \). Then, \( C = \frac{2400}{56250000} = 4.27 \times 10^{-5} \mathrm{~F} \).
06

Convert Capacitance to Microfarads

Since capacitance is usually given in microfarads, convert the result: \( 4.27 \times 10^{-5} \mathrm{~F} = 42.7 \mathrm{~\mu F} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Storage
Capacitors are essential components for energy storage in electronic circuits. They temporarily hold electric charge, storing energy for various applications.
When a capacitor is charged, it accumulates energy in an electric field between its plates. The amount of energy stored is determined by its capacitance and the voltage across it. This makes capacitors crucial for devices like cardiac defibrillators, where rapid energy discharge is needed.
The stored energy (E) in a capacitor is calculated using the formula: \( E = \frac{1}{2} C V^2 \). Here, \( C \) represents capacitance and \( V \) is voltage.
This equation illustrates how energy varies with both capacitance and voltage. A larger capacitance or higher voltage increases energy storage ability. This concept of energy storage is vital in many technological and medical devices.
Capacitance Calculation
Calculating capacitance is crucial in understanding a capacitor's ability to store charge.
Capacitance (C) indicates how much electric charge a capacitor can hold at a given voltage. For a successful calculation, we rearrange the energy formula to solve for \( C \).
The formula becomes: \( C = \frac{2E}{V^2} \), where \( E \) is energy and \( V \) is voltage in volts.
  • First, ensure the voltage is in volts before calculation.
  • Substitute the known values of energy and converted voltage into the formula.
  • Solve the equation to find the capacitance.

For the cardiac defibrillator, after substituting the values, the capacitance is found to be \( 4.27 \times 10^{-5} \, \mathrm{F} \), or \( 42.7 \, \mu \mathrm{F} \). This conversion into microfarads helps convey a meaningful and practical measurement.
Cardiac Defibrillator
A cardiac defibrillator is a medical device used to restore a normal heartbeat. It delivers a dose of electric current to the heart, effectively "shocking" it back into a regular rhythm.
In this application, capacitors play a pivotal role. They store the electrical energy needed for the defibrillator to function efficiently.
The rapid release of stored energy helps achieve the sudden burst necessary to restart the heart.
For this, the defibrillator's capacitor is charged to a high voltage of 7500 volts, sufficient to store 1200 Joules of energy.
  • Capacitance and the ability to efficiently deliver energy are key to the effectiveness of the defibrillator.
  • A well-designed capacitor ensures the correct amount of energy is released in a controlled manner.
  • This precise energy delivery is critical for the device's success in medical emergencies.

Understanding the role of capacitors in such life-saving technology underscores their importance in modern medicine.

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Most popular questions from this chapter

(I) What is the capacitance per unit length (F/m) of a coaxial cable whose inner conductor has a 1.0 -mm diameter and the outer cylindrical sheath has a \(5.0-\mathrm{mm}\) diameter? Assume the space between is filled with air.

(II) \(\mathrm{A} 2.70-\mu \mathrm{F}\) capacitor is charged to 475 \(\mathrm{V}\) and a \(4.00-\mu \mathrm{F}\) capacitor is charged to 525 \(\mathrm{V}\) . (a) These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected?

(II) Six physics students were each given an air filled capacitor. Although the areas were different, the spacing between the plates, \(d,\) was the same for all six capacitors, but was unknown. Each student made a measurement of the area \(A\) and capacitance \(C\) of their capacitor. Below is a Table for their data. Using the combined data and a graphing program or spreadsheet, determine the spacing \(d\) between the plates.

A multilayer film capacitor has a maximum voltage rating of \(100 \mathrm{~V}\) and a capacitance of \(1.0 \mu \mathrm{F}\). It is made from alternating sheets of metal foil connected together, separated by films of polyester dielectric. The sheets are \(12.0 \mathrm{~mm}\) by \(14.0 \mathrm{~mm}\) and the total thickness of the capacitor is \(6.0 \mathrm{~mm}\) (not counting the thickness of the insulator on the outside). The metal foil is actually a very thin layer of metal deposited directly on the dielectric, so most of the thickness of the capacitor is due to the dielectric. The dielectric strength of the polyester is about \(30 \times 10^{6} \mathrm{~V} / \mathrm{m}\). Estimate the dielectric constant of the polyester material in the capacitor.

(1) 2200 \(\mathrm{V}\) is applied to a \(2800-\mathrm{pF}\) capacitor. How much electric energy is stored?

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