A defibrillator is used to restart a person's heart after it stops beatingEnergy is delivered to the heart by discharging a capacitor through the body tissues near the heartIf the capacitance of the defibrillator is 9.40 and the energy delivered is to be 251 Jto what potential difference must the capacitor be charged? Kv
A defibrillator is used to restart a person's heart after it stops beatingEnergy is delivered to the heart by discharging a capacitor through the body tissues near the heartIf the capacitance of the defibrillator is 9.40 and the energy delivered is to be 251 Jto what potential difference must the capacitor be charged? Kv
College Physics
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ISBN:9781305952300
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A defibrillator is used to restart a person's heart after it stops beatingEnergy is delivered to the heart by discharging a capacitor through the body tissues near the heartIf the capacitance of the defibrillator is 9.40 and the energy delivered is to be 251 Jto what potential difference must the capacitor be charged?
Kv
![**Potential Difference in a Defibrillator Capacitor**
A defibrillator is a crucial medical device used to restart a person's heart after it stops beating. The energy necessary to restart the heart is delivered by discharging a capacitor through body tissues near the heart. Understanding the electrical properties involved is important for the effective use of the defibrillator.
**Given Information:**
- Capacitance (C) of the defibrillator: 9.40 μF (microfarads)
- Energy (E) delivered: 251 J (joules)
**Problem:**
To determine the potential difference (V) to which the capacitor must be charged in order to deliver the specified energy.
**Formula:**
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
Where:
- \(E\) is the energy (in joules, J),
- \(C\) is the capacitance (in farads, F),
- \(V\) is the potential difference (in volts, V).
**Calculating Potential Difference:**
Rearrange the formula to solve for \(V\):
\[ V = \sqrt{\frac{2E}{C}} \]
Plugging in the given values:
\[ V = \sqrt{\frac{2 \times 251 \, J}{9.40 \times 10^{-6} \, F}} \]
\[ V = \sqrt{\frac{502}{9.40 \times 10^{-6}}} \]
\[ V \approx \sqrt{53.40 \times 10^6} \]
\[ V \approx 7309 \, V \]
Thus, the potential difference required is approximately 7.31 kV.
In the provided image, the potential difference is incorrectly identified as 6.96 kV.
**Visual Explanation and Summary:**
The image features a calculation box where users can input their answers. Below the input field for the potential difference, the user's incorrect answer of 6.96 kV is marked with an 'X' indicating an error. The necessary potential difference calculation shows that the capacitor should be charged up to approximately 7.31 kV to deliver 251 joules of energy effectively to restart the heart.
Providing accurate calculations in such contexts ensures that medical devices like defibrillators work correctly, potentially saving lives.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b578b6a-d8d1-4da1-b505-9cebf317d6ed%2F1c445b06-a3e9-4f28-a2bd-230a730489c4%2Ftmj456p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Potential Difference in a Defibrillator Capacitor**
A defibrillator is a crucial medical device used to restart a person's heart after it stops beating. The energy necessary to restart the heart is delivered by discharging a capacitor through body tissues near the heart. Understanding the electrical properties involved is important for the effective use of the defibrillator.
**Given Information:**
- Capacitance (C) of the defibrillator: 9.40 μF (microfarads)
- Energy (E) delivered: 251 J (joules)
**Problem:**
To determine the potential difference (V) to which the capacitor must be charged in order to deliver the specified energy.
**Formula:**
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
Where:
- \(E\) is the energy (in joules, J),
- \(C\) is the capacitance (in farads, F),
- \(V\) is the potential difference (in volts, V).
**Calculating Potential Difference:**
Rearrange the formula to solve for \(V\):
\[ V = \sqrt{\frac{2E}{C}} \]
Plugging in the given values:
\[ V = \sqrt{\frac{2 \times 251 \, J}{9.40 \times 10^{-6} \, F}} \]
\[ V = \sqrt{\frac{502}{9.40 \times 10^{-6}}} \]
\[ V \approx \sqrt{53.40 \times 10^6} \]
\[ V \approx 7309 \, V \]
Thus, the potential difference required is approximately 7.31 kV.
In the provided image, the potential difference is incorrectly identified as 6.96 kV.
**Visual Explanation and Summary:**
The image features a calculation box where users can input their answers. Below the input field for the potential difference, the user's incorrect answer of 6.96 kV is marked with an 'X' indicating an error. The necessary potential difference calculation shows that the capacitor should be charged up to approximately 7.31 kV to deliver 251 joules of energy effectively to restart the heart.
Providing accurate calculations in such contexts ensures that medical devices like defibrillators work correctly, potentially saving lives.
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