The elements of a simple defibrillator
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The elements of a simple defibrillator
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[post_date] => 2025-01-09 07:59:57
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[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Defibrillators are medical devices used to treat life threatening cardiac arrhythmias. Defibrillators deliver electrical energy to the heart with the goal of depolarizing the heart muscle completely, and reverting the heart back to a normal sinus rhythm. A simple defibrillator consists of a power supply, a capacitor, an inductor, and a set of paddles. Figure 1 illustrates a circuit diagram of a simple defibrillator.
Figure 1: Schematic of a Simple Defibrillator
When the switch is connected to point A, the capacitor is connected to the power supply. The electrical current begins to flow through the circuit until the voltage at the terminals of the capacitor is equal to the voltage of the power supply. At this point, the capacitor has reached its total charging capacity and the current flowing through the circuit has stopped.
A graph depicting how the current changes with respect to time as the capacitor is being charged is illustrated below. Mathematically, the graph is represented by the equation
, where RC is the product of the circuit resistance and the capacitor’s capacitance. RC is also known as the time constant. The greater the time constant, the longer it takes for the current to approach zero.
Graph 1: Current vs Time graph for a Charging Capacitor
When the switch is connected to point B, the capacitor begins to discharge and electrical energy is delivered through the paddles to the heart. The current delivered to the heart must last for several milliseconds in order for the heart to completely depolarize. However, the current of a discharging capacitor decreases exponentially. An inductor is needed to prolong the duration of the current flow by inducing a voltage that opposes current flow. The tendency to resist current flow is called inductance. The amount of electrical energy delivered to the heart can be calculated by the equation: E=(Q × Vₒ) /2 where Q is the charge on either plate of the capacitor and Vₒ is the voltage of the power supply.
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[question] => Which of the following graphs correctly depicts how the charge (Q) on the capacitor varies with respect to time (t) when the capacitor is charging?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
When the capacitor is charging, it continues to store charges until it reaches its total charging capacity.
The capacitor will no longer store any additional charges when it is fully charged.
The graph depicting a charging capacitor will rise quickly and then plateau.
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[each_answer] => A.
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[each_answer] => B.
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[each_answer] => C.
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[each_answer] => D.
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => At t = RC, what is the current in the circuit when the capacitor is charging?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
The current can be calculated from the equation provided in the passage:
Substitute RC or t in the equation.
Simplify the equation to determine the correct answer.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Vₒ/Re
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[1] => Array
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[each_answer] => B. Vₒ/2R
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[2] => Array
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[each_answer] => C. Vₒ/R
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[each_answer] => D. Vₒ/RC
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => Which of the following correctly depicts the direction the charge (Q) flows through the circuit i) when the capacitor is charging; ii) when the capacitor is discharging?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Electrons move from the negative terminal of the battery to the positive terminal.
Current is the flow of positive charges, which is opposite to the flow of electrons
The capacitor plates separate and store positive and negative charges.
As shown in the diagram above, the current flows clockwise while the capacitor is charging, and the current also flows clockwise when the capacitor is discharging. The correct answer is i) clockwise; ii) clockwise.
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[each_answer] => A. i) clockwise; ii) clockwise
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[each_answer] => B. i) counterclockwise; ii) counterclockwise
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[2] => Array
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[each_answer] => C. i) counterclockwise; ii) clockwise
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[each_answer] => D. i) clockwise; ii) counterclockwise
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[3] => Array
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[quiz_unique_key] => 2261298308
[question] => Which of the following will shorten the time required for the capacitor to become fully charged?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
The time constant, which is the product of the circuit resistance and the capacitor’s capacitance RC, determines how long it takes for the current to stop flowing.
Anything that decreases R or C will decrease the time it takes for the capacitor to become fully charged.
Capacitance ( C) equals Q /Ed. Increasing the distance between the parallel plates of the capacitor will decrease C and hence the time constant.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Increasing the inductance of the inductor
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[1] => Array
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[each_answer] => B. Increasing the distance between the parallel plates of the capacitor
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[2] => Array
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[each_answer] => C. Using a transformer to increase the voltage of the power supply
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[each_answer] => D. Using a capacitor with plates that have a greater surface area
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[quiz_unique_key] => 574431310
[question] => If a 2 x 10⁻⁵ farad capacitor and a 5000 volt battery are used in the circuit depicted in Figure 1, then what is the total electrical energy delivered through the paddle when the capacitor is discharging?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
The amount of electrical energy delivered to the heart can be calculated by the equation: E = (Q Vo) /2
C = Q/ Vo Rearrange this equation to solve for Q
Q = C Vo
Substitute the second equation into the first one to get the following equation: E = 12 (C V2)
Plug the values provided in the question stem into this newly derived equation to calculate the total electrical energy discharged from the capacitor.
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[answers] => Array
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[0] => Array
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[each_answer] => A. 750 Joules
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[1] => Array
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[each_answer] => B. 250 Joules
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[2] => Array
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[each_answer] => C. 500 Joules
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[each_answer] => D. 1000 Joules
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[558272|1] => C
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