Cardiac dysrhythmia and defibrillators
- Dashboard
- Cardiac dysrhythmia and defibrillators
Most Recent Score Report
| Questions Correct | Questions Answered | Time Spent | Status | Attempt Date | |
|---|---|---|---|---|---|
| -- | -- | -- | -- | -- |
Previous Score Reports
| Questions Correct | Questions Answered | Time Spent | Status | Attempt Date | |
|---|---|---|---|---|---|
| -- | -- | -- | -- | -- |
Cardiac dysrhythmia and defibrillators
Array
(
[passage] => WP_Post Object
(
[ID] => 559030
[post_author] => 12815
[post_date] => 2025-01-09 08:03:59
[post_date_gmt] => 2025-01-09 13:03:59
[post_content] => Practice Passage (Question 1-4)
*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.
Cardiac dysrhythmia, or irregular heartbeat, is a condition that can have a number of causes and manifestations. One example of this is fibrillation, a condition in which the heart erratically contracting due to spontaneous electrical impulses from the cardiac nodes.
A defibrillator is a device that charges up and can deliver a large amount of electrical energy in a short amount of time to someone suffering from cardiac dysrhythmia. This burst of electrical energy has the potential to briefly stop a heart that is beating irregularly. The heart will then immediately restart, and will often revert to a steady, normal beating pace again, as the spontaneous impulses have now been eliminated. One of the main components in a defibrillator is a capacitor.
Figure 1. Defibrillator placement on a patient with cardiac dysrhythmia.
The hospital you work for uses a defibrillator that has a single capacitor with a capacitance of 10-6 F. When it is about to be used, the defibrillator is charged to a voltage of 4000 V.
[post_title] => Cardiac dysrhythmia and defibrillators
[post_excerpt] =>
[post_status] => publish
[comment_status] => closed
[ping_status] => closed
[post_password] =>
[post_name] => cardiac-dysrhythmia-and-defibrillators
[to_ping] =>
[pinged] =>
[post_modified] => 2025-01-09 08:06:15
[post_modified_gmt] => 2025-01-09 13:06:15
[post_content_filtered] =>
[post_parent] => 0
[guid] => https://medlifemastery.com/?post_type=passage&p=559030
[menu_order] => 0
[post_type] => passage
[post_mime_type] =>
[comment_count] => 0
[filter] => raw
)
[questions] => Array
(
[0] => Array
(
[quiz_unique_key] => 578908434
[question] => How much charge is stored on the defibrillator ‘s capacitor when it has been fully charged?
[value] => Array
(
[answer] => 3
[description] => Reason for the Correct Answer:
The definition of capacitance is C = Q/V.
We’re told that the capacitor is subjected to a voltage of 4000V.
Plug in the capacitance of 10-6 F for C, and the voltage of 4000V for V. Then solve for Q.Q = C x V, so Q = ( 10-6 F) x (4000 V) = 0.004 C
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 10 C
)
[1] => Array
(
[each_answer] => B. 4 C
)
[2] => Array
(
[each_answer] => C. 0.004 C
)
[3] => Array
(
[each_answer] => D. 0.001 C
)
)
)
[1] => Array
(
[quiz_unique_key] => 3873426850
[question] => If you were to connect an additional 3 x 10-6F capacitor in parallel with the original 10-6F capacitor in the defibrillator, what would be the equivalent capacitance?
[value] => Array
(
[answer] => 1
[description] => Reason for the Correct Answer:
For parallel capacitors, you can add individual capacitors to find the equivalent capacitance
The capacitors you are adding in parallel are 3 x 10-6 F and 10-6
Adding 3 x 10-6 F + 10-6 F gives 4 x 10-6 F
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 4 x 10-6 F
)
[1] => Array
(
[each_answer] => B. 0.1 x 10-6 F
)
[2] => Array
(
[each_answer] => C. 2 x 10-6 F
)
[3] => Array
(
[each_answer] => D. 0.5 x 10-6 F
)
)
)
[2] => Array
(
[quiz_unique_key] => 83407773
[question] => Increasing the separation between the capacitor plates while hooked up to the voltage source will decrease the charge stored for which of the following reasons?
[value] => Array
(
[answer] => 4
[description] => Reason for the Correct Answer:
The capacitor is hooked up to the voltage source, so the voltage across the plates will remain the same.
The capacitance of a parallel plate capacitor is C=eoA/D
The separation between plates is the variable d, and in the formula : C=eoA/d, as d increases the capacitance decreases.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. It increases the capacitance of the capacitor.
)
[1] => Array
(
[each_answer] => B. It increases the voltage across the capacitor plates.
)
[2] => Array
(
[each_answer] => C. It decreases the voltage across the capacitor plates.
)
[3] => Array
(
[each_answer] => D. It decreases the capacitance of the capacitor.
)
)
)
[3] => Array
(
[quiz_unique_key] => 2261298308
[question] => How much energy is stored in the defibrillator ‘s capacitor when it has been fully charged?
[value] => Array
(
[answer] => 1
[description] => Reason for the Correct Answer:
The energy stored in a capacitor is E=1/2 CV2.
We’re told that the capacitor is subjected to a voltage of 4000V.
Plug in the capacitance of 10-6 F for C, and the voltage of 4000V for V. Then solve for E to get ½(10-6)(4000V)2 = 8J
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 8 J
)
[1] => Array
(
[each_answer] => B. 2 J
)
[2] => Array
(
[each_answer] => C. 0.002 J
)
[3] => Array
(
[each_answer] => D. 10 J
)
)
)
[4] => Array
(
[quiz_unique_key] => 2377279144
[question] => If you were to connect an additional 10-6 F capacitor in series with the original 10-6 F capacitor in the defibrillator, what would be the new equivalent capacitance?
[value] => Array
(
[answer] => 2
[description] => Reason for the Correct Answer:
For series capacitors, you can use 1/Ceq = 1/C1 + 1/C2 + … to find the equivalent capacitance.
Don’t forget to take “one over” the value you get from 1/C1 + 1/C2 in order to find the equivalent capacitance.
1/(10-6F) + 1/(10-6F) = 2/(10-6F), and the inverse gives 0.5 x 10-6 F
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 2.5 x 10-6 F
)
[1] => Array
(
[each_answer] => B. 0.5 x 10-6 F
)
[2] => Array
(
[each_answer] => C. 7 x 10-6 F
)
[3] => Array
(
[each_answer] => D. 0.70 x 10-6 F
)
)
)
)
[total_question] => 5
[correct_answers] => Array
(
[559030|1] => C
[559030|2] => A
[559030|3] => D
[559030|4] => A
[559030|5] => B
)
[hide_display_feedback_settings] =>
[hide_solutions] =>
)
