Electromyography and passive filters
- Dashboard
- Electromyography and passive filters
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 | |
|---|---|---|---|---|---|
| -- | -- | -- | -- | -- |
Electromyography and passive filters
Array
(
[passage] => WP_Post Object
(
[ID] => 558925
[post_author] => 12815
[post_date] => 2025-01-09 08:02:06
[post_date_gmt] => 2025-01-09 13:02:06
[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.
Electromyography, or EMG, is a means of measuring electric potentials produced by a person’s muscles in order to determine if there are any abnormalities within the muscle, such as neuromuscular disease. Various electric properties can be measured via electrodes placed in a muscle, including the amplitude and the duration of an action potential. Additionally, an EMG will not record any electrical activity when a muscle is at rest, so the presence of electrical activity can be a sign of pathology as well.
To perform the test, two electrodes can be inserted into muscle tissue which measure the difference in voltage between the two points. The signals generated during muscle contraction are then passed through an amplifier to boost the signal, a filter to remove noise from the amplified signal, and then an analog-to-digital converter.
Though the filtering process is now digital itself, the original design of an EMG involved an analog system of fundamental circuit components, the simplest being a resistor and capacitor, known as a passive low pass filter (Figure 1). As long as the current passes through the circuit below the cut-off frequency, it behaves essentially like a normal circuit. However, frequencies above the cut-off frequency of the filter will cross the capacitor and go to the ground. The cut-off frequency is determined by both the resistance and capacitance of the circuit; the higher the capacitance or resistance, the lower the cut-off frequency will be.
Figure 1. Passive, first order low-pass RC filter
[post_title] => Electromyography and passive filters
[post_excerpt] =>
[post_status] => publish
[comment_status] => closed
[ping_status] => closed
[post_password] =>
[post_name] => electromyography-and-passive-filters
[to_ping] =>
[pinged] =>
[post_modified] => 2025-01-09 08:02:06
[post_modified_gmt] => 2025-01-09 13:02:06
[post_content_filtered] =>
[post_parent] => 0
[guid] => https://medlifemastery.com/?post_type=passage&p=558925
[menu_order] => 0
[post_type] => passage
[post_mime_type] =>
[comment_count] => 0
[filter] => raw
)
[questions] => Array
(
[0] => Array
(
[quiz_unique_key] => 578908434
[question] => While DC current does not work with an RC filter, it will charge up the capacitor until it is fully charged, and afterward the current will go through Vout instead (no current will go to the ground). If the capacitance is 3 Farads, and there is a charge of 150 Coulombs at the precise moment the DC current passes through Vout; what is the voltage?
[value] => Array
(
[answer] => 2
[description] => Reason for the Correct Answer:
While the capacitor will essentially act as a break in the circuit upon being fully charged, but at the precise moment before the current flows through Vout the capacitor will satisfy normal capacitance equations.
Use the equation C= Q/V to solve for V.
With a charge (Q) of 150 Coulombs, and a capacitance (C) of 3 Farads, we can check our math to see that voltage still must be 50 Volts. (CxV=Q)
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 30 Volts
)
[1] => Array
(
[each_answer] => B. 50 Volts
)
[2] => Array
(
[each_answer] => C. 100 Volts
)
[3] => Array
(
[each_answer] => D. 150 Volts
)
)
)
[1] => Array
(
[quiz_unique_key] => 3873426850
[question] => If a dielectric is placed between the plates of the capacitor, how will the cut-off frequency of the filter be affected?
[value] => Array
(
[answer] => 4
[description] => Reason for the Correct Answer:
C = εA/d
The dielectric is directly proportional to capacitance. Therefore, placing a dielectric between the plates will increase the capacitance.
The cut-off frequency is inversely related to the resistance and capacitance of the circuit. So, a dielectric will decrease the cut-off frequency of the filter.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. The cut-off frequency will increase
)
[1] => Array
(
[each_answer] => B. Not enough information is provided
)
[2] => Array
(
[each_answer] => C. The cut-off frequency will not change
)
[3] => Array
(
[each_answer] => D. The cut-off frequency will decrease
)
)
)
[2] => Array
(
[quiz_unique_key] => 83407773
[question] => Guillain–Barré syndrome is an autoimmune condition that results in the demyelination of peripheral nerves. If only the myelin of a nerve is affected, how might the loss of myelin affect the voltage seen in an EMG?
[value] => Array
(
[answer] => 1
[description] => Reason for the Correct Answer:
Myelin acts to insulate the nerve fiber innervating muscle.
A loss of insulation would result in a decrease of transmembrane resistance.
Resistance is directly related to voltage by Ohms law (V=IR) therefore a decrease in resistance implies a decrease in voltage as well.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. The voltage of the action potential would decrease
)
[1] => Array
(
[each_answer] => B. There would be no electric signal at all
)
[2] => Array
(
[each_answer] => C. The voltage of the action potential would increase
)
[3] => Array
(
[each_answer] => D. The voltage would stay the same
)
)
)
[3] => Array
(
[quiz_unique_key] => 2261298308
[question] => A charge of 12 x10-6 C exists on the capacitor with 6×10-6 capacitance. If the plates are 1mm apart, what is the strength of the electric field?
[value] => Array
(
[answer] => 1
[description] => Reason for the Correct Answer:
The strength of an electric field is given by the formula V=Ed, where E is the electric field strength, in V/m
V= Q/C So, 12 x10-6 / 6×10-6 = 2 V
V = Ed So, 2 = V x 0.001m or E=2000 V/m
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 2000 V/m
)
[1] => Array
(
[each_answer] => B. 200 V/m
)
[2] => Array
(
[each_answer] => C. 2 V/m
)
[3] => Array
(
[each_answer] => D. 20 V/m
)
)
)
[4] => Array
(
[quiz_unique_key] => 2377279144
[question] => Nerve conduction tests are similar to EMGs, but measure the speed of signal transduction rather than the voltage change. Which of the following nerves should have the fastest signal transduction?
[value] => Array
(
[answer] => 1
[description] => Reason for the Correct Answer:
Conductivity is inversely related to resistivity.
The formula for resistance in terms of resistivity is R = ρL/A. Where p is the resistivity of the material, L is the length, and A is the cross sectional area.
Since the length of all the nerves is the same, and the nerves should have the same amount of resistivity (they are the same material) then the deciding factor should be the diameter of the nerve.
The nerve with the largest diameter (A-alpha fibers with a 15 µm diameter) will have the lowest resistance, and therefore the highest conductance, leading to the fastest signal transduction.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. A 5cm A-alpha fiber nerve for muscle innervation with a diameter of 15 µm
)
[1] => Array
(
[each_answer] => B. A 5cm B fiber nerve for autonomic response with a diameter of 3 µm
)
[2] => Array
(
[each_answer] => C. A 5cm A-delta fiber nerve for temperature sensation with a diameter of 5 µm
)
[3] => Array
(
[each_answer] => D. A 5cm C fiber nerve for pain sensation with a diameter of 0.5 µm
)
)
)
)
[total_question] => 5
[correct_answers] => Array
(
[558925|1] => B
[558925|2] => D
[558925|3] => A
[558925|4] => A
[558925|5] => A
)
[hide_display_feedback_settings] =>
[hide_solutions] =>
)
