In a myelinated nerve fiber, refractive period is 1/2500 seconds. What is the impulse rate?
Correct Answer: 2500 per sec
Description: A new action potential cannot occur in an excitable fiber as long as the membrane is still depolarized from the preceding action potential. The only condition that will re-open them is for the membrane potential to return either to or almost to the original resting membrane potential level. That period is called refractory period.Refractory period for large myelinated nerve fibers is 1/2500 second; hence such a fiber can carry a maximum of about 2500 impulses a second.Ref: Ganong&;s review of medical physiology, 23rd edition, Page no:87
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