Solutions for transients in arbitrarily branching cables: III. Voltage clamp problems
نویسندگان
چکیده
منابع مشابه
Solutions for transients in arbitrarily branching cables: II. Voltage clamp theory.
Analytical solutions are derived for arbitrarily branching passive neurone models with a soma and somatic shunt, for synaptic inputs and somatic voltage commands, for both perfect and imperfect somatic voltage clamp. The solutions are infinite exponential series. Perfect clamp decouples different dendritic trees at the soma: each exponential component exists only in one tree; its time constant ...
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Transient responses of a space-clamped squid axon membrane to step changes of voltage or current are often approximated by exponential functions of time, corresponding to a series resistance and a membrane capacity of 1.0 muF/cm(2). Curtis and Cole (1938, J. Gen. Physiol.21:757) found, however, that the membrane had a constant phase angle impedance z = z(1)(jomegatau)(-alpha), with a mean alpha...
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6 = 2(cos( (π+2kπ) 6 ) + i sin( (π+2kπ) 6 )) where k = 0, 1, 2, · · · , 5. Let rk = 2(cos( (π+2kπ) 6 ) + i sin( (π+2kπ) 6 )) where k = 0, 1, 2, · · · , 5. Therefore r0 = 2(cos( π 6 )+i sin( 6 )) = √ 3+i, r1 = 2(cos( π 2 )+i sin( 2 )) = 2i, r2 = 2(cos( 5π 6 )+ i sin( 6 )) = − √ 3 + i, r3 = 2(cos( 7π 6 ) + i sin( 6 )) = − √ 3 − i, r4 = 2(cos( 2 ) + i sin( 2 )) = −2i and r5 = 2(cos( 6 ) + i sin( 1...
متن کاملSolutions to Review Problems for Midterm III
6 = 2(cos( (π+2kπ) 6 ) + i sin( (π+2kπ) 6 )) where k = 0, 1, 2, · · · , 5. Let rk = 2(cos( (π+2kπ) 6 ) + i sin( (π+2kπ) 6 )) where k = 0, 1, 2, · · · , 5. Therefore r0 = 2(cos( π 6 )+i sin( 6 )) = √ 3+i, r1 = 2(cos( π 2 )+i sin( 2 )) = 2i, r2 = 2(cos( 5π 6 )+ i sin( 6 )) = − √ 3 + i, r3 = 2(cos( 7π 6 ) + i sin( 6 )) = − √ 3 − i, r4 = 2(cos( 2 ) + i sin( 2 )) = −2i and r5 = 2(cos( 6 ) + i sin( 1...
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ژورنال
عنوان ژورنال: Biophysical Journal
سال: 1993
ISSN: 0006-3495
DOI: 10.1016/s0006-3495(93)81039-7