problem26_93

# problem26_93 - 10 4 3 10 4 1 1 1 4 2000 3 2000 2000 2000-×...

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26.93: a) The circuit can be re-drawn as follows: Then 1 / 2 1 2 eq 1 eq 1 eq + = + = R R V R R R V V ab ab cd and T T R R R R R + = 2 2 eq . But β + = = + = 1 1 2 ) ( 2 eq 1 2 2 1 ab cd T T V V R R R R R R R . b) Recall n n n V V V V V V V V ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( 0 1 2 0 1 2 0 1 + = + = + = + = + = - . If ) 3 1 ( 2 1 1 1 2 1 1 2 1 + = + + = = R R R R R R R R T and 73 . 2 3 1 ) 3 2 ( 2 = + + = . So, for the n th segment to have 1% of the original voltage, we need: 0 4 005 . 0 : 4 01 . 0 ) 73 . 2 1 ( 1 ) 1 ( 1 V V n n n = = + = + . c) 2 1 2 1 1 2 R R R R R T + + = . 10 0 . 4 ) 10 0 . 8 ( ) 10 2 . 3 ( ) 10 0 . 8 10 2 . 3 ( ) 6400 ( 2 . 10 2 . 3 ) 10 0 . 8 ( ) 6400 ( 2 ) 6400 ( 6400 3 8 6 8 6 6 8 2 - × = × × × + × = × = × + + = β R T d) Along a length of 2.0 mm of axon, there are 2000 segments each 1.0 m μ long. The voltage therefore attenuates by:
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Unformatted text preview: . 10 4 . 3 ) 10 . 4 1 ( 1 ) 1 ( 4 2000 3 2000 2000 2000--× = × + = ⇒ + = V V V V e) If Ω × = ⇒ Ω × = 8 12 2 10 1 . 2 10 3 . 3 T R R and . 10 2 . 6 5-× = . 88 . ) 10 2 . 6 1 ( 1 2000 5 2000 = × + = ⇒-V V 1...
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