Database esercizi Ingegneria dei sistemi biologici

ESERCIZI TSB 2

2 D

D  λ=

=

z m 4 z

4 λ m

c

=

f

 λ R × Ru

1

=R

R × R × R = =300

R Ω



1 4 3 2 2 R 3

( )

( )

=R −T

R 1+ a T 0

t 0 R1’=

( )

( )

−T =158.4

R 1+ a T 0 Ω

1

R1’ = R1 dopo l’aumento di

temperatura Δ R

A 1

⋅

=V =0.112

V ¿

0 2 R

( )

1+ A 1

R R

2 4

=

A=

Con R R 3

1 √

− 3

( )=1 ( )=

ϑ ϑ

=kH

V cos cos

I 2

( ) ( )−kH ( ) ( )=0

ϑ ϑ

=KH

V cos 60 cos sin 60 sin

II √

−1 3

( )

+V =

aVR= V I II

2 4

PER RICAVARE VII USO SEMPRE L’ANGOLO 60

PER RICAVARE VIII USO SEMPRE L’ANGOLO 120 −t −t /

T T

( ) ( ) ( )

ⅇ ⅇ

=M =M

M t 0 sen a

2 2

xy xy 0

c G = zm/F zm



−4

=4.4

λ= ×10

f z m =gF=200

z

= g*F g=  m

F

2

D 

=

z m 4 λ

√

D= z 4 λ=18,7616

m

λF

= =0.125

W D 

=R

R × R × R

1 4 2 3

R × R

1 4

= =400

R Ω

2 R 3

ⅆ ⅆ

R l 

=

G= R l

ⅆ ⅆ

R l

=G =0.1 Ω

R l ⅆ

A R

⋅

=V =0.32V

V ¿ R

2

( )

1+ A

R R

2 4

= =4

A= R R

1 3

1 zp c



=2 =110

zp= mm

F c 2 f

R c c



zp ≥ f≤

2 f 2 zp

C

=

f

- con zp in metri

MAx 2 zp 2

c D



λ= =

z m

f 4 λ

2 d

t= d = t*c/2

 

c ( )

π 

( )=M =1

=1 M

Mxy 0 sen 0

0 2

1 1

−t ∕ T 2

( ) ( ) ⅇ

=Mxy =1⋅ = =0.169

Mxy t 0 t ∕ T 80 ∕ 45

ⅇ ⅇ

2

−t 1

( )

−t ∕ T T

( )=M ( ) ⅇ =0,28346

+ =1−

M t 1−ⅇ Mz 0

1 1

z 0 80 ∕ 240

ⅇ

( ) ( )

Mxy 0.109 0

=atan =atan =30

α Mz 0.710 ϑ

=kH

V cos

I √

1 3

ϑ− ϑ

=

V kH cos kH sin

II 2 2

v √

1 3 

II = − tanϑ

v 2 2

I ( ) ( )

v 2 v

−2 1 1

II II

ϑ= − =

tan 1−

√ √

v 2 v

3 3

I I

( )

( )

2 v

1 II

ϑ=atan 1−

√ v

3 I