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WITCHER [35]
3 years ago
7

In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit

ion 1, position 2, and position 3.
A) The gain 1formula58.mml at the position 1 is _____.
B) The gain 1formula58.mml at the position 2 is _____.
C) The gain 1formula58.mml at the position 3 is _____.
Engineering
1 answer:
Elenna [48]3 years ago
3 0

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

a)

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

so

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

║Vo/Vi ║ = 3.4    ( negative sign phase inversion)

b)

R2 = 74kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

so

(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0

-Vi/5kΩ + -Vo/74kΩ = 0

Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8  ( negative sign phase inversion)

c)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0           ( 5MΩ = 5000kΩ )

so

-Vi/5kΩ + -Vo/5000kΩ = 0

Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

║Vo/Vi ║ = 1000  ( negative sign phase inversion)

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Greeley [361]

Answer:

The bending stress increases linearly away from the neutral axis until the maximum values at the extreme fibers at the top and bottom of the beam. The maximum bending stress is given by: where c is the centroidal distance of the cross section (the distance from the centroid to the extreme fiber).

Explanation:

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8 0
3 years ago
The gage pressure measured as 2.2 atm, the absolute pressure of gas is 3.2 bar. Please determine the local atmospheric pressure
LiRa [457]

Answer:

97.085\ \text{kPa}

Explanation:

P_{g} = Gauge pressure = 2.2 atm = 2.2\times 101325=222915\ \text{Pa}

P_{abs} = Absolute pressure = 3.2\ \text{bar}=3.2\times 10^5\ \text{Pa}

P_{atm} = Local atmospheric pressure

Absolute pressure is given by

P_{abs}=P_{atm}+P_g\\\Rightarrow P_{atm}=P_{abs}-P_g\\\Rightarrow P_{atm}=3.2\times 10^5-222915\\\Rightarrow P_{atm}=97085\ \text{Pa}=97.085\ \text{kPa}

The local atmospheric pressure is 97.085\ \text{kPa}.

8 0
3 years ago
A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa and 400°C. It is cooled to 40°
leonid [27]

Answer:

heat transfer for the process is - 643.3 kJ

Explanation:

given data

mass m = 2 kg

pressure p1 = 500 kPa

temperature t1 = 400°C = 673.15 K

temperature t2 = 40°C = 313.15 K

pressure p2 = 300 kPa

to find out

heat transfer for the process

solution

we know here mass is constant so

m1 = m2

so by energy equation

m ( u2 - u1 ) = Q - W

Q is heat transfer

and in process  P =  A+ N that is linear spring

so

W = ∫PdV

= 0.5 ( P1+P2) ( V1 - V2)

so for case 1

P1V1 = mRT

put here value

500 V1 = 2 (0.18892) (673.15)

V1 = 0.5087 m³

and

for case 2

P2V2 = nRT

300 V2 = 2 (0.18892) (313.15)

V2 = 0.3944 m³

and

here W will be

W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )

W = -45.72 kJ

and

Q is here for Cv = 0.83 from ideal gas table

Q = mCv ( T2-T1 ) + W

Q = 2 × 0.83 ( 40 - 400 ) - 45.72

Q = - 643.3 kJ

heat transfer for the process is - 643.3 kJ

7 0
3 years ago
Air at 27°C and a velocity of 5 m/s passes over the small region As (20 mm × 20 mm) on a large surface, which is maintained at T
JulsSmile [24]

Answer:

a) The maximum possible heat removal rate = 2.20w

b) Fin length = 37.4 mm

c) Fin effectiveness = 89.6

d) Percentage increase = 435%

Explanation:

See the attached file for the explanation.

5 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is
Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

α  =  \frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0
3 years ago
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