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FinnZ [79.3K]
3 years ago
11

For the given nonlinear circuit, ????x = [???? ????x−1]/ 10 in which the units of ????x and ????x are V and A respectively. The

source voltage is 3V and the source resistance is 1Ω.
Use graphical load-line techniques to solve for ????x and ????x. [Use a computer program to plot the characteristic curve and load line.]
Engineering
1 answer:
notsponge [240]3 years ago
7 0

V_{x} = 0.75 V, I_{x}= 0.56 A

<u>Explanation:</u>

Given data:

V_{s} = 3 V

R_{s} = 1 Ω

I_{x} = \frac{3V_{X} }{4}

Solution:

Using Kirchhoff's voltage law.

The law states that the algebraic addition of potential difference in loop is similar to zero.

V_{s} - I_{x}R_{s} - V_{x} = 0

V_{s} - I_{x}R_{s} = V_{x}

3 - \frac{3V_{X} }{4} × 1 = V_{x}

12 - 3V_{x}  = 4

3 - 3V_{x}  =

V_{x} = 3/4

V_{x} = 0.75 V

I_{x} = \frac{3V_{X} }{4}

I_{x} = \frac{3}{4} × 0.75

I_{x} = \frac{2.25}{4}

I_{x} = 0.56 A.

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4 years ago
Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload
Vinil7 [7]

Answer:

1) V_o = 10 liters

2) V_o = 12.26 liters

Explanation:

For isothermal process n =1

V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}

V_o = 10 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.03

actual \ volume = c1\times 10 = 10.3 liters

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}

V_o = 12.26 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

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3 0
3 years ago
Read 2 more answers
A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50
anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure  = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

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normal stress = ( 50×119.5 ) / 2 × 0.5

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and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

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