Use the node-voltage method to find vo and the power delivered by the current source in the circuit in figure, if the v = 30V and the i = 1.9A . Use the node a as the reference node to find vo. Use the node b as the reference node to find vo. Find the power delivered by the 1.9A current source.

Answers

Answer 1

p=61.75W

(a)
use node a as reference and is shown in Figure 1

Apply KCL at super node  between b and c
=1.9+V^b/50+v^c/150+V^C?20+55=0
=v ^b/50+v^c/150+V^C/75=-1.9
=v^ b + v^ c=-1.9(50)
=v^ b +V^ C=-95.....(1)

From super node b and c,
v^ b-v ^c=30.....(2)

by solving equation 1 and 2
V^ b=-32.5V
V^ c=-62.5V

therefore the voltage V^ 0=-32.5V

(B) Apply KCL at node a.
-1.9+v^ a/50+v^ a- v^ c/150+v^a-v^c+20+55+0
V^ a/50+v^a-V^c/150+v^a-v^c/75=1.9
v^ a/50+v^a-v^c/50=1.9
v^ a+ v^ a - v^ c=1.9(50)
2V^ a -v^ c=95....(1)

put V^ c=-30v in equation(1)
2V^ a-(-30)=95
2V^ a=(95-30)
v^ a= 32.5V

From figure 2, the voltage v^ 0=-v^ a
therefore v^ 0=-32.5 V

(c)  power delivered by the 1.9 A current source is
p=iv^ 0
=(1.9)(-32.5)
=-61.75W
The negative sign indicates the power delivering by a source,

P=61.75W  

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The program:

#include<iostream>

using namespace std;

int main()

{

int quarters, dimes, nickels, total_cents;

  cout<<"Enter the number of quarters:\n";

  cin>>quarters;        

  cout<<"Enter the number of dimes:\n";

  cin >>dimes;

   cout <<"Enter the number of nickels:\n";

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   cout<<total_cents;  

 return 0;

}

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}

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Compute the elastic modulus for the following metal alloys, whose stress-strain behaviors may be observed in the "Tensile Tests" module of Virtual Materials Science and Engineering (VMSE): (a) titanium, (b) tempered steel, (c) aluminum, and (d) carbon steel. How do these values compare with those presented in Table 6.1 for the same metals?

Answers

The elastic modulus for steel given in the table is 207 GPa, which is in reasonable agreement with this value E = 200.75 GPa

The elastic modulus is the slope in the linear elastic region:

б2 = б1 / Э2 - Э1

Since stress-strain curves for all of the metals/alloys pass through the o

rigin, and if we take σ1 = 0 then ε1 = 0. Determinations of σ2 and ε2 are possible by moving the cursor to some arbitrary point in the linear region of the curve and then reading corresponding values in the “Stress” and “Strain” windows that are located below the plot.

(a) A screenshot for the titanium alloy in the elastic region is shown below

Here the cursor point resides in the elastic region at a stress of 492.4 MPA (which is the value of σ2) at a strain of 0.0049 (which is the value of ε2). Thus, the elastic modulus is equal to

| 492.4-0

0.0049-0

E=100489.79 MPA

E=100.5 GPA

The elastic modulus for titanium given in the table is 107 GPA, which is in reasonably good agreement with this value.

(b) A screenshot for the tempered steel alloy in the elastic region is shown below

Here the cursor point resides in the elastic region at a stress of 916.7 MPA (which is the value of σ2) at a strain of 0.0045 (which is the value of ε2). Thus, the elastic modulus is equal to :

1916.7-0

0.0045 -0

E = 203711.11 MPA

E = 203.7 GPA

The elastic modulus for steel given in the table is 207 GPA, which is in good agreement with this value.

(c) A screenshot for the aluminum alloy in the elastic region is shown below.

Here the cursor point resides in the elastic region at a stress of 193.6 MPa (which is the value of σ2) at a strain of 0.0028 (which is the value of ε2). Thus, the elastic modulus is equal to:

193.6 – 0

0.0028 - 0

E = 69142.85 MPa

E = 69.14 GPa

The elastic modulus for aluminum given in the table is 69 GPa, which is in excellent agreement with this value.

(d) A screenshot for the carbon steel alloy in the elastic region is shown below

Here the cursor point resides in the elastic region at a stress of 160.6 MPa (which is the value of σ2) at a strain of 0.0008 (which is the value of ε2). Thus, the elastic modulus is equal to:

160.6 – 0

0.0008 - 0

E = 200750 MPa

E = 200.75 GPa

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Estimate the incremental cancer risk for a 60-kg worker exposed to a
particular carcinogen under the following circumstances. Exposure time is 5
days per week, 50 weeks per year, over a 25-year period of time. The worker is
assumed to breathe 20m3 of air per day. The carcinogen has a potency factor
of 0.02 (mg/kg-day)-1 and its average concentration is 0.05mg/m3. Average Life
is 70-year.
1year=365 day

Answers

Answer:

Carcinogen conc. = 0.05 mg/m3

; Lung absorption factor =0.8; Breathing rate = 1m3

/hr

Carcinogen potency factor = 0.02 (mg/kg-day)

-1

Total exposure time = (2 hours/work day)×(5 days/week)×(50 weeks/year)×(20 years)

= 10000 hours

Exposed carcinogen concentration = (0.05 mg/m3

)×[ 1 m3

/h]× (10000 hours) = 500 mg

Absorbed carcinogens in lung = (0.8)×500 mg = 400 mg

Chronic daily intake of carcinogen through lung (CDI)

= (Exposed concentration)/(body weight× averaging time)

Here given body weight = 60 kg; Averaging time = 70 years

So, CDI= (400 mg)/(60 kg×70 years×365days/year) = 2.61 ×10-4 mg/(kg×day)

Lifetime incremental risk of cancer through inhalation of air = CDI×PF

=[2.61 ×10-4 mg/(kg×day)]×[ 0.02 (mg/kg-day)

-1

] =5.22×10-6 (> than the allowable lifetime

incremental risk of cancer, i.e., 10-6, and thus there is a concern.)

Explanation:

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