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2 years ago

Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,

Engineering

1 answer:

iren [92.7K]2 years ago

4 0

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA, IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters IE IC IB VE VB Vc

Unit mA mA μA V V V

Value 4.65 4.6039 46.039 10.7 10 4.6039

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Answer:

C. Have your hazard lights on

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2 years ago

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Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr

mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

$P _1= 15 psia$

$P _1= 100 psia$

We know that work for isothermal process

$W=mRT\ln \dfrac{P_1}{P_2}$

Lets take mass is 1 kg.

So work per unit mass

$W=RT\ln \dfrac{P_1}{P_2}$

We know that for air R=0.287KJ/kg.K

$W=RT\ln \dfrac{P_1}{P_2}$

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W= - 169.28 KJ/kg

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$\Delta S=-0.287\ln \dfrac{100}{15}$

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3 years ago

Some organizations prohibit workers from bringing certain kinds of devices into the workplace, such as cameras, cell phones, and

gulaghasi [49]

Answer:

Personal computers:

Personal computers may be useful and lead to productivity as using a computer an employee familiars with is a good thing. However, the disadvantages in some facilities especially ones dealing with customer and information security can include data theft, unauthorized data sharing, uses of internet connection for personal purposes, as this can slow down internet connection at the facility, distraction at work place etc.

Hard drive:

Due to large amount of data that can be stored on a hard drive, it might not be allowed in some facilities to avoid data theft and unauthorized transfer.

Music players:

This might be restricted to avoid distraction at work. Noice in places such as libraries would cause abnormality and poor service delivery. An employee with loud speaker at work would not only distracts himself but also other staffs and customers.

PSP Game Device and other game devices:

Playing games during working hour may jeopardize the productivity and therefore might be resctrited in some facilities and working places.

Electronic digital notepad:

Carrying a handheld electronic digital notepad to the work place can cause lack of concentration and division of attention on work and other personal activities. These can harm working harmony and business productivity.

Video recorder:

In some facilities, this device might not be allowed due to facility privacy and protection from unwanted navigation.

Explanation:

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2 years ago

Water is the working fluid in an ideal regenerative Rankine cycle with one open feedwater heater. Superheated vapor enters the f

Zepler [3.9K]

Answer:

a) The net powered developed will be 145715 KW.

b) The rate of Heat Transfer to the steam passing through the boiler will be 326340 KW.

c) The thermal efficiency will be 0.45.

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Explanation:

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2 years ago

Estimate the chronic daily intake of toluene from exposure to a city water supply that contains a toluene concentration equal to

vredina [299]

Answer: the total chronic daily intake is 0.03296 mg/ kg.d

Explanation:

First of we have to determine the CDI due to ingestion of drinking water

CDI = (C x IR x EF x ED) / (BW x AT) let leave this as equation 1

where C is the concentration of the life time risk of drinking water(1.0mg/L)

ED is the risk dying per year((70),

EF is the number of days per year(365 day/year)

BW is the weight of body(65.4)kg

AT is the average tenure time of life(75years)

IR = 2.3 L/days

so we substitute

CDI₁ = (1(mg/L) x 2.3(L/day) x 365 x 70) / (65.4 x 365 x 75)

= 3.29 x 10⁻² mg/kg.d

next we determine the CDI due to dermal contact with water.

find the ingestion rate which is equal to the exposure time in hour/day.

IR = ET

(20 min/day) / (60 min/hour)

= 0.333 h/day

now lets substitute for CD1₂

1.0 mg/L for C , 0.333 h/day for m , 365 day/year for EF, 70 year for ED, 65.4 kg for Bw , and 75 year for AT

0.8 for submergence and 1.69m² for area of skin of adult female in our equation 1

CDI₂ = [(1 mg/1) × (1.69 m²) × (9.0 x 10⁻⁶(m/h)) × 0.333(h/day) x 365 x 70)) / (65.4 x 365 x 75)] × ( 0.8 x 10³ L/m³)

CDI₂ = 5.41 x 10⁻⁵ mg/kg.d

now we find the CDI due to inhalation during bath

we substitute 1.0 mg/L for C, 11.3m³/day for IR , 365 day/year for EF , 70 year for ED. 65.4 kg for BW , and 75 year for AT in our equation 1

CDI₃ = [( 1ug/m³ × 1mg/10³ug) x (11.3m³/day x 1day/24 hr) x 365 x 70)] / (65.4 kg x 365 x 75 )

= 6.71 x 10⁻⁶ mg/kg.d

finally we Calculate the total chronic daily intake value

CDI_total = CDI₁ + CDI₂ + CDI₃

we substitute

CDI_total = 3.29 x 10⁻² + 5.41 x 10⁻⁵ + 6.71 x 10⁻⁶

= 0.03296 mg/kg.d

so the total chronic daily intake is 0.03296 mg/ kg.d

4 0

2 years ago