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vladimir1956 [14]
3 years ago
5

You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(

57,000)t + 18.3°) Determine Vout(t). Please express Vout(t) in the same format as Vin(t)
Engineering
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

Output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Explanation:

Given:

dc gain A = 23.6 dB

Input signal V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)

Now convert gain,

A = 10^{\frac{23.6}{20} } = 15.13

DC gain at frequency f = 0 is given by,

  A = \frac{V_{out} }{V_{in} }

V_{out} =AV_{in}

V_{out} = 15.13 \times   0.18 \cos (2\pi (57000)t +18.3)

At zero frequency above equation is written as,

V_{out} = 2.72 \times \cos 18.3

V_{out} = 2.72

Now we write output voltage as input voltage,

V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Therefore, output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

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An irreversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP
Shkiper50 [21]

Answer:

Carnot heat pump

Explanation:

Carnot heat pump is an ideal heat pump in which all processes are reversible and that consume minimum amount of work to and produces maximum amount of heating effect compare to all real engine.And that is why COP of Carnot heat pump is more as compare to real heat pump.

All real heat pump are not perfectly reversible heat pump So this is also called irreversible heat pump .Due to irreversibility the COP of  irreversible heat pump is always  less than the COP of  Carnot heat pump.

6 0
3 years ago
Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,
iren [92.7K]

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

4 0
3 years ago
The undisturbed soil at given borrow pit is found to have the following property:
tankabanditka [31]

Answer:

Check the explanation

Explanation:

Determine the weight o ids in the each truck °slim the relation,  

W,=\frac{W}{1+w}

Here, W is net weight of soil and water on the truck and w is water content  

substitute 72 7 kN for W and 15% for w.

you will need to also determine the number of truck loads required using the relation:

Number of truck loads required = \frac{W_{sc} }{W_{s}}

Kindly check the attached image below for the full explanation to the question above.

7 0
3 years ago
How may acres of land are estimated to be lost each day to development?
Law Incorporation [45]
175 acres per hour of agricultural land lost to development – 3 acres per minute.
3 0
3 years ago
Explain the process of fractional distillation (include cracking)
Grace [21]

Answer:

Explanation:

Fractional distillation is the separation of a mixture into its component parts, or fractions. Chemical compounds are separated by heating them to a temperature at which one or more fractions of the mixture will vaporize. It uses distillation to fractionate

it is the separation of a liquid mixture into fractions differing in boiling point (and hence chemical composition) by means of distillation, typically using a fractionating column.

cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either: heated to 600-700°C.

I hope this was helpful

5 0
2 years ago
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