1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
VashaNatasha [74]
3 years ago
10

What are the desired characteristics or values for the following parameters of an ideal amplifier?o Phase change as a function o

f the log of the frequency o Common Mode Rejection Ratioo Input Resistanceo Output Resistance
Engineering
1 answer:
kvv77 [185]3 years ago
4 0

Answer and Explanation:

- Phase change as a function of the log of the frequency

The phase shift should be independent of the frequency of the input signal. It should have an Infinite bandwidth with zero phase shift

- Common Mode Rejection Ratio

An ideal amplifier has Infinite common-mode rejection ratio (CMRR). A perfect operational amplifier amplifies only the voltage difference between its two inputs, completely rejecting all voltages that are common to both.

- Input Resistance

Infinite input impedance and so, zero input current & Zero input offset voltage.

The differential input impedance of the operational amplifier is defined as the impedance between its two inputs; the common-mode input impedance is the impedance from each input to ground.

And the impedance between its inputs should be infinite.

- Output Resistance

Zero output impedance and Infinite output voltage range.

Output Resistance leads to voltage drop. Thee voltage drop across the output impedance effectively reduces the open loop gain. So, the optimal value for this should be zero.

N.B - For AC circuits such as this one being discussed, resistance is known as impedance.

You might be interested in
Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb
Nezavi [6.7K]

Answer:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- \xi_1 and \xi_2: extent of the reactions (2).

- F_{O_2}^2, F_{CH_4}^2, F_{H_2O}^2, F_{HCHO}^2 and F_{CO_2}^2: Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- F_{O_2}^2=50mol/s-\xi_1-2\xi_2: oxygen mole balance.

- F_{CH_4}^2=50mol/s-\xi_1-\xi_2: methane mole balance.

- F_{H_2O}^2=\xi_1+2\xi_2: water mole balance.

- F_{HCHO}^2=\xi_1: formaldehyde mole balance.

- F_{CO_2}^2=\xi_2: carbon dioxide mole balance.

Thus, the degrees of freedom are:

DF=7unknowns-5equations=2

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

0.900=\frac{\xi_1+\xi_2}{50mol/s}

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

0.860=\frac{F_{HCHO}^2}{50mol/s}

In such a way, one realizes that the output formaldehyde's molar flow is:

F_{HCHO}^2=0.860*50mol/s=43mol/s

Which is equal to the first reaction extent \xi_1, therefore, one computes the second one from the fractional conversion of methane as:

\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s

Now, one computes the rest of the output flows via:

- F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s

- F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s

- F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s

- F_{HCHO}^2=43mol/s

- F_{CO_2}^2=2mol/s

The total output molar flow is:

F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s

Therefore the output stream composition turns out into:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Best regards.

7 0
2 years ago
A(n) _____________ is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when
Alenkasestr [34]

A <u>centrifugal switch</u> is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when the motor reaches approximately 75% of its rated speed.

Hope that helps!

7 0
2 years ago
The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C
s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325  P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.15 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K  

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0
3 years ago
Write a statement to print the data members of InventoryTag. End with newline. Ex: if itemID is 314 and quantityRemaining is 500
Advocard [28]

Answer:

#include <stdio.h>

typedef struct InventoryTag_struct {

int itemID;

int quantityRemaining;

} InventoryTag;

int main(void) {

InventoryTag redSweater;

redSweater.itemID = 314;

redSweater.quantityRemaining = 500;

/* Your solution goes here */

printf("Inventory ID: %d, Qty: %d\n",redSweater.itemID,redSweater.quantityRemaining);

getchar();

return 0;

}

Explanation:

7 0
3 years ago
The interior wall of a building is made from 2×4 wood studs, plastered on one side. If the wall is 13 ft high, determine the loa
Elanso [62]

Answer:

load  = 156 lb/ft

Explanation:

given data

interior wall of a building = 2×4 wood studs

plastered = 1 side

wall height =  13 ft

solution

we get here load so first we get wood stud load  and that is  

we know here from ASCE-7 norm

dead load of 2 x 4 wood studs with 1 side plaster  = 12 psf

and we have given height 13 ft

so load will be =  12 psf × 13 ft

load  = 156 lb/ft

7 0
3 years ago
Other questions:
  • Use the convolutional integral to find the response of an LTI system with impulse response ℎ(????) and input x(????). Sketch the
    8·1 answer
  • Q5. A hypothetical metal alloy has a grain diameter of 2.4 x 10-2 mm. After a heat treatment at 575°C for 500 min, the grain dia
    7·1 answer
  • A PMOS device with VT P = −1.2 V has a drain current iD = 0.5 mA when vSG = 3 V and vSD = 5 V. Calculate the drain current when:
    12·1 answer
  • A thermistor is a temperature‐sensing element composed of a semiconductor material, which exhibits a large change in resistance
    13·1 answer
  • Should you ever grab a tool with expose wiring
    13·2 answers
  • typedef struct bitNode { int data; struct bitNode *left; struct bstNode *right; } bstNode; int solve(bstNode* root) { if (root =
    15·1 answer
  • Which one of the following is not an economic want?
    6·1 answer
  • . H<br> Kijwhayhwbbwyhwbwbwgwwgbwbwhwh
    6·2 answers
  • Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.​
    7·1 answer
  • What are the main causes of injuries when using forklifts?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!