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aleksklad [387]

2 years ago

15

Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:

1 answer:

natta225 [31]2 years ago

7 0

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

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Describe what V1-V4 is

photoshop1234 [79]

Answer:

it is wave inversion

Explanation:

8 0

3 years ago

For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta

Crank

Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

$c=\sqrt(\gamma*R*T)$

Where R is the gas's particular constant defined in terms of Cp and Cv:

$R=Cp-Cv$

For particular values given:

$R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}$

$R=188.9 \frac{J}{Kg-K}$

The gamma undimensional constant is also expressed as a function of Cv and Cp:

$\gamma=Cp/Cv$

$\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}$

$\gamma=1.29$

And the variable T is the temperature in Kelvin. Thus for the known temperature:

$c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)$

$c=\sqrt(91867.73 \frac{J}{Kg})$

The Jules unit can expressing by:

$J=N.m=\frac{Kg.m}{s^2}* m$

$J=\frac{Kg.m^2}{s^2}$

Replacing the new units for the speed of the sound:

$c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})$

$c=\sqrt(91867.73 \frac{m^2}{s^2})$

$c=313 m/s$

3 0

3 years ago

Read 2 more answers

How do Consumers sometimes interact with a producers?

Vika [28.1K]

I really don’t know good luck

7 0

3 years ago

A __________ is an added note showing additional or more specific information.

xxMikexx [17]

Answer:

awnsers should be added to know to show additional

8 0

3 years ago

The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.

ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

So

we can say that acceleration= 0.6 x 9.81

$acceleration,a=5.88\ \frac{m}{s^2}$

We know that

$a=\omega ^2A$

So now by putting the values

$a=\omega ^2A$

$5.88=\omega ^2 \0.15\times 10^{-3}$

$\omega =198.09\ \frac{rad}{s}$

We know that

ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

4 0

2 years ago