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2 years ago

What is meant by infinite slew rate

1 answer:

3 0

Am infinite slew rate means that the changes in the output voltage occur immensely when the input voltage changes.
Slew rate is measurement of the response of an operational amplifier. For an ideal operational amplifier, time delay is negligible. Hence it has an infinite slew rate.
In simpler terms it means that it can provide output voltage simultaneously with the input voltage changes.

Hope this helps :)

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HELP!!!

Vitek1552 [10]

We shall consider two properties:
1. Temperature difference
2. Thermal conductivity of the material

Use a cylindrical rod of a given material (say steel) which is insulated around its circumference.

One end of the rod is dipped in a large reservoir of water at 100 deg.C and the other end is dipped in water (with known volume) at 40 deg. C. The cold water if stored in a cylinder which is insulated on all sides. A thermometer reads the temperature of the cold water as a function of time.

This experiment will show that
(a) heat flows from a region of high temperature to a region of lower temperature.
(b) The thermal energy of a body increases when heat is added to it, and its temperature will rise.
(c) The thermal conductivity of water determines how quickly its temperature will rise. If mercury replaces water in the cold cylinder, its temperature will rise at a different rate because its thermal conductivity is different.

5 0

3 years ago

Read 2 more answers

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

Countries with traditional economies are often less developed because:

3241004551 [841]

People are resistant to change, People have political freedom, Farmers institute new farming techniques and there is little need for medical facilities.

7 0

3 years ago

Read 2 more answers

If a moving car speeds up until it is going twice as fast, how much kinetic energy doe s it have compared with its initial kinet

dezoksy [38]

The kinetic energy of an object of mass m and velocity v is given by
$K= \frac{1}{2} mv^2$

Let's call $v_i$ the initial speed of the car, so that its initial kinetic energy is
$K_i = \frac{1}{2} mv_i^2$
where m is the mass of the car.

The problem says that the car speeds up until its velocity is twice the original one, so
$v_f = 2 v_i$
and by using the new velocity we can calculate the final kinetic energy of the car
$K_f = \frac{1}{2} mv_f^2 = \frac{1}{2}m (2 v_i)^2 = 4 ( \frac{1}{2} mv_i^2)=4 K_i$
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.

6 0

3 years ago

1) draw a simple circuit with a voltage source and four resistors wired in series

Norma-Jean [14]

Answer:

In this circuit (see attachment #1), we have:

- A voltage source: in this case, we choose a battery. A voltage source is a device producing an electromotive force (in a battery, this is done by means of a chemical reaction), which is responsible for "pushing" the electrons along the circuit and creating a current. The electromotive force (emf) of the battery is also called voltage, and it is indicated with the letter V.

- Four resistors: a resistor is a device which opposes to the flow of current. The property that describes by "how much" the resistor "opposes" to the flow of current is called "resistance", and it is indicated with the letter R.

- In this circuit, the 4 resistors are in series. Resistors are said to be in series when they are connected along the same branch of the circuit, so that the same current flow across each of them.

- For resistors in series, the equivalent resistance of the circuit is given by the sum of the individual resistances:

$R=R_1+R_2+...+R_n$

In this circuit (see attachment #2), we have:

- A voltage source: as before, we have chosen a battery, providing an electromotive force to the circuit

- Three resistors wired in parallel. Resistors are said to be connected in parallel when they are connected along different branches, but with their terminals connected to the same point, so that each of them has the same potential difference across it.

- For resistors in parallel, the equivalent resistance of the circuit is calculated using the formula:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

In this circuit (see attachment #3), we have:

- A voltage source (again, we have choosen a battery)

- Three resistors, of which:

-- 2 of them are connected in parallel with each other

-- the 3rd one it is in series with the first two

If we call $R_1,R_2$ the resistances of the first 2 resistors in parallel, their equivalent resistance is:

$\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}\\\rightarrow R_{12}=\frac{R_1 R_2}{R_1+R_2}$

Then, these two resistors are connected in series with resistor $R_3$ ; and so, the total resistance of this circuit will be:

$R=R_{12}+R_3=\frac{R_1R_2}{R_1+R_2}+R_3=\frac{R_1R_2+R_3(R_1+R_2)}{R_1+R_2}$

In this circuit (see attachment #4), we have:

- A voltage source (again, a battery)

- We have 6 resistors, which are arranged as follows:

-- Two branches each containing 3 resistors

-- The two branches are in parallel with each other

So, the total resistance of the two branches are:

$R_{123}=R_1+R_2+R_3$

$R_{456}=R_4+R_5+R_6$

And since the two branches are in parallel, their total resistance will be:

$\frac{1}{R}=\frac{1}{R_{123}}+\frac{1}{R_{456}}\\\rightarrow R=\frac{R_{123}R_{456}}{R_{123}+R_{456}}=\frac{(R_1+R_2+R_3)(R_4+R_5+R_6)}{R_1+R_2+R_3+R_4+R_5+R_6}$

4 0

3 years ago