When attempting to develop better decision making skills, communication with peers should be kept to a minimum. True or false

When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4

daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

$V_{0} = A_{1} V_{1} + A_{2} V_{2} \\$

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6 and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

$CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |$

$CMRR = |\frac{48}{0.4} |\\CMRR = 120$

4 0

3 years ago

What does it mean to say the mass is conserved during a physical change ?

Olin [163]

Mater doesn't just appear or disappeared. Chemical elements are still there just the connections and how it combines changes.

So what goes into your chemical eqation must still exist after the change.

4 0

3 years ago

Read 2 more answers

A forward horizontal force of 3 lb is used to pull a 60 lb sled at constant velocity on a frozen pond. the coefficent of (kineti

puteri [66]

It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m

7 0

3 years ago

A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find

pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

$I=1.41 A$ (current in the wire)

$B=5.61\mu T=5.61\cdot 10^{-6} T$ (strength of magnetic field)

Solving for r, we find the distance from the wire:

$r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m$

4 0

3 years ago

Many minerals, such as silver, are good conductors of heat and electricity. How could technology use this property to help ident

Aleks [24]

Answer:

This property could be used to create technologically-advanced tools or machines that could easily locate the mineral deposits.

Explanation:

Mineral deposits are hard to find, unless you have the skill or the proper tools in locating them. This is the reason why many people are mining in order to explore the different areas where they could find these deposits.

If one would consider the property of minerals, such as being good conductors of heat and electricity, then they could create a tool or machine that would aid in their exploration. Inventors could probably come up with a sensitive detector which signals when it reaches an area of high heat and electric conductivity. Since most minerals such as gold, silver, copper, galena, bornite and the like have this property, then miners will have a lesser amount of time looking for them.

If this technology will be implemented, though, regulation policy must be strictly implemented because it might lead to over-mining thus leading to the depletion of mineral deposits.

3 0

3 years ago