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Ilia_Sergeevich [38]
2 years ago
9

If you measure the voltage across a diode and it measures 0.3 V, the diode is probably made of __________ and is __________ bias

ed.
Engineering
1 answer:
fgiga [73]2 years ago
3 0

Answer:

  • germanium
  • forward

Explanation:

The forward-biased voltage drop of a semiconductor diode depends on the material it is made of and the method of fabricating the junction.

<h3>Voltage drops</h3>

A silicon diode will have a voltage drop that can be estimated as 0.7 V. The actual value will depend on current and temperature.

A germanium diode will have a voltage drop closer to 0.3 V, again depending on current and temperature.

A Schottky diode may have a voltage drop between 0.2 and 0.3 V, so may be difficult to distinguish from a germanium diode.

If it measures 0.3 V, the diode is probably made of <u>germanium</u>, and is <u>forward</u> biased.

__

The reverse-biased voltage reading will generally match the open-circuit voltage of the testing device.

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Tech A says that in some cases, the electronic brake control module can be programmed with a new tire size to restore proper ele
Vlad [161]

Answer:

Both Techs A and B

Explanation:

Electronic braking systems are controlled by the electronic brake control module. It is a microprocessor that processes information from wheel-speed sensors and the hydraulic brake system to determine when to release braking pressure at a wheel that's about to lock up and start skidding  and activates the anti lock braking system or traction system when it detects it is necessary.

Some electronic brake control modules can be programmed to the size of the vehicle's new tires to restore proper electronic brake control performance. While others may require replacing the module to match the module's programming to the installed tire size. So, both technicians A and B are correct.

3 0
3 years ago
Mohr's circle represents: A Orientation dependence of normal and shear stresses at a point in mechanical members B The stress di
blsea [12.9K]

Answer:

The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members

Explanation:

Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

\sigma _{x'x'}=\frac{\sigma _{xx}+\sigma _{yy}}{2}+\frac{\sigma _{xx}-\sigma _{yy}}{2}cos(2\theta )+\tau _{xy}sin(2\theta )

And \tau _{x'x'}=-\frac{\sigma _{xx}-\sigma _{yy}}{2}sin(2\theta )+\tau _{xy}cos(2\theta )

Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.

The Mohr circle is graphically displayed in the attached figure.

4 0
3 years ago
The entire system of components that produces power and transmits it to the road is called the vehicle's _____.
IrinaK [193]

Answer:

Powertrain

Explanation:

6 0
3 years ago
The minimum safe working distance from exposed electrical conductors
SVEN [57.7K]

Answer:

b

Explanation:

4 0
3 years ago
Find E[x] when x is sum of two fair dice?
Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

​

P(X=3)=P(1,2)+P(2,1)=

36

2

​

=

18

1

​

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

​

=

12

1

​

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

​

=

9

1

​

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

​

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

​

=

6

1

​

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

​

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

​

=

9

1

​

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

​

=

12

1

​

P(X=11)=P(5,6)+P(6,5)=

36

2

​

=

18

1

​

P(X=12)=P(6,6)=

36

1

​

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

​

⋅P(X

i

​

)

=2×

36

1

​

+3×

18

1

​

+4×

12

1

​

+5×

9

1

​

+6×

36

5

​

+7×

6

1

​

+8×

36

5

​

+9×

9

1

​

+10×

12

1

​

+11×

18

1

​

+12×

36

1

​

=

18

1

​

+

6

1

​

+

3

1

​

+

9

5

​

+

6

5

​

+

6

7

​

+

9

10

​

+1+

6

5

​

+

18

11

​

+

3

1

​

=7

E(X

2

)=∑X

i

2

​

⋅P(X

i

​

)

=4×

36

1

​

+9×

18

1

​

+16×

12

1

​

+25×

9

1

​

+36×

36

5

​

+49×

6

1

​

+64×

36

5

​

+81×

9

1

​

+100×

12

1

​

+121×

18

1

​

+144×

36

1

​

=

9

1

​

+

2

1

​

+

3

4

​

+

9

25

​

+5+

6

49

​

+

9

80

​

+9+

3

25

​

+

18

121

​

+4

=

18

987

​

=

6

329

​

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

​

=

5.833

​

=2.415

4 0
3 years ago
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