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3 years ago

What is the purpose of a rectifier assembly

Engineering

1 answer:

Elden [556K]3 years ago

6 0

Answer:

Explanation:

I assume you are talking about electronics. A rectifier changes Alternating current into direct current.

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Technician A says that rear-wheel drive vehicles usually get better traction than front-wheel drive vehicles. Technician B says

cluponka [151]

C is the correct answer

3 0

3 years ago

A multiple choice question on an economic quiz contains 10 questions with five possible answers each. Compute the probability of

Zepler [3.9K]

Answer:

Explanation:

I can tell you rn, without doing the math, that ur chances are very slim. That almost never works. I took a math quiz like that once, it had 12 questions and I made a 5%...that 5/100.

A piece of advice...study

4 0

3 years ago

Consider a step pn junction made of GaAs at T = 300 K. At zero bias, only 20% of the total depletion region width is in the p-si

Nat2105 [25]

Answer:

0.31 μm

Explanation:

this question wants us to Determine the depletion region width, xn, in the n-side in unit of μm. using the information below.

density in the p-side = 5.68x10^16

density in the n-side = 1.42x10^16

$\sqrt{\frac{2*12.7*8.85E-10}{1.6E-14}(\frac{1}{5.68E16}+\frac{1}{1.42E16} )(1.2) }$

= √(1.42x10⁵)(1.76056335x10⁻¹⁷ + 7.042253521x10⁻¹⁷)(1.2)

= √150.74x10⁻¹¹

= 3.882x10⁻⁵

approximately 0.39μm

xn = 0.39 x 0.8

= 0.31μm

0.31 um is the depletion region width. thank you!

3 0

3 years ago

Two semiconductor materials have exactly the same properties except material A has a bandgap energy of 0.90 eV and material B ha

AlekseyPX

Answer: hello your question is incomplete attached below is the complete question

answer : Ac = 5° , A $_{f}$ = 2.5°

Explanation:

Bandgap energy for material A = 0.90 eV

Bandgap energy for material B = 1.10 eV

<u>Calculate the ratio of ni for Material B to Material B </u>

Total derivation ( d ) = d1 + d2

d = A $_{c}$ ( μ $_{c}$ - 1 ) + A $_{f}$ ( μ $_{f}$ - 1 ) ---- ( 1 )

where : d = 1° , μ $_{c}$ = 1.5 , μ $_{f}$ = 1.6

Input values into equation 1 above

1° = 0.5Ac + 0.6Af ---- ( 2 )

also d = d1 [ 1 - w/ w1 ] ------ ( 3 )

∴ d = Ac ( μ $_{c}$ - 1 ) ( 1 - w/w1 )

1° = Ac ( 1.5 - 1 ) ( 1 - 0.06/0.1 ) --- ( 4 )

resolving equation ( 4 )

Ac = 5°

resolving equation ( 2 )

A $_{f}$ = 2.5°

8 0

3 years ago

A Carnot heat engine absorbs 235 KW of heat from a heat source and rejects 164 KW to the atmosphere. Determine the thermal effic

nevsk [136]

Answer:

43.2%

Explanation:

Given that,

Heat absorbed by a carnot heat engine, $Q_1=235\ kW$

Heat rejected to the atmosphere, $Q_2=164\ kW$

We need ti find the thermal efficiency of the heat engine. It is equal to the ratio of output work to the energy supplied. Its mathematical form is given by :

$\eta=1-\dfrac{Q_1}{Q_2}\\\\\eta=1-\dfrac{235}{164}\\\\\eta=-0.432$

$\eta=-43.2\%$

The egative value of efficiency shows work is done by the engine.

5 0

3 years ago