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Vlad [161]
2 years ago
9

What is the purpose of a rectifier assembly

Engineering
1 answer:
Elden [556K]2 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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A PMOS device with VT P = −1.2 V has a drain current iD = 0.5 mA when vSG = 3 V and vSD = 5 V. Calculate the drain current when:
Ksju [112]
The answer is b ! Hope I helped
7 0
2 years ago
A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp
iragen [17]

Answer:

Elastic modulus of steel  = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm  = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × 10^{-4} m

solution

we know that Elastic modulus is express as

Elastic modulus = \frac{stress}{strain}    ................1

here stress is

Stress = \frac{Force}{area}       .................2

Area = (0.022)²

and  

Strain = \frac{extension}{length}       .............3

so here put value in equation 1 we get

Elastic modulus = \frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }  

Elastic modulus of steel = 202.27 × 10^9 Pa

Elastic modulus of steel  = 202.27 GPa

3 0
3 years ago
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t
LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

 pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.  

Using the given mass m, molar mass M, we can get the following equation:  

 pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

 

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

 U=-1.848*10^6 J

Note, that the energy was taken away from the system.  

5 0
3 years ago
3. Low-voltage conductors rarely cause<br> injuries.
hram777 [196]
True
the answer to this would be true
8 0
2 years ago
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