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shepuryov [24]
3 years ago
6

The AGC control voltage: ___________

Engineering
1 answer:
lyudmila [28]3 years ago
3 0

Answer:

The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.

Explanation:

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. The job of applications engineer for which Maria was applying requires (a) excellent technical skills with respect to mechanic
Hitman42 [59]

Answer:

  1. Tell us about your self
  2. Are your confident that you are the right candidate for this position
  3. why should i hire you
  4. Do you like working under supervision
  5. How do you like to work ( in a group or individually )
  6. What is your ultimate workplace goal
  7. what are your future plans
  8. What do you expect from the Organization when given the job
  9. Do you like taking on critical problems
  10. How long can you work in this position

Explanation:

For a job of applications engineer which require excellent technical skills, commitment  to working , ability to deal well and confidently with customers a willingness to travel and very intelligent and well-balanced personality.

The ten questions you should ask Maria to determine if she is qualified for the job are :

  1. Tell us about your self ( functions you have )
  2. Are your confident that you are the right candidate for this position
  3. why should i hire you
  4. Do you like working under supervision
  5. How do you like to work ( in a group or individually )
  6. What is your ultimate workplace goal
  7. what are your future plans
  8. What do you expect from the Organization when given the job
  9. Do you like taking on critical problems
  10. How long can you work in this position
5 0
3 years ago
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

8 0
3 years ago
A motorist is driving his car at 60km/hr when he observes that a traffic light 250m ahead turns red. The traffic light is
Alecsey [184]

Explanation:

Okay soo-

Given-

u = 60 km/hr = 60×1000/3600=50/3 m/s

t = 20 s

s = 250 m

a = ?

v = ?

Solution -

Here, acceleration is uniform.

(a) According to 2nd kinematics equation,

s = ut + ½at^2

250 = 50/3 ×20 + 0.5×a×20×20

250-1000/3=200a

(750-1000)/3=200a

a = -250/(3×200)

a = -5/12

a = 0.4167 m/s^2

The required uniform acceleration of the car is 0.4167 m/s^2.

(b) According to 1st kinematics equation

v = u + at

v = 50/3 + (-5/12)×20

v = 50/3-25/3

v = 25/3

v = 8.33 m/s

The speed of the car as it passes the traffic light is 8.33 m/s.

Good luck!

5 0
3 years ago
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th
Ilia_Sergeevich [38]

Answer:

Flow velocity

50.48m/s

Pressure change at probe tip

1236.06Pa

Explanation:

Question is incomplete

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively

solution

In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe

please check attachment for complete solution and step by step explanation

8 0
3 years ago
Ihjpr2 ywjegnak'evsinawhe2'qwmasnh ngl,;snhy
WITCHER [35]

Answer:

ummm ok?

Explanation:

6 0
3 years ago
Read 2 more answers
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