A 100 MHz generator with Vg= 10/00 v and internal resistance 50 ohms air line that is 3.6m and terminated in a 25+j25 ohm load

Answers

Answer 1

The value of Vz at point z from the generator  = 17.748 ∠ -107.62°  V

Given data :

Internal resistance = 50 ohms

Vg = 10 ∠ 0° v

length of air line = 3.6 m

Terminating resistance = 25 + j25 Ω

First step : Determine the Total impedance

Total Impedance ( z ) = Rin + Rline + Rl + jXl

                                   = 50 + 50 + 25 + j25  

                                   = 125 + j25  ≈ 127.47 ∠ 11.3°

Next step : Determine the current in the circuit

current ( I ) = Vg / z

                  =  ( 10∠0° v ) / ( 127.47 ∠ 11.3° )

                 = 0.0784 ∠ -11.3  amp

Final step : Determine the value of Vz at point z from the generator

Vz = ( Vg + I * Ri ) - ( RI * I + Rline * I )

    = ( 10∠0° + 0.0784 ∠ -11.3  * 50 ) - ( 25 + j25  + 50 * 0.0784 ∠ -11.3 )

    = -5.37 - j16.91  ≈  17.748 ∠ -107.62°  V

Hence we can conclude that the value of Vz at point z form the generator 17.748 ∠ -107.62°  V

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Hello your question is incomplete below is the complete question

A 100 MHz generator with Vg =10< 0 degree (v) and internal resistance 50-ohm is connected to a lossless 50-ohm air line that is 3.6m long and terminated in a 25+j25 (ohm) load.

Find (a) V(z) at a location z from the generate.


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Answer:

the answer is Letter C

Explanation:

DON'T TRULY TRUST ME PLEASE..

For flow of a liquid metal through a circular tube, the velocity and

temperature profiles are u(r)=2Um[1-(r/ro)2] and T(r) - Ts

= C2 [1 – (r/r0)2],respectively, where C2 is constant. What is the value of the Nusselt

number NuD at this location?

Answers

Answer:

Known: Form of the velocity and temperature profiles at a particular axial location for flow in a circular tube. Find: Nusselt number at the prescribed.

Explanation:

A double-threaded Acme stub screw of 2-in. major diameter is used in a jack having a plain thrust collar of 2.5-in. mean diameter. Coefficients of running friction are estimated as 0.10 for the collar and 0.11 for the screw.

a. Determine the pitch, lead, thread depth, mean pitch diameter, and lead angle of the screw

b. Estimate the starting torque for raising and for lowering a 5000-lb load.

c. What is the efficiency of the jack when raising the load?

d. Would the screw overhaul if a ball thrust bearing of negligible friction were used in place of the plain thrust washer?

Answers

This is the answer for the question

Determine the mass density of an oil if 0.3 tonnes of the oil occupies a volume of 4m.

Answers

Answer:

Do you mean 4m^3 and 3.0 tones?

Explanation:

solution:

Mass = m = 3.0 tones

- 1 ton = 1,000 kg

= 3.0 × 1,000

= 3,000 kg

volume = v = 4m^3

Required:

Mass density of oil = p = ?

We know that;

[tex]p = \frac{mass}{volume} = \frac{m}{v} = \frac{3000}{4} = 750kg |m^{3} ans [/tex]

The answer is:

750kg / m^3

All forms of energy can be transformed into what form of energy, often times as waste?

Answers

I think thermal energy. It tends to be the final form off all energy equations in physics.

In a modern industrial plant the most likely polyphase power system in use is a/an blank system

Answers

Answer:

  three-phase

Explanation:

2-phase, 3-phase, and 6-phase systems are in use. The most common of these is 3-phase.

A modern industrial plant most likely uses a 3-phase system.

__

Additional comment

Serious research began on 6-phase power transmission systems in 1973. They have been shown to increase power transfer, and to reduce corona, magnetic fields, and audio noise compared to 3-phase systems. However, they are not in widespread industrial use.

A periodic digital waveform has a pulse width 25 and a period of 150 . Determine the frequency and the duty cycle

Answers

The frequency  is 6.67 kHz and the duty cycle is:16.67%.

Frequency and duty cycle

Given:

Pulse width=25

Period=150

Frequency:

Frequency=1/(150×10^-6)

Frequency=1/0.00015

Frequency=6.666 kHz

Frequency=6.67 kHz (Approximately)

Duty cycle:

Duty cycle=(25×10^-6)/ (150×10^-6)×100%

Duty cycle=16.67%

Therefore the frequency  is 6.67 kHz and the duty cycle is:16.67%.

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