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2 years ago

What is voltage drop?

Engineering

1 answer:

lord [1]2 years ago

6 0

Answer:

Explanation:

It is the voltage a voltmeter would read when connected across something that has resistance.

___________0_________O______O_______

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|_____________________| |_____________ |

The diagram above is supposed to represent 3 lightbulbs connected in series. The vertical lines in the middle are supposed to be a battery which powers the three light bulbs. If you put a voltmeter across one of the lightbulbs, it will read a voltage that is 1/3 of the voltage of the battery.

Answer

That reading you get across the one light bulb is The Voltage Drop.

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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

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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

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= ( 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

Learn more about voltage : brainly.com/question/11288970

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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