1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bonufazy [111]
3 years ago
9

How to find the voltage(B Aab) in series parallel circuit? ​

Engineering
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

  Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

  R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

  Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

  Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

  (1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

  V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

  Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

  (Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

  (V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

  (V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

You might be interested in
2)Prueba de presión Cuando a una persona se le somete a una prueba de presión, por lo general se le indica que, al llegar el rit
Olin [163]

Usando la ecuación de regresión que modela la frecuencia cardíaca máxima, podemos obtener el valor predicho de los problemas dados así:

La ecuación lineal que modela la frecuencia cardíaca máxima permitida en función de la edad del paciente está relacionada con la fórmula:

  • m = - 0,875x + 190

<em>x = edad del paciente; m = máx. frecuencia cardíaca permitida</em>

1.) <u>Frecuencia cardíaca máxima permitida para una persona de 50 años:</u>

Sustituye x = 50 en la ecuación:

m = -0,875 (50) + 190

m = 146,25

Por lo tanto, la frecuencia cardíaca máxima permitida es de aproximadamente 146 latidos / min.

2.) <u>Edad para una persona con frecuencia cardíaca máxima de 160 latidos / min</u>:

Sustituye m = 160 en la ecuación:

160 = -0,875x + 190

0,875x = 190 - 160

0,875x = 30

x = 30 / 0,875

x = 34,28

Por tanto, la edad de la persona sería de unos 34 años.

Más información: brainly.com/question/25395533

3 0
2 years ago
Which statement best describes the lower vc-turbo engine compression ratios?.
kobusy [5.1K]

The statement that best describes the lower vc-turbo engine compression ratios is that its generate an increased amount of power with the average fuel consumption.

<h3>Vc-turbo engine</h3>

In the engine, lower temperatures allow to run more ignition timing advance which makes significantly more power than increasing the compression ratio in a turbo engine.

Hence, the turbo engines run lower static compression ratios to increase the amount of power they can reliable generate on pump gas.

Therefore, the statement that best describes the lower vc-turbo engine compression ratios is that its generate an increased amount of power with the average fuel consumption.

Read more about turbo engines

<em>brainly.com/question/26409491</em>

7 0
2 years ago
1. A piston having a diameter of 5.48 inches and a length of 9.50 in slides downward with a velocity, V, through a vertical pipe
const2013 [10]

Answer:

V = 0.00459 ft/s  

Explanation:

Since the Piston is moving downwards with a constant velocity V, from the first Newton’s law we know that all vertical forces, must have zero resultant (their sum over vertical axis must equal to zero). Therefore, force that pulls the piston down, is equalized by force of viscous friction Fd= Fvf = 0.5lb (lb here is the pound-force unit). We will relate F ѵ f  with τ and from that derive the equation for V.

Fѵf = τ  . A

Where τ  = µ. du/dy = µ . V/b  , and A = π . D . l from this Follows:

Fѵf= (V.  A .µ )/b     V= ( Fѵf .b )/(A.µ)    

Placing all the known values in the equation ( remember  to transform inches to feet, by multiplying inches values with the factor 1/12), we obtain :  

ft2

V = ((0.5lb)   .   (0.002/12 ft))/(π   .   (5.48/12 ft)  .  (9.50/12 ft)  .  (0.016 lb.s/(ft^2 )))

V = 0.00459 ft/s  

Download docx
3 0
3 years ago
1. Use the charges to create an electric dipole with a horizontal axis by placing a positive and a negative charge (equal in mag
White raven [17]

Answer:

2)

a)  to the right of the dipole    E_total = kq [1 / (r + a)² - 1 / r²]

b)To the left of the dipole      E_total = - k q [1 / r² - 1 / (r + a)²]

c) at a point between the dipole, that is -a <x <a  

      E_total = kq [1 / x² + 1 / (2a-x)²]

d) on the vertical line at the midpoint of the dipole (x = 0)

E_toal = 2 kq 1 / (a ​​+ y)² cos θ

Explanation:

2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.

This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point

Let's analyze each point separately.

The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.

a) to the right of the dipole

The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity

           E₊ = k q / (r + a)²

where 2a is the distance between the charges of the dipole and the field is to the right

the negative charge creates an incoming field of magnitude

           E₋ = -k q / r²

The field is to the left

therefore the total field is the sum of these two fields

           E_total = E₊ + E₋

           E_total = kq [1 / (r + a)² - 1 / r²]

we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.

b) To the left of the dipole

The result is similar to the previous one by the opposite sign, since the closest charge is the positive one

E₊ is to the left and E₋ is to the right

          E_total = - k q [1 / r² - 1 / (r + a)²]

We see that this field is also directed to the left

c) at a point between the dipole, that is -a <x <a

In this case the E₊ field points to the right and the E₋ field points to the right

                      E₊ = k q 1 / x²

                      E₋ = k q 1 / (2a-x)²

                      E_total = kq [1 / x² + 1 / (2a-x)²]

in this case the field points to the right

d) on the vertical line at the midpoint of the dipole (x = 0)

    In this case the E₊ field points in the direction of the positive charge and the test charge

    in E₋ field the ni is between the test charge and the negative charge,

the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)

                      E₊ = kq 1 / (a ​​+ y) 2

                      E₋ = kp 1 / (a ​​+ y) 2

                      E_total = E₊ₓ + E_{-x}

                      E_toal = 2 kq 1 / (a ​​+ y)² cos θ

e) same as the previous part, but on the negative side

                        E_toal = 2 kq 1 / (a ​​+ y)² cos θ

When analyzing the previous answer there is no point where the field is zero

The different configurations are outlined in the attached

3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive

a) to the right

in this case the two field goes to the right

           E_total = kq [1 / (r + a)² + 1 / r²]

b) to the left

            E_total = - kq [1 / (r + a)² + 1 / r²]

c) between the two charges

E₊ goes to the right

E₋ goes to the left

            E_total = kq [1 / x² - 1 / (2a-x)²]

d) between vertical line at x = 0

             

E₊ salient between test charge and positive charge

           E_total = 2 kq 1 / (a ​​+ y)² sin θ

In this configuration at the point between the two charges the field is zero

8 0
3 years ago
A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35
lukranit [14]

Answer:

P_O = 0.989 watt = 19.9 dBm

Explanation:

Given data:

P_1 power = 20 dBm  = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor = 10 log \frac{P_1}{P_f}

solving for  final power

20 = 10 log\frac{P_1}{P_f}

2 = log \frac{P_1}{P_f}

100 = \frac{0.1}{P_f}

P_f = 0.001 watt = 0 dBm

Directivity D =  10 \frac{P_f}{P_b}

35 = 10 \frac{0.001}{P_b}

P_b = 3.162 \times 10^{-7} wattt

output Power  = P_1 -P_f - P_b

                       = 0.1 - 0.001 - 3.162 \times 10^{-7}

P_O = 0.989 watt = 19.9 dBm

6 0
3 years ago
Other questions:
  • What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00
    5·1 answer
  • As you discovered in lab last week, the advantage of CMOS logic is that no drain current flows through the MOSFETs when the outp
    14·1 answer
  • A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T
    6·1 answer
  • A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig
    8·1 answer
  • Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals:
    5·1 answer
  • Can you list three ways that real life earthquake conditions may differ from those made by a shaking table
    12·1 answer
  • If an older multimeter is set to display its reading in megohms, what must you do to determine the correct value? Please help.
    5·1 answer
  • A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int
    6·1 answer
  • Engineers will redesign their products when they find flaws. TRUE O False​
    10·1 answer
  • Select the correct answer. which process involves creating a product by heating metals and changing their shape through the appl
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!