can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and angle. thanks
can anyone help me with this please.i have the current - 1

Answers

Answer 1
Answer:

Answer:

  • branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
  • branch 2: i = 21.466∠63.435°; pf = 0.447 leading
  • total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

  X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915*10^(-3)\approx j4.99984\,\Omega

The capacitive reactance is ...

  X_C=(1)/(j\omega C)=(-j)/(100\pi\cdot 318.31*10^(-6)F)\approx -j10.00000\,\Omega

Branch 1

The impedance of branch 1 is ...

  Z1 = 8 +j4.99984 Ω

so the current is ...

  I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

  Z2 = 5 -j10 Ω

so the current is ...

  I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

  It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.


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or a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 164 mm2 without plastic deformation

Answers

Answer:

The maximum load (in N) that may be applied to a specimen with this cross sectional area is F=43788 N

Explanation:

We know that the stress at which plastic deformation begins is 267 MPa.

We are going  to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:

P=(F)/(A)    (equation 1)

We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before  plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.

Before apply the equation, we need to convert the units of area in m^(2). So,

A[m^(2)]=A[mm^(2) ]((1 m)/(1000 mm)) ^(2)\nA[m^(2)]=164 mm^(2)((1 m)/(1000 mm))^(2) \nA[m^(2)]=0.000164 m^(2)

And then, from the equation 1,

F=PA\nF=(267 MPa)(0.000164 m^(2))\nF=(267x10^(6) Pa)(0.000164 m^(2))\nF=(267x10^(6) N/m^(2))(0.000164 m^(2))\nF=43788 N

Write a Python program that does the following. Create a string that is a long series of words separated by spaces. The string is your own creative choice. It can be names, favorite foods, animals, anything. Just make it up yourself. Do not copy the string from another source. Turn the string into a list of words using split. Delete three words from the list, but delete each one using a different kind of Python operation. Sort the list. Add new words to the list (three or more) using three different kinds of Python operation. Turn the list of words back into a single string using join. Print the string.

Answers

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

\n Then I used methods append(), insert() and extend() for adding elements to the list.

\n Finally I converted list into a string using join() and adding space in between the elements of the list.

List the RTL (Register Transfer Language) sequence of micro-operations needed to execute the instruction STORE X

from the MARIE instruction set architecture. Then write the corresponding signal sequence to perform these micro-operations and to reset the clock cycle counter.

You may refer to the provided "MARIE Architecture and Instruction Set" file in the Front Matter folder.

Answers

Answer:

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Explanation:

STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.

1. First of all the address X has to be tranfered on to the Memory Address Register MAR.  

MAR<----X

2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR

MBR<-----AC

3. Store the MBR into memory where MAR points to.

M[MAR]<------MBR

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Why do we care about a material's ability to resist torsional deformation?(A) Because the angle of twist of a material is often used to predict its shear toughness
(B) Because rotating shafts are used in engineering applications
(C) We don't care - simply an academic exercise.
(D) Because it can determine G and inform us of a materials ability to resist shear deformation

Answers

Answer:

(A) Because the angle of twist of a material is often used to predict its shear toughness

Explanation:

In engineering, torsion is the solicitation that occurs when a moment is applied on the longitudinal axis of a construction element or mechanical prism, such as axes or, in general, elements where one dimension predominates over the other two, although it is possible to find it in diverse situations.

The torsion is characterized geometrically because any curve parallel to the axis of the piece is no longer contained in the plane initially formed by the two curves. Instead, a curve parallel to the axis is twisted around it.

The general study of torsion is complicated because under that type of solicitation the cross section of a piece in general is characterized by two phenomena:

1- Tangential tensions appear parallel to the cross section.

2- When the previous tensions are not properly distributed, which always happens unless the section has circular symmetry, sectional warps appear that make the deformed cross sections not flat.

A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool 5 large watermelons, 10 kg each, to 8 C. If the watermelons are initially at 28 C, determine how long it will take for the refrigerator to cool them.

Answers

Answer:

\Delta t = 5866.667\,s\,(97.778\,m)

Explanation:

The specific heat for watermelon above freezing point is 3.96\,(kJ)/(kg\cdot K). The heat liberated by the watermelon to cool down to 8°C is:

Q_(cooling) = (5)\cdot (10\,kg)\cdot (3.96\,(kJ)/(kg\cdot K) )\cdot (20\,K)

Q_(cooling) = 3960\,kJ

The heat absorbed by the household refrigerator is:

\dot Q_(L) = COP\cdot \dot W_(e)

\dot Q_(L) = 1.5\cdot (0.45\,kW)

\dot Q_(L) = 0.675\,kW

Time needed to cool the watermelons is:

\Delta t = (Q_(cooling))/(\dot Q_(L))

\Delta t = (3960\,kJ)/(0.675\,kW)

\Delta t = 5866.667\,s\,(97.778\,m)

 

Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month, a slash, and the year. End with newline. The program will be tested with inputs 1 2000 and then with inputs 5 1950. Ex: If the input is 1 2000, the output is: 1/2000

Answers

Answer:

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       System.out.println(birthMonth+"/"+birthYear);

   }

}

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       // Get input values for birthMonth and birthYear from the user

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       // Output the month, a slash, and the year

       System.out.println(birthMonth + "/" + birthYear);

   }

}

When the program is tested with inputs 1 2000, the output will be:

1/2000

And when tested with inputs 5 1950, the output will be:

5/1950

Know more about java program:

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