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Musya8 [376]
3 years ago
11

zener shunt regulator employs a 9.1-V zener diode for which VZ = 9.1 V at IZ = 9 mA, with rz = 40 and IZK= 0.5 mA. The available

supply voltage of 15 V can varyas much as ±10%. For this diode, what is the value of VZ0?For a nominal load resistance RL of 1 k and a nominal zenercurrent of 10 mA,what current must flow in the supply resistorR? For the nominal value of supply voltage, select a valuefor resistor R, specified to one significant digit, to provideat least that current. What nominal output voltage results?For a ±10% change in the supply voltage, what variationin output voltage results? If the load current is reduced by50%, what increase in VO results? What is the smallest valueof load resistance that can be tolerated while maintainingregulation when the supply voltage is low? What is the lowestpossible output voltage that results? Calculate values for theline regulation and for the load regulation for this circuit usingthe numerical results obtained in this problem.
Engineering
1 answer:
gulaghasi [49]3 years ago
5 0

Answer:

V_z=9.1v

V_{zo}=8.74V

I=10mA

R=589 ohms

Explanation:

From the question we are told that:

Zener diode Voltage V_z=9.1-V

Zener diode Current I_z=9 .A

Note

rz = 40\\\\IZK= 0.5 mA

Supply Voltage V_s=15

Reduction Percentage P_r= 50 \%

Generally the equation for Kirchhoff's Voltage Law is mathematically given by

V_z=V_{zo}+I_zr_z

9.1=V_{z0}+9*10^{-3}(40)

V_{zo}=8.74V

Therefore

At I_z-10mA

V_z=V_{z0}+I_zr_z

V_z=8.74+(10*10^{-3}) (40)

V_z=9.1v

Generally the equation for Kirchhoff's Current Law is mathematically given by

-I+I_z+I_l=0

I=10mA+\frac{V_z}{R_l}

I=10mA+\frac{9.1}{0}

I=10mA

Therefore

R=\frac{15V-V_z}{I}

R=\frac{15-9.1}{10*10^{-3}}

R=589 ohms

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A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0
3 years ago
A. A 3-kg plastic tank that has a volume of 0.2 m^3 is lled with liquid water. Assuming the density of water is 1000 kg=m^3, det
Tpy6a [65]

Answer:

The answer is below

Explanation:

a) The weight of the combined system is the sum of the weight of the water and the weight of the tank

m_{water}=V_{tank}.\rho_{wtaer}\\\\m_{water}=0.2m^3*1000kg/m^3\\\\m_{water}=200 \ kg\\\\m_{total} = m_{water}+m_{tank}\\\\But\ m_{tank}=3kg,therefore:\\\\m_{total} =200kg+3kg\\\\m_{total} =203\ kg\\\\weight_{total}=m_{total}g\\\\weight_{total}=203kg*9.81m/s^2\\\\weight_{total}=1991.43\ N

b) Since the weight of a system can be divided into smaller portions, hence weight is an extensive property.

c) When analyzing the acceleration of gases as they flow through a nozzle, the geometry of the nozzle which is an open system can be chosen as our system.

d) Given that:

\rho_{water}=1000kg/m^3\\\\1kg/m^3=0.062428lb/ft^3\\\\1000kg/m^3=1000kg/m^3*\frac{0.062428lb/ft^3}{kg/m^3}=62.43lb/ft^3\\ \\\rho=SG*\rho_{water}=1.03*62.43=64.272lb/ft^3\\\\P=P_{atm}+\rho g H\\\\P=14.7\ psia+64.272\ lb/ft^3*32.2\ ft/s^2*175\ ft*\frac{1\ ft^2}{12^2\ in^2}*\frac{1\ lbf}{32.2\ lbm.ft/s^2}  \\\\P=92.8\ psia

6 0
3 years ago
Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No
Degger [83]

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is 1.557\times10^{-7}m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

Re=\frac{vd}{\nu}

Here, v is velocity, \nu is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

Re=\frac{vd}{\nu}

Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

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3 years ago
The insulation resistance of a motor operated by an electronic drive is to be tested using a megger. What precaution should you
EleoNora [17]
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4 0
3 years ago
In this problem set, you will implement multidimensional scaling (MDS) from scratch. You may use standard matrix/vector librarie
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Features of Multidimensional scaling(MDS) from scratch is described below.

Explanation:

Multidimensional scaling (MDS) is a way to reduce the dimensionality of data to visualize it.  We basically want to project our (likely highly dimensional) data into a lower dimensional space and preserve the distances between points.  

If we have some highly complex data that we project into some lower N dimensions, we will assign each point from our data a coordinate in this lower dimensional space, and the idea is that these N dimensional coordinates are ordered based on their ability to capture variance in the data.  Since we can only visualize things in 2D, this is why it is common to assess your MDS based on plotting the first and second dimension of the output.  

If you look at the output of an MDS algorithm, which will be points in 2D or 3D space, the distances represent similarity. So very close points = very similar, and points farther away from one another = less similar.

Working of MDS

The input to the MDS algorithm is our proximity matrix.  There are  two kinds of classical MDS that we could use:  Classical (metric) MDS is for data that has metric properties, like actual distances from a map or calculated from a vector .Nonmetric MDS is for more ordinal data (such as human-provided similarity ratings) for which we can say a 1 is more similar than a 2, but there is no defined (metric) distance between the values of 1 and 2.

Uses

Multidimensional scaling (MDS) is a means of visualizing the level of similarity of individual cases of a dataset. MDS is used to translate "information about the pairwise 'distances' among a set of n objects or individuals" into a configuration of n points mapped into an abstract Cartesian space.

8 0
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