Ask question

Ask question

All categories

3 years ago

10

. For each of the following situations, determine if the diode below is in foward or reversebias(a) (2 points) V1 = 0 V, V2= 2V(

b) (2 points) V1 = 4.5 V, V2=2.8V(c) (2 points) V1=-1V, V2=-1.3V

1 answer:

Mazyrski [523]3 years ago

6 0

Assuming V1 is the anode and v2 the cathode (Anode is P region and Cathode is N)

Answer:

a) Reverse bias

b) Forward bias

c) Forward bias

Explanation:

Forward bias: It happens whenever the N region of the diode is more positive than the P region. Hence, the depletion zone increase ceasing the current through the circuit -> V1 -V2 < 0

Reverse bias: It happens whenever the P region of the diode is more positive than the N region. In this case, the depletion zone begins to shrink, if enough voltage is applied current could go through the circuit -> V1 - V2 > 0

a) V = V1 - V2 = 0 - 2 = -2 -> -2 is smaller than zero therefore, we have reverse bias

b) V = V1 - V2 = 4.5 - 2.8 = 1.7 -> 1.7 is greater than zero therefore, we have forward bias

c9 V = V1 - V2 = -1 - -1.3 = 0.3 -> 0.3 is greater than zero therefore, we have forward bias

You might be interested in

The moisture content in air (humidity) is measured by weight and expressed in pounds or ____________________.

VikaD [51]

Moisture content is measured in terms of pounds of water per pound of air (lb water/lb air) or grains of water per pound of air (gr. of water/lb air).

Hope this helps❤

3 0

2 years ago

A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t

Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

$tan\phi=\frac{\bigtriangleup l}{h}$

$tan\phi=\frac{0.08}{1.38}$

$tan\phi= 0.05797$

For small angle of $\phi$ , $tan\phi$ can take as $\phi$ .

$\phi = 0.05797 rad.$

Thus, the shear strain is 0.05797 rad.

7 0

3 years ago

Zachary finished an internship as a Software Quality Assurance Engineer. For which job is he best qualified?

astra-53 [7]

Answer:

B. A software development firm needs someone to find and fix bugs on multiple computer platforms.

Explanation:

A software quality assurance engineer is someone who monitors every phase of the software development process so as to ensure design quality, making sure that the software adheres to the standards set by the development company. Finding bugs would make this intern a amazing bug finder

6 0

3 years ago

Read 2 more answers

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

Rufina [12.5K]

Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

6 0

2 years ago

From an aerial photograph, one observes that on a level section of a (multilane) highway, 25% of the vehicles are trucks, 75% ar

egoroff_w [7]

Answer:

(a) Flow rate of vehicles = No of vehicles per mile * Speed

=No of cars per mile * Speed +No of trucks per mile * Speed

= 0.75*50*60 + 0.25*50*40

=2750 vehicles / hour

(b) Let Density of vehicles on grade = x

Density on flat * Speed =Density on grade * Speed

So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25

So, x= 57.89

So, Density is around 58 Vehicles per Mile.

(c) Percentage of truck by aerial photo = 25%

(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %

4 0

3 years ago