UFMFHT-30-1 Applied Electronics 3 Level 1 - BJT as a Switch One application of BJT is in switching-type circuits, where a load is either switched OFF or ON. ON in this context means that the load current has the nominal value, while OFF signifies no or insignificant current flow through the load. A typical switching application is one where a BJT turns ON or OFF an LED depending on a logic-level input voltage, le, a voltage that is either OV or +5V. The appropriate circuit diagram is shown in Figure 1. UPPLY Figure 1 Twitch The input voltage VIN in Figure hereby controls the BJT, which is either in the cutoff region (OFF) or in the Saturation region (ON). IF VIN=0, then the Base current must be zero since the BE-voltage is zero. Hence, the BIT is in the cut-off region, also said to be turned-off. IF VIN=SV, then the circuit needs to be designed so that the BJT is in the Saturation region. For this we need to know the following Transistor parameters • Vaam: the Base-Emitter voltage when the BIT is on this is usually 0.7V Vow the Collector-Emitter saturation voltage; this is given in Transistor datasheets and assumed here to be 0.2V • B: the current gain in the forward-active region, assumed here to be 100 In order to design for correct operation, we must also know the characteristic of the LED, LA we must know the nominal LED current and voltage. This largely depends on the color of the LED and is shown in the following table. UPMEHT-30-1 Applied Electronics CORO Forward Voltage Ultraviolet Materia Nitride AIN Buminimalium Nitride (Gat19 AmiGuminium Nitride Indium Gamit GON Violet 28-4 Blue 25-37 indium Gallium Nitride Sicon Carbide Green 19-40 Galium Phosphide Blumn Galluminium Phosphide AG Alumn Gallium Phosphide GP) Galium Arsenide Phosphide GP Aluminium Gallium Indium Phosphide A Yellow 21-22 Orange/be 20-21 Gallum Arsenide Phosphide Blumn Gallium Indium Phosphide A విజయ 15-20 Alumio Galium Arsenidee Gabun Arsenide Phosphidea lumia Galuminium Phosphide AG Galium Phosphide Gallium Arsenidea! om Galium Arsenide Infrared >9 For this laboratory assignment we choose a red LED. eared LED. A typical excerpt of adatasheet is shown in Figure 2 Symbol Auteng 20 Forward Current Par For Current Sagestion Current Reverse Voltage Power Dis Operation Temer Storage Tempe Lead Seleng Temperature 10 40 40-100 Figure 2. LED datashee To increase the lifetime of the LED, we choose an LED current = 10mA. Also, V-125 chosen We then can calculate the required values for the Base resistor Re and the Collector resistor Reas follows: UFMFHT-30-1 5 Applied Electronics C-Eloop: -Vol+Ve+Ic"Rc+Va=0 Since the transistor in the ON-case is in the Saturation region, we replace Veswith V. Also, the LED is ON, hence, Viis chosen to be 1.8V and the Collector current (which is the same as the LED current) is 10mA. We then can solve for the Collector resistor: Reet - Vesa) / ltp = 1K0 To ensure that the BJT is in the Saturation region, we choose a Base current I, which is somewhat higher than necessary: >>[/B = 10mA/100 = 100HA A typical factor here is 10, so that the Base current 1, becomes 10*1004A = ImA. We then can calculate the required Base resistor by observing the Base-Emitter loop: -VX+R*4 + Vajon = 0 Solving this equation for Releads to: R$ = (V-Venl/=(5V-0.7V)/1mA = 4,360 We can simulate this circuit using a 2N3904 as the BJT and a red LED shown in Figure 3. LEDI R2 Q1 2N3904 Vin RI w 43ΚΩ VI JOV 5V 10ms 20ms s Hole 12V Figure 3: BIT as a Switch The current gain of the used 2N3904 transistor must be changed to 100. We can do this by double-clicking on the BJT, which opens up the dialog box for changing the BIT parameters as shown in Figure 4: UFMFHT-30-1 Applied Electronics 7 (Note: to show two separate Windows within the same Grapher View, Copy and then Paste the View; you then can select which traces to display and can independently zoom each graph) Figure 5 clearly shows that the LED current in the ON state is 10mA. We can also look at the Base Current, shown in Figure 7. + Figure 7 Base Current it clearly can be seen that the Base current is ImA in the ON case. (Note: It must be pointed out here that the negative spike in the Base current originates from discharge of the parasitic Base-to-Emitter and Base to-Collector capacitance) Design a BJT-as-a-Switch (such as shown in Figure 3) having the following parameters: uno = 24V V SV (ON) or OV (OFF) Vam - 0.7V V0.2V B-200 V 1.8V kr = 10mA (a) Show the calculations for all circuit components (b) Simulate your circuit and show Input Voltage and LED current in a transient (c) simulation showing at least two periods (d) Build your circuit on a breadboard . (e) Measure the input and output voltage using an oscilloscope Your submission must include the following (see the template for this level):

Passphrases can be used to create strong passwords that are easy to remember. Option B is the correct answer.

Passphrases are a type of password that involves using multiple words, often unrelated, to create a phrase or sentence. These passwords can be longer and more complex than traditional passwords and are often easier for people to remember because they involve a memorable phrase. Using mnemonics, birthdates, or numbers may also be helpful for creating strong passwords, but passphrases are often considered the most effective method for creating strong, memorable passwords.

Option B is the correct answer.

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From the question above, :Width of the channel = 4 m

Depth of the flow = 2 m

Discharge = 20 m³/sec

Change in channel width = 1 mAs the discharge is constant, we have the equation of continuity as;

Q = A₁V₁ = A₂V₂

Where,Q = Discharge in m³/s

V₁ = Velocity of fluid in the original channel

A₁ = Area of the flow in the original channel

V₂ = Velocity of fluid in the new channel

A₂ = Area of the flow in the new channel

As the channel is rectangular in shape, we can write the equation as;

Q = W₁ * D₁ * V₁ = W₂ * D₂ * V₂

Where, D₁ = D₂ = D (Depth of flow)

W₁ = 4 m

W₂ = 4 + 1 = 5 m

V₁ = Velocity of fluid in the original channel

V₂ = Velocity of fluid in the new channel∴

4 * 2 * V₁ = 5 * D₂ * V₂∴ 8V₁ = 5D₂V₂∴ V₂ = 8/5 * V₁

The velocity of fluid in the new channel = 1.6V₁

Type of flow in the original channel can be determined using the Froude number as;

Fr = V₁/ √gD

Where g is the acceleration due to gravity

Fr₁ = V₁/ √gD = V₁/ √(9.81 * 2) = V₁/ 4.429

Fr₂ = V₂/ √ gD = (1.6V₁)/ √(9.81 * 2) = 1.6V₁/ 4.429

Fr₁ = V₁/ √gD = V₁/ 4.429

Fr₂ = 1.6V₁/ 4.429

Fr₁ < 1 → Subcritical flow

Fr₂ < 1 → Subcritical flow

As the value of Fr is less than 1, the type of flow is Subcritical flow

.Change in depth of the flow;

D₂ - D₁ = (W₁/W₂ - 1) * D₁D₂ - 2 = (4/5 - 1) * 2

D₂ - 2 = -0.4∴ D₂ = 1.6 m

Change in depth of the flow = D₂ - D₁ = 1.6 - 2 = -0.4 mA

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Answer:

Maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration. Therefore it is not possible.

Explanation:

let's first determine the mass fraction of the phases as follows

\(W_{a}=\frac{m_{a} }{m_{a}+m_{p_{g2}p_{b} } }\)

Given that masses are \(m_{a}=7.39kg\) and \(m_{p_{g2}p_{b} }=3.81kg\)

Thus, \(W_{a} =\frac{7.39}{7.39+3.81}=0.659\)

\(W_{M_{g2Pb} } =1-W_{a}=1-0.659=0.341\)

Now determine the concentration

\(W_{a}=\frac{C_{0}-C_{a} }{C_{b}-C_{a} }\)

At 81% PB and 70% Pb only point at which \(M_{g2Pb}\) exists, we can solve

\(0.659=\frac{70-81}{C_{b}-81 }\)

\(C_{b}=\)64.3%

Refer text book page no. 326, figure 9.20, from figure we can say maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration.

Given: Size of the column = 300 × 300 mm Dead Load (DL) = 961 kNLive Load (LL) = 769 kN Allowable Soil Bearing Pressure = 260 kPa Soil Unit Weight = 18.2 kN/m³Concrete Unit Weight = 23.4 kN/m³Total depth of footing = 577 mm Effective depth of footing = 462 mm.

The objective is to determine the dimension of the square footing in meters "m".The formula to calculate the area of the footing is:A = (DL + LL)/p Where A = Area of the footing DL = Dead Load LL = Live Load p = Allowable Soil Bearing Pressure (260 kPa)The area of the footing A = (961 + 769)/260 = 7.31 m².

The dimension of the footing can be determined using the area as follows:

A = L²L = sqrt (A)The dimension of the footing L = sqrt (7.31) = 2.70 m The dimension of the square footing is 2.70 meters.

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Answer:

transmission bandwidth required is very large.

Explanation:

Answer:B

Explanation: Dead Grass more flammable than wood and Green grass contains water.

Answer:

No it is not permitted

Explanation:

It is not permitted  because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.

The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.

Answer:

r = 1.922 mm

Explanation:

We are given;

Yield stress; σ = 250 MPa = 250 N/mm²

Force; F = 29 KN = 29000 N

Now, formula for yield stress is;

σ = F/A

A = F/σ

Where A is area = πr²

Thus;

r² = 2900/250π

r² = 3.6924

r = √3.6924

r = 1.922 mm

An example of blocking is repressurizing valves. The correct option is a.

Blocking means stopping something from coming or going somewhere.  Controlling the pressure and flow within a piping system. Regulating the flow's direction within a piping system. Reducing the flow rates in a system of pipes. Enhancing security by reducing vacuum or pressure in a pipe system.

Valves have a wide range of applications, including controlling water flow for irrigation, industrial process control, and residential uses like on/off and pressure control for dishwashers, washers, and faucets.

Therefore, the correct option is a. Repressurizing valves.

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Any compressor's discharge temperature cannot exceed 225 degrees F. if the discharge temperature rises above 225 degrees.

However, many compressor packages will contain components that limit discharge temperature to no more than 148.9°C (300°F). In general, it is advised to never exceed operating discharge temperatures of 176.7°C (350°F).

The primary cause of high condensing pressure is high discharge pressure. High condensing pressure can be caused by insufficient condenser heat dissipation, condenser fouling, insufficient cooling air or water volume, and high cooling water or air temperature.

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Answer:

If you are driving below 40 mph, you should leave at least one second for every 10 feet of vehicle length.

The speed of the rotor with respect to the stator is 2,856 rpm, and the speed of the rotor with respect to the stator magnetic field is 2,860 rpm.  

The synchronous speed of a 3-phase induction motor is given by the formula: Ns = 120f/p, where Ns is the synchronous speed in rpm, f is the frequency of the power supply, and p is the number of poles. In this case, since the motor has 8 poles, the synchronous speed is Ns = 120f/8 = 15f.

The speed of the rotor with respect to the stator is given by the formula: Nr = (1 - s)Ns, where Nr is the rotor speed, and s is the slip. The slip is given as 0.05, so the rotor speed is Nr = (1 - 0.05)15f = 14.25f.

The speed of the rotor with respect to the stator magnetic field is given by the formula: Nrm = Nr - Ns = 14.25f - 15f = -0.75f. This indicates that the rotor is rotating in the opposite direction to the stator magnetic field, with a speed of 0.75 times the frequency.

The speed of the rotor magnetic field with respect to the rotor is the slip speed, which is given as Nsr = sNs = 0.05*15f = 0.75f.

The speed of the rotor magnetic field with respect to the stator is the sum of the rotor speed and the rotor magnetic field speed, which is Ns + Nsr = 15f + 0.75f = 15.75f.

The speed of the rotor magnetic field with respect to the stator magnetic field is the difference between the rotor speed and the rotor magnetic field speed, which is Nr - Nsr = 14.25f - 0.75f = 13.5f.

Therefore, the calculated speeds are as follows: i) the speed of the rotor with respect to the stator is 14.25f or 2,856 rpm (assuming a 50 Hz power supply), ii) the speed of the rotor with respect to the stator magnetic field is -0.75f or -150 rpm, iii) the speed of the rotor magnetic field with respect to the rotor is 0.75f or 150 rpm, iv) the speed of the rotor magnetic field with respect to the stator is 15.75f or 3,150 rpm, and v) the speed of the rotor magnetic field with respect to the stator magnetic field is 13.5f or 2,700 rpm.

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The entropy change of helium during this process is -9 Btu/R

The formula is given as

\(Change in S = m[Cvln\frac{T_{2} }{T_{1} } + Rln\frac{V_{2} }{V_{1} }\)

Where the variables are M = mass = 39

Cv = specific heat of helium = 0.753

t2 = 240 + 460

T1 = 60 + 460

R = gas constant = 0.4961

V1 = initial state = 50

V2 = final state = 20

we would have to put these values in the formula that we have above

\(39[0.753\frac{240+460}{60+460} +0.4961ln\frac{20}{50}]\)

= -9 Btu/R.

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Energy transfer is factor that affects unit operation this is because each processing of change requires energy transfer.

Unit operations involves reactions that lead to physical change or chemical transformation.

It include separating of mixtures which can be done by filtering, crystallization and polymerization.

Unit operation often results into changes that can be seen or visible changes.

Energy transfer is one the factors that affect unit operation.

Therefore, Energy transfer is factor that affects unit operation this is because each processing of change requires energy transfer.

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Answer:

SpaceX

Explanation:

SpaceX successfully launches 60 more Starlink satellites as it continues towards 2020 service debut. SpaceX has launched another big batch of Starlink satellites, the low Earth orbit spacecraft that will provide connectivity for its globe-spanning high-bandwidth broadband internet network.

Answer:

Power= Heat energy\ time

heat energy= mc∆T

specific heat of water = 4180

1.5*4180*(60-25)

=219450

time = 10× 60= 600 secs

power = 219450/600

Power= 365.75 Watt

=0.37 KW

Explanation:

the actual has to be bigger because the heater might be required to handle more ...also to accommodate extra energy.

To combine the two sensor readings while minimizing noise, we need to amplify and filter each signal separately before adding them together. The first sensor can be amplified using a non-inverting amplifier with a gain of 10, which will bring its output range to 0-2 V and also reduce the impact of high frequency noise.

The second sensor can be amplified using an inverting amplifier with a gain of 2, which will bring its output range to -2 to 2 V and also reduce high frequency noise.Next, we can filter each signal using a low-pass filter with a cut-off frequency of 10 kHz for the first sensor and a cut-off frequency of 40 kHz for the second sensor. These filters will eliminate any high frequency noise while allowing signals of interest to pass through.Finally, we can add the two signals together using an op-amp summing amplifier with equal resistor values for each input. This will ensure that both signals are added in equal proportion in the final output. To prevent the output from ever going negative, we can add a diode in series with the output that only allows current to flow in the positive direction.Overall, the circuit will consist of two amplifiers, two low-pass filters, and a summing amplifier with a diode in series with the output. By amplifying and filtering each signal separately, we can minimize noise and combine the two signals in equal proportion without ever producing a negative output.

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Answer:

True!! took the test

Explanation:

Answer:

Explanation:

......................

Answer:

a)  316 stainless steel

b)  316 stainless steel

c)  Carbon steel

d)  stainless steel 316

e)  stainless steel 316

f )   Carbon steel

Explanation:

A) A 10,000 M' storage tank for Toluene

The best material for building this tank is 316 stainless steel this is because it is relatively cheaper than other materials such as Hest Alloy C while still maintaining  the required tensile strength needed to build the storage tank for Toluene

B) A 5.0 m' tank for storing a 30% w/w aqueous solution of sodium chloride

The suitable material for building this tank is 316 stainless steel as it is cheaper than Titanium and Hest Alloy C and it is compatible with sodium chloride

C) A 2 m diameter, 20 m high distillation column, distilling acrylonitrile

The best material for constructing this column is  Carbon steel and this is because  it has a very high compressive strength which is approximately . 1320 MPa

D)  A 100 m storage tank for strong nitric acid.

The best material is stainless steel 316 and this is due to its high tensile strength

E) 316 stainless steel

F) Carbon steel is suitable for building this absorption column because  it has a very high compressive strength which is approximately . 1320 MPa

Cranes and derricks installed on floating surfaces, such as barges or ships, must have a designated location where they can be easily seen by operators.

This requirement is crucial for ensuring the safety and efficiency of lifting operations in maritime settings.

The designated location for visibility is typically a control cab or an elevated platform situated near the crane or derrick. This position provides operators with an unobstructed view of the entire lifting area, including the load, the surrounding environment, and any potential obstacles or hazards.

Clear visibility is essential because it allows operators to monitor the lifting operation closely, ensuring that it is conducted in a safe and controlled manner.

By having a direct line of sight, operators can make informed decisions, react promptly to any emergencies or malfunctions, and communicate effectively with other personnel involved in the operation.

Compliance with this visibility requirement is a vital safety measure, as it helps prevent accidents, reduces the risk of collisions, and enhances overall operational efficiency.

Additionally, it enables operators to adhere to industry regulations and guidelines governing the safe use of cranes and derricks on floating surfaces.

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ᴄᴏᴍᴘʟᴇᴛᴇɴᴇꜱꜱ ᴍᴇᴀɴꜱ ᴛʜᴇ ꜱᴛᴀᴛᴇ ᴏꜰ ʙᴇɪɴɢ ᴄᴏᴍᴘʟᴇᴛᴇ ᴀɴᴅ ᴇɴᴛɪʀᴇ; ʜᴀᴠɪɴɢ ᴇᴠᴇʀʏᴛʜɪɴɢ ᴛʜᴀᴛ ɪꜱ ɴᴇᴇᴅᴇᴅ.

ʙᴇᴀᴜᴛʏ ᴍᴇᴀɴꜱ combination of qualities, such as shape, colour, or form, that pleases the aesthetic senses, especially the sight.

According to the question of theorem in which we have to convert, then the PDA is given below:

A theorem is a statement in mathematics that has been proven to be true using accepted methods of proof. It can be used to make predictions about various mathematical concepts and to prove the validity of other mathematical theorems. A theorem is the result of a logical argument and is often comprised of previously established facts or laws.  A theorem is a fundamental building block in mathematics, as it is used to prove other theorems.

PDA: Q={q0,q1,q2,q3,q4,q5}

Σ={a,b}

Γ={a,b,X,R,S,T,ε}

q0=q0

q5=accept state

δ:(q0,a,ε)->(q1,X,RST)

   (q1,a,R)->(q1,X,RS)

   (q1,b,R)->(q1,X,ST)

   (q1,a,S)->(q2,ε,aTb)

   (q1,b,S)->(q3,ε,bTa)

   (q2,a,T)->(q2,ε,a)

   (q2,b,T)->(q2,ε,ε)

   (q3,a,T)->(q3,ε,ε)

   (q3,b,T)->(q3,ε,b)

   (q2,a,X)->(q4,ε,ε)

   (q2,b,X)->(q4,ε,ε)

   (q3,a,X)->(q4,ε,ε)

   (q3,b,X)->(q4,ε,ε)

   (q4,a,ε)->(q1,X,RS)

   (q4,b,ε)->(q1,X,ST)

   (q1,ε,X)->(q5,ε,ε)

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Answer:

Electrical energy

Answer:

  (i) 12 V in series with 18 Ω.

  (ii) 0.4 A; 1.92 W

  (iii) 1,152 J

  (iv) 18Ω — maximum power transfer theorem

Explanation:

As seen by the load, the equivalent source impedance is ...

  10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

  (36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

The load current is ...

  (12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

  P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

  (600 s)(1.92 J/s) = 1,152 J

__

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

  ((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

If the function is called with a single parameter of 50, the resulting iterable will include integers from 0 through 49, inclusive.

In most programming languages, when specifying a range or interval, the starting value is inclusive, while the ending value is exclusive. This means that the range will include all values starting from the initial value and ending one element before the specified end value.

In this case, since the parameter is 50, the function will generate an iterable that includes integers starting from 0 and ending at 49 (inclusive). The number 50 will not be included in the resulting iterable.

It is important to clarify the behavior of the function, especially when dealing with range or interval operations, as the inclusivity or exclusivity of the start and end values can affect the expected output.

By specifying that the range is inclusive for 0 through 49, it ensures that all integers within that range will be included in the resulting iterable, while excluding the value of 50.

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