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alukav5142 [94]
2 years ago
7

What are two ways you can see that there is a smaller load going to the base of the transistor?

Engineering
1 answer:
Lyrx [107]2 years ago
7 0

Answer:

<h3><em>Transistor switches can be used to switch a low voltage DC device (e.g. LED’s) ON or OFF by using a transistor in its saturated or cut-off </em><em>state</em></h3>

  1. <em>. Cut-off </em><em>Region</em>

<em>Here the operating conditions of the transistor are zero input base current ( IB ), zero output collector current ( IC ) and maximum collector voltage ( VCE ) which results in a large depletion layer and no current flowing through the device. Therefore the transistor is switched “Fully-OFF”.</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>2</em><em>.</em><em>Saturation </em><em>Region</em>

<em>Here the transistor will be biased so that the maximum amount of base current is applied, resulting in maximum collector current resulting in the minimum collector emitter voltage drop which results in the depletion layer being as small as possible and maximum current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.</em>

Explanation:

hope it helps

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An airplane travels 1000 miles at velocity of 500 miles in every hour. solve for the time it took the plane to arrive.
alukav5142 [94]

Answer:

Time taken by airplane to cover given distance = 2 hour

Explanation:

Given:

Distance cover by airplane = 1,000 miles

Velocity of airplane = 500 miles per hour

Find:

Time taken by airplane to cover given distance

Computation:

Time taken = Distance / Velocity(Speed)

Time taken by airplane to cover given distance = Distance cover by airplane / Velocity of airplane

Time taken by airplane to cover given distance = 1,000 / 500

Time taken by airplane to cover given distance = 2 hour

7 0
3 years ago
Problem 4.041 SI Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at -26oC with a vo
Rom4ik [11]

Answer:

0.0297M^3/s

W=68.48kW

Explanation:

Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

state 1

X=quality=1

T=-26C

density 1=α1=5.27kg/m^3  

entalpy1=h1=234.7KJ/kg

state 2

T2=70

P2=8bar=800kPa

density 2=α2=31.91kg/m^3  

entalpy2=h2=306.9KJ/kg

Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.

m1=m2

(Q1)(α1)=(Q2)(α2)

\frac{(Q1)(\alpha 1) }{\alpha 2} =Q2\\Q2=\frac{(0.18)(5.27) }{31.91} =0.0297M^3/s

the volumetric flow rate at the exit is 0.0297M^3/s

To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation

W=m(h2-h1)

m=Qα

W=(0.18)(5.27)(306.9-234.7)

W=68.48kW

the compressor power is 68.48kW

4 0
3 years ago
What is the thermal efficiency of this reheat cycle in terms of enthalpies?
schepotkina [342]

Answer:

   \eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}

Explanation:

For close gas turbine:

       Gas turbine works on Brayton cycle.Gas turbine have lots of applications like ,it is use in aircraft,in land applications etc.

Reheating is the method to improve the efficiency of the gas turbine.In reheating gas is expanding in two turbine instead of one turbine alone.Two turbine like high pressure turbine and low pressure turbine are used for expansion.

In the above diagram 1-2 is a compressor,2-3 heat addition,3-4 high pressure turbine,4-5 reheating of cycle 5-6 low pressure turbine,6-1 heat rejection,

We know that    \eta =\frac{W_{net}}{Q_{s}}

Now take h_{1},h_{2},,h_{3},h_{4},h_{5},h_{6} represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.

So total heat supplied Q_S=

\left (h_3-h_2\right )+\left (h_5-h_4\right )

Net work out put

W_{net}=\left (h_5-h_6\right )-\left (h_2-h_1\right )

So efficiency   \eta =\frac{W_{net}}{Q_{s}}

      \eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}

5 0
4 years ago
Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1
babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A     ⇒    P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E)    ⇒  ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒  ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0
4 years ago
calculate how much black eyes seeds are necessary to plant a 6- hectare( 14.425 acres) field. given that the weight of 1000 blac
11111nata11111 [884]

Answer: 1.38g

Explanation:

Width of planting area = 100m

Field size = 6-hectares(14.425 acres)

Weight of 1000 black eye seed = 230g

1 lb = 453.4g

1 black eye seed = 230g/1000 = 0.23g = 0.00023kg

1 hectare = 10,000sq metre

6 hectare = 60,000sq metre

(Weight/Area) kg/m2

0.00023kg / 10,000 = 2.3×10^-8kg/m^2

And field size = 6hectares = 60,000m^2

(2.3×10^-8) × 60,000 = 0.00138kg of black eye seed

0.00138kg × 1000 = 1.38g

6 0
3 years ago
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