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3 years ago

A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?

Engineering

1 answer:

Illusion [34]3 years ago

3 0

Answer:

0.0659 A

Explanation:

Given that :

$I_{0} = 6nA$ ( saturation current )

at 25°c = 300 k ( room temperature )

n = 2 for silicon diode

Determine the saturation current at 100 degrees = 373 k

Diode equation at room temperature = I = Io $\frac{V}{e^{0.025*n} }$

next we have to determine the value of V at 373 k

q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1

Given that I is constant

Io = $\frac{e^{0.025*2} }{31.08}$ = 0.0659 A

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Carbon dioxide contained in a piston–cylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volu

Lunna [17]

Answer:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$

Explanation:

For this case we know the following info :

$V_i = 0.3 m^3$ represent the initial volume

$V_f = 0.1 m^3$ represent the final volume

We know that the pressure and volume are related with the following expression:

$P = aV^{-2}= \frac{a}{V^2}$

Where a is a constant given $a = 6.5 Kpa m^6$

And we need to calculate the work associated to this process.

We have a compression, and by definition the work is defined with the following general expression:

$W = \int_{V_i}^{V_f} P dV$

If we replace the expression for P we got:

$W = \int_{V_i}^{V_f} a V^{-2} dV$

If we integrate we got:

$W = -a (\frac{1}{V}) \Big|_{V_i}^{V_f}$

Using the fundamental theorem of calculus we have:

$W = -a (\frac{1}{V_f} -\frac{1}{V_i})$

And replacing the values we got:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$

5 0

4 years ago

What has alternating electric and magnetic fields that travel in the form of a wave?

ad-work [718]

Answer:

a electromagnet

Explanation:

8 0

3 years ago

Calculate the scale and speed of the pattern in order to gain useful results for a turbine operate at 150 rev/min at height diff

Rzqust [24]

Answer:

first mark me as a brainleast

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3 years ago

Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: L

Travka [436]

Answer:

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Explanation:

Load 1: Active power P_1 = 20 HP = 14.91kW;

Reactive power $Q1 = P tan(\phi)$

$= 14.91\times tan(cos^{-}0.8) = 11.18 kvar
$

Load 2: Active power P_2 = 20 kW;

Reactive power Q2 = 0 since the load is purely resistive.

Load 3: Active power $P_3$ = 0 due to purely capacitiveload

Reactive power $Q_3$ = -20 Var

a) since all three loads are connected in parallel therefore

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar

Since Q = 0, the power factor is unity.

Supply current per phase is given by

$I = \frac{P}{\sqrt{3}V_{L}}$

$= \frac{34910}{\sqrt{3}\times 480} = 41.99 A$

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3 years ago

Technician A says that for any fuel to burn efficiently, it must be vaporized and fully mixed with the proper amount of air. Tec

Ira Lisetskai [31]

Both technicians are correct

Explanation:

They are each correct for separate reasons. Tech A is correct due to pressure and temperature both being important factors and quantities to vaporize anything. Tech B is correct due to fuel particles being easier to burn due to their chemical structure.

8 0

2 years ago