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Minchanka [31]
2 years ago
14

The current flowing into the collector lead of a certain bipolar junction transistor (BJT) is measured to be 1 nA. If no charge

was transferred in or out of the collector lead prior to t = 0, and the current flows for 1 min. calculate the total charge which crosses into the collector.
Engineering
1 answer:
Sever21 [200]2 years ago
5 0

Answer:

the total charge which crosses into the collector is 60 nC

Explanation:

Given the data in the question;

current flowing into the collector lead of the bipolar junction transistor (BJT); i = 1 nA = 10⁻⁹ A

no charge was transferred in or out of the collector lead prior to t = 0

the current flow time t = 1 min = 60 sec

Now we write the relation between current, charge, and time;

i = dq / dt

where i is current, q is charge and t is time. { d refers to change }

Now,

q=\int\limits^t_{t=0} {i(t)} \, dt

q=\int\limits^{t=60}_{t=0} { (10^{-9}) } \, dt

q = ( 10^{-9}) (t)_0^{60

q = ( 10^{-9}) ( 60 - 0 )

q = 60 × 10⁻⁹ C

q = 60 nC

Therefore, the total charge which crosses into the collector is 60 nC

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victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
3) Explain how dc machines Can work as motor and generator​
weeeeeb [17]

The working principle of a DC machine is when electric current flows through a coil within a magnetic field, and then the magnetic force generates a torque that rotates the dc motor. The DC machines are classified into two types such as DC generator as well as DC motor.

5 0
1 year ago
A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat
ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000  

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n  ..........1

and

total cost of capital intensive  = $1,244,000 + $92.5 n   ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n  <  $33800 + $143 n

solve we get

n > \frac{906000}{50.5}

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > \frac{1244000}{104.5}  

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0
3 years ago
After replacing a vacuum booster, the brakes lock up on a road test. Technician A says there is air trapped inside the brake lin
vitfil [10]

Answer:

Technician B

Explanation:

The brakes can lockup due to the following reasons

1) Overheating break systems

2) Use of wrong brake fluid

3) Broken or damaged drum brake backing plates, rotors, or calipers

4) A defective ABS part, or a defective parking mechanism or proportioning valve

5) Brake wheel cylinders, worn off

6) Misaligned power brake booster component

5 0
3 years ago
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
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