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Lera25 [3.4K]
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
Engineering
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

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kap26 [50]

Answer:

Certamente você conhece três dimensões: comprimento, largura e profundidade. Além disso, quando se pensa um pouco fora da caixa também seria possível adicionar a dimensão do tempo.

Provavelmente, algumas pessoas viajam na maionese quando toca-se nesse assunto. Vem em suas mentes universos paralelos e até mesmo realidades alternativas. Mas também não se trata disso.

Explanation:

Basicamente as dimensões são as facetas do que nós percebemos a ser realidade. Existem muitos debates sobre dimensões na física. Um dos que mais chamam a atenção se chama Teoria das Cordas.

r

5 0
3 years ago
A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. W
irakobra [83]

Answer:

P_p = 27000 psf

Explanation:

given,

height of the retaining wall = h = 12 ft

internal angle of friction (∅)= 30°

unit weight = 125 pcf

Rankine passive earth pressure = ?

k_p is the coefficient of passive earth pressure

k_p = \dfrac{1 + sin\phi}{1 - sin\phi}

k_p = \dfrac{1 + sin30^0}{1 - sin30^0}

k_p = 3

Passive earth pressure

P_p = \dfrac{1}{2}k_p \gamma H^2

P_p = \dfrac{1}{2}\times 3\times 125 \times 12^2

      P_p = 27000 psf

Rankine passive earth pressure on the wall is equal to P_p = 27000 psf

7 0
3 years ago
What amount of force is being exerted in the horizontal direction?
ZanzabumX [31]

Explanation:

Use trigonometry.  The horizontal force is:

Fₓ = F cos θ

Fₓ = 20 N cos 60°

Fₓ = 10 N

7 0
4 years ago
Consider a solid circular shaft subjected to bending and torsion so that the state of stress of interest involves only a normal
Alex17521 [72]

Answer:

The detailed explanation of answer is given in attached file.

Explanation:

7 0
4 years ago
Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton
SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude q_{1},q_{2} the magnitude of force between them is given by

F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}

where

k_{e} is constant

r is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}

Applying value of the constant we get

F_{1}=\frac{62.208}{0.7^{2}}

Thus F_{2}=126.955Newtons

5 0
3 years ago
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