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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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Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e
a_sh-v [17]

Answer:

Outside temperature =88.03°C

Explanation:

Conductivity of air-soil from standard table

   K=0.60 W/m-k

To find temperature we need to balance energy

Heat generation=Heat dissipation

Now find the value

We know that for sphere

q=\dfrac{2\pi DK}{1-\dfrac{D}{4H}}(T_1-T_2)

Given that q=500 W

so

500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)

By solving that equation we get

T_2=88.03°C

So outside temperature =88.03°C

6 0
4 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat
Yuki888 [10]

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2  (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

4 0
4 years ago
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