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3 years ago

6

Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o

f this silicon piece? For an applied voltage of 5 V, how much current would flow though it?

1 answer:

lukranit [14]3 years ago

7 0

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.

The resistance R of a given resistor, is expressed by the following formula:

R = ρ L / A

Replacing by the values for resistivity, L and A, we have

R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2

R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA

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Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch

Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 $\frac{ft^{3}}{sec}$

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4 0

3 years ago

zener shunt regulator employs a 9.1-V zener diode for which VZ = 9.1 V at IZ = 9 mA, with rz = 40 and IZK= 0.5 mA. The available

gulaghasi [49]

Answer:

$V_z=9.1v$

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$R=589 ohms$

Explanation:

From the question we are told that:

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Zener diode Current $I_z=9 .A$

Note

$rz = 40\\\\IZK= 0.5 mA$

Supply Voltage $V_s=15$

Reduction Percentage $P_r= 50 \%$

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$V_z=V_{z0}+I_zr_z$

$V_z=8.74+(10*10^{-3}) (40)$

$V_z=9.1v$

Generally the equation for Kirchhoff's Current Law is mathematically given by

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3 years ago

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Dovator [93]

Explanation:

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True

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True

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d. The output of an ideal half wave rectifier is zero volts only when the input is zero volts.

False

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e. A turn-on voltage of a diode (y,) greater than zero can cause the output of a full wave rectifier to be zero volts even when the input is not zero volts.

False

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f. An advantage of the half wave rectifier is that is can use a smoothing capacitor, while a full wave rectifier cannot.

False

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8 0

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