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2 years ago

The p-type or N-type semiconductor is not of practical use. Why?

Physics

1 answer:

Savatey [412]2 years ago

6 0

Explanation:

The p-type semiconductor doesn't have greater energy than the n-type semiconductor. This is because the semiconductor itself doesn't really have an energy. Charge carriers (electrons/holes) have energies. The relative positioning of the bands tell you how much energy an electron or a hole must have for it to exist.

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Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places

Olenka [21]

Answer:

The magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage $V_{out} = 17.33$

Resistance $R_{1} = 3$ kΩ

Voltage gain $A_{v} = 4.33$

For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

$A_{v} = 1+\frac{R_{f} }{R_{1} }$

$4.33 = 1+ \frac{R_{f} }{3 k }$

$R_{f}$ ≅ 10 kΩ

For finding input voltage we use,

$A_{v} = \frac{V_{out} }{V_{2} }$

$V_{2} = \frac{17.33}{4.33}$

$V_{2} = 4$ V

The Phase of $V_{2}$ is zero because output voltage phase is 360°

Therefore, the magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

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3 years ago

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sergiy2304 [10]

Answer:

A kilowatt (kW) is a unit of power.

Explanation:

The power of an object is given by :

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Here,

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The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.

Option (1) : A kilojoule (kJ) is a unit of power is incorrect.

Option (2) : A gigawatt (GW) is a unit of energy is incorrect.

Option (3) : A watt (W) is a unit of energy is incorrect.

Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.

Option (4) : A kilowatt (kW) is a unit of power is correct.

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Answer:

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3 years ago

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3 years ago

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Answer:

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Explanation:

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3 years ago

The p-type or N-type semiconductor is not of practical use. Why?​

The p-type or N-type semiconductor is not of practical use. Why?