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dem82 [27]
3 years ago
11

How do we find percentage error in measuring voltage across a resistor​

Engineering
2 answers:
Black_prince [1.1K]3 years ago
6 0

Answer:

  use the percentage error relation

Explanation:

The percentage error in anything is computed from ...

  %error = ((measured value)/(accurate value) -1) × 100%

__

The difficulty with voltage measurements is that the "accurate value" may be hard to determine. It can be computed from the nominal values of circuit components, but there is no guarantee that the components actually have those values.

Likewise, the measuring device may have errors. It may or may not be calibrated against some standard, but even measurement standards have some range of possible error.

maksim [4K]3 years ago
3 0

Answer:

use the percentage error relation

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Briefly describe the function of the thermostatic expansion valve in a vapour compression refrigeration system
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Answer:

Explanation:

Thermostatic expansion valve is mainly a throttling device commonly used in air conditioning systems and refrigerators.

It is an automatic valve that maintains proper flow of refrigerant in the evaporator according to  the load inside the evaporator. When the load in the evaporator is higher the valve opens and  allows the increase in flow of refrigerant and when the load reduces the valve closes a bit and  reduces the flow of refrigerant. This process leads to higher efficiency of compressor as well as the whole refrigeration system.  Thus TEV works to reduce the pressure of refrigerant from higher condenser pressure to the lower evaporator pressure. It also keeps the evaporator active.      

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On aircraft equipped with fuel pumps, when is the auxiliary electric driven pump used?.
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2 years ago
The AGC control voltage: ___________
lyudmila [28]

Answer:

The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.

Explanation:

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2 years ago
What can be the main disadvantage of pulse amplitude modulation?​
Feliz [49]

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2 years ago
A small hot surface at temperature Ti-430K having an emissivity 0.8 dissipates heat by radiation into a surrounding area at T2-4
Nat2105 [25]

Answer:

389.6 W/m²

Explanation:

The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K

So, P = σεA(T₁⁴ - T₂⁴)

h = P/A = σε(T₁⁴ - T₂⁴)  

Substituting the values of the variables into the equation, we have

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴

h = 38955213360 × 10⁻⁸ W/m²

h = 389.55213360 W/m²

h ≅ 389.6 W/m²

5 0
2 years ago
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