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abruzzese [7]
3 years ago
7

Consider the null-type Wheatstone bridge. Suppose 3 4 R R =   =   100 2 , 200 2 , R2 is variable calibrated resistor, and 1

R =   100 1 is the transducer resistance. Calculate the means and uncertainty of R2 that is required to balance the bridge?

Engineering
1 answer:
Lilit [14]3 years ago
4 0

Answer:

Means of R2 is 200 ohms

The uncertainty is 8

Explanation:

Firstly, please check the first attachment for the complete version of the question including the diagram of the Wheatstone bridge. Thanks!

In this question, we are asked to calculate the means of the resistor R2 and its uncertainty required to balance the bridge.

Please check attachment for complete solution.

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A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal
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Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

  V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X_{L}, which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX_{L}

X_{L} = \frac{V}{I} ................ 2

Given that; V = 240V, f = 50Hz, \pi = \frac{22}{7}, I = 15A, so that;

From equation 2,

X_{L}= \frac{240}{15}

    = 16Ω

To determine the inductance of the inductor,

X_{L} = 2\pifL

L = \frac{X_{L} }{2 \pi f}

  = \frac{16}{2*\frac{22}{7}*50 }

 = 0.05091

The inductance of the inductor is 0.051H.

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A mercury thermometer has a cylindrical capillary tube with an internal diameter of 0.2 mm. If the volume of the thermometer and
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To solve this problem we will proceed to calculate the specific volume from the area of the cylinder and the sensitivity. Later we will calculate the volumetric coefficient of thermal expansion and finally we will be able to calculate the volume through the relation of the two terms mentioned above. Our values are

\text{Sensitivity}= 2mm/\°C

\text{Internal diameter } d= 0.2mm

\text{Differential expansion of Hg } \lambda_L = 1.82*10^{-4}/\°C

Let's start by calculating the specific volume which is given by

v = \pi (\frac{d}{2})^2 \gamma

Here,

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\gamma = Sensitivity

Replacing our values we have

v = (\frac{\pi}{4})(0.2mm)^2(2mm/\°C)

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Now we will obtain the value of the volumetric coefficient of thermal expansion of mercury through the differential expansion coefficient of Hg whic is three times, then

\lambda_V = 3\lambda_L

\lambda_V = 3(1.82*10^{-4}/\°C)

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Finally the relation to calculate the volume the bulb must is

\text{Specific volume} = \text{Bulb Volume} \times \text{Volumetric Coefficient}

v = v_B \times \lambda_V

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Therefore the volume that the bulb must have is 115mm^3

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