As I promised last time, I will going to show you the solutions to teasers on our previous topic about the Factors Affecting Resistance here in Electrical Engineering. By the way, this topic is still covered on our review of basic Physics for this is very important one when we reached our major topics in this Electrical Engineering course.

Last time, I leaved you four worded problems in order for you to analyze and understand the principles fully. But if you still running out of time to solve it, I will show to you it now.

The first problem given was:

The first problem given was:

**Problem 1 :**The resistance of a copper wire 2, 500 cm long and a 0.090 cm in diameter is 0.67 ohm at 20 degree celcius. What is the resistivity of copper at this temperature?

The solution here is quiet simple. We just have to substitute the given values from the formula since we have uniform units:

p = RA/ l = 0.67 ohm x [ TT ( 0.090 cm) ^2/ 4 ] / 2, 500 cm

= 1.7 x 10 ^-6 ohm.cm - answer

The unit of resistivity in the British engineering system of units differs from that just given in that different units of length and area are employed. The unit of area is the circular mil., the area of the circle 1 mil (0.001 in) in diameter, and the unit of length is the foot. Since the areas of two circles are proportional to the squares of their diameters, the area of a circle in circular mils is equal to the square of its diameter in mils. In this system of units the resistivity of a substance is numerically equal to the resistance of a sample of that substance 1 ft long and 1 circular mil in area, and is expressed in ohm-circular mils per foot.

The abbreviation CM is often used for circular mils. This should not be confused with the abbreviation used for centimeters (cm). We will use the more standard cmil.

Let's solve another problem using the principles above.

Problem 2 : Find the resistance of 100 ft of copper wire whose diameter is 0.024 in and whose resistivity is 10.3 ohm.cmils/ft.

Convert first the given diameter in mils. Since, 1 mil = 0.001 in as mentioned above.

Therefore, d= 0.024 in = 24 mils.

Then getting the area we have, A = d^2 = 24 ^2 cmils. Substituting the values:

R = pl / A = (10.3 ohm.cmils/ft)(100 ft) / 24 ^ 2 cmils.

R = 1.8 ohm - answer

Problem 3 : A silver wire has a resistance of 1.25 ohm at 0 degree celcius and the temperature coefficient of resistance of 0.00375 per degree celcius. To what temperature must the wire be raised to double the resistance?

Since we are asking for the temperature, just derive the formula of EQ . 1 in our previous post here in Electrical Engineering topics. It will be:

**t = Rt - R0 / Ro oo**= ( 2.50 -1.25 ) ohm / 1.25 ohm x 0.00375 /C

t = 266 degree celcius - answer

It should be clearly understood that R0 in the above equation ordinarily refers to the resistance at 0 degree celcius and not to the resistance of any other temperature. A value of oo based upon the resistance at room temperature, for example, is appreciably different from the value based upon 0 degree celcius. This may be made clearer by the graphic analysis of the variation of resistance with temperature. ( Click the image to enlarge)

In the above illustration, the resistance Rt of a conductor at any temperature t is plotted. For a pure metal, this curves gives a linear relation ( approximately). Note the fact that the curve does not pass through the origin; i.e. at 0 degree celcius the resistance is not zero. Hence, we cannot say that R oo t. The slope of the curve delta R/delta t is constant . Since,

oo = delta R / delta t /Ro = slope / Ro

it is clear that the value of oo depends upon the base temperature chosen for Ro. In computations involving temperature variation of resistance, the value of Ro must be obtained by using the equation below:

oo = delta R / delta t /Ro = slope / Ro

it is clear that the value of oo depends upon the base temperature chosen for Ro. In computations involving temperature variation of resistance, the value of Ro must be obtained by using the equation below:

**Rt = Ro + Ro oo t = Ro( 1 + oo t)**

Understanding the principles of effect of temperature in resistance.

**Problem 4 :**A tungsten filament has a resistance of 133 ohm at 150 degree celcius. If oo = 0.0045/C., what is the resistance of the filament at 500 degree celcius?

From the EQ 1: ( of our previous post)

Ro = Rt / 1 + oo t = 133 ohm / 1 + (0.00450/C ) x 150 degree celcius

Ro = 79.4 ohms

Getting the resistance at 500 degree celcius,

R500 = R0( 1 + oo t500)

R500 = 79.4 ohms [ 1 + (0.00450/C) x 500 degree celcius]

R500 = 258 ohms - answer

Since it is the resistivity factor that changes with temperature. The equations of EQ 1 and EQ 2 from our previous lecture may be written with p in place of R.

I will not give the table for resistivities and temperature coefficients of resistivity of materials for it is always given in the problem. You don't have to memorize it.

Since it is the resistivity factor that changes with temperature. The equations of EQ 1 and EQ 2 from our previous lecture may be written with p in place of R.

**pt = p0( 1 + oot) ------------------------------------- EQ. 3**I will not give the table for resistivities and temperature coefficients of resistivity of materials for it is always given in the problem. You don't have to memorize it.

I think this is now over for the basic review....

On my next post, we will now begin to discuss the real scope of Electrical Engineering.

Cheers!

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