Solutions To Brain Teasers Number 2

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:

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:

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.

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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Factors Upon Which The Resistance Of Conductor Depends

Let's continue our discussion with our basic concepts here in Electrical Engineering. I guess most of you are familiar with this topic but this should not be neglected because it has a big role during board exam of Electrical Engineering.

What is the topic for today?

Let's begin....

We all know that Georg Simon Ohm who formulated the law that the resistance of a conductor varies directly with its length, inversely with its cross-sectional area, and depends upon the material of which it is made.

From the study of resistors in series, one would expect that the resistance of a piece of uniform wire is directly proportional to its length, since it can be thought of as a series of small pieces of wire whose total resistance is the sum of the resistances of the individual pieces.

Let's have an example. Consider a wire 1 ft in length and having a cross- sectional area of 0.3 in^2. By thinking of this as equivalent to three wires ( 1 ft in length) each having a cross-sectional area of 0.1 in^2 connected in parallel, we may infer that

1/R = 1/R1 + 1/R2 + 1/R3

or since R1 = R2 = R3,

1/R = 3/R1 and R1 = 3R

showing that the resistance of one of the small wires is three times as great as that of the large wire. This suggests that the resistance of a wire is inversely proportional to the cross-section, a fact that was verified experimentally by Ohm.

Using R varies as l and R varies as 1/A, as mentioned above, we can write R varies as l/A where l is the length and A the cross sectional area of a uniform conductor. This relation can be written in the form of equation

R = p l/A

where p is a quantity, characteristics of the material of the conductor, called the resistivity of the substance. The term specific resistance is sometimes used instead of resistivity.

From the equation above:

p = RA/l

If A and l are given values of unity, it is seen that p is numerically equal to the resistance of a conductor having unit cross section and unit length.

If R is in ohms, A in square centimeters, and l in centimeters, then p is in ohm-centimeters. This unit is somewhat more convenient than the mks unit the ohm-meter.

Conductance and Conductivity

Since the reciprocal of the resistance, 1/R occurs often in parallel circuits, it is frequently convenient to designate this concept as the conductance of the resistor. The symbol used for conductance is G, and the unit is mho. In a parallel circuit the total conductance is given by G = G1 + G2 + G3. Less often the reciprocal of resistivity 1/p is used, and this concept is called the conductivity of the material. The symbol for conductivity is o (not exactly the symbol) and the unit is mho/cm.

Change of Resistance With Temperature

The electric resistance of all substances is found to change more or less with the changes of temperature. Three types of changes are observed. The resistance may increase with increasing temperature. This is true of all pure metals and most alloys. The resistance may decrease with increase of temperature. This is true of a semiconductor like carbon and of glass and many electrolytes. The resistance may be independent of temperature. This is approximately true of many special alloys, such as manganin ( Cu 0.84, Ni 0.12, Mn 0.04).

Experiments have shown that, for moderate temperature range, the change of resistance with temperature of metallic conductors can be represented by the equation.

Rt = Ro + Ro oo t = Ro( 1 + oo t) -----------------EQ. (1)

where Rt is the resistance at temperature t, Ro is the resistance at 0 degree celcius, and oo is a quantity characteristic of the substance and known as the temperature coefficient of resistance. The defining equation for oo is obtained by solving Eq 1, giving,

oo = Rt - Ro/ Rot ----------------------------------EQ.(2)

The temperature coefficient of resistance is defined as the change in resistance per unit resistance per degree rise in temperature, based upon the resistance at 0 degree celcius.

Although Eq 1 is only approximate, it can be used over medium ranges of temperature for all but very precise work.

Since Rt - Ro and Ro have the same units, their units will cancel in the fraction in Eq 2. Hence, the unit of oo depends only upon the unit of t. For instance, for copper oo = 0.004/C, but only 5/9 x 0.004/F.

For clear understanding of the principles above. I will show you on my next post some illustrative problem solving. For you to review in advance I will show it now.

A Teaser

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?

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

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?

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?

I will going to reveal the solutions on my next post with additional discussions again for this topic here in Learn Electrical Engineering for Beginners.

Cheers!

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