6.7 Example from Office of Academic Technologies on Vimeo.

**6.07 Example**

Now, let’s do an example associated with the resistance of an object with a certain resistivity. Let’s assume that we have a truncated right circular cone shape of object. In other words, it is something like this. A resistor in this shape.

Let’s give some dimensions to this resistor. The radius over here is *a*, and the radius over here is *b*. Also, let’s say the length of this resistor is equal to *L*. Let’s assume that the taper is very small, so that we can assume the current density is uniform across any cross-section.

Now, we’d like to determine first the resistance of this object, of course, in terms of these given quantities. Also, in the second part of the problem, we’d like to show that the answer reduces to *R* is equal to *ρ* times *L* over *A*, length divided by the cross-section of the area, for this special case of 0 taper.

So, we have a resistor of this shape, a truncated cone, with the radius at this end is equal to *a*, and the radius on the other end of the tapered corner is *b*, with length *L*. And this object has the resistivity *ρ*, which is also given, (resistivity of the medium), and we’d like to determine the resistance of this object.

Well, as it is given over here, we have seen that the relationship between the resistance and resistivity was such that resistivity times the length of the object, divided by its cross-sectional area gave us the resistance of that object. Of course, this expression is true if the cross-sectional area of the object remains constant along its length. Whereas in our case, we don’t have that situation. In other words, as we move along *x*-axis, the cross-sectional area of the object changes. And then, arbitrary location, let’s say somewhere over here, we have the cross-section, which will have a certain radius. Let’s call that one as *y*. As this distance *x* away from the origin, both in terms of this coordinate system.

So, we’d like to figure out, or write down an expression for the radius of the medium for this specific *x* distance. We can’t easily do that since this is a tapered cone. If you look at over here, that we have this line, and that line equation will indeed give us how *y* is varying (*y*-coordinate is varying) with respect to *x*-coordinate. Now, if you recall line equation from geometry, and that is *y* is equal to *mx* plus the intercept — and let’s denote the intercept by *c* over here, in order to avoid confusion with this radius *b*. So, here it is the intercept of the line and here *m* is the slope of the line.

Well, in our case, we see that our conditions are such that at *x* is equal to 0, which is at the origin, corresponding *y* is equal to *a*. On the other hand, at *x* is equal to *L*, which is the length of this cone, corresponding *y* is equal to *b*. So if we use these conditions in our general line equation — because our purpose is to determine these constants, the slope, and as well as the intercept of that line — the first condition (let’s mark this as the first condition and this one as the second condition) is going to give us *y* is equal to *a*, when *x* is equal to 0, so *m* times 0 plus *c*. So this is going to give us just 0. From here, we will see that the intercept will be equal to *a*.

The second condition says that when *x* is equal to *L*, *y* is equal to *b*. Therefore, we can write this condition as *b* is equal to *m* times, for *x* we will substitute *L*, and we have already determined what the intercept is, and that is *a*. So *b* is equal to *ml* plus *a*. From here we can write down the slope of this line as *b* minus *a* over *L*.

Now we know the equation, how *y* is changing with respect to *x*. Therefore, we can write down the general equation as *y* is equal to the slope, and that is *b* minus *a* over *L*, times *x* plus the intercept, which is equal to *a*.

Okay. If we therefore, go to our truncated cone, what we can do is we can choose an incremental length of this cone. Let’s say, such that the thickness is *dx*. And for this length, the radius is *y*, and this *dx* is so small, we can assume that we will have the same radius over here. Then we can express the resistance associated with this unit as the incremental resistance, *dR*. And once we determine the resistance of that unit, then we can go to the next segment with a length of *dx* and calculate the resistance of that unit. And do this throughout this cone, and eventually, once we obtain all those incremental resistances, just add them along this length to be able to get the overall resistance.

So, then the incremental resistance, *dR*, is going to be equal to resistivity of the medium, times an incremental length of *dx* divided by the corresponding cross-sectional area. And since cross-sectional area will be the area of this circle, and that will be *π* times its radius squared. And the radius is the corresponding *y* over here. And the square of *y* (let’s just write it down as *y* squared) and in expanded form. Therefore, this is going to be equal to *ρ* times *dx* divided by *π*, times square of *y*, which is *b* minus *a* over *L*, *x* plus *a* quantity squared.

Now, to be able to get the total resistance, then we will add all these incremental resistances to one another, and the addition process over here is the integration. So if you take the integral of both sides, then we will end up with the total resistance of the medium. And of course, our variable over here is *x*. The boundaries of the integral will be by adding all these incremental cross-sectional regions with length of *dx*. Therefore, *x* is going to vary from 0 to *L*. Therefore, the boundaries will go from 0 to *L*.

Okay. Now, to be able to take this integral, we will say that net of the quantity in this square bracket in the denominator, *b* minus *a* over *L* times *x* plus *a*, *b* equal to *u*. Then, *b* minus *a* over *L* *dx* is going to be equal to *du*. Since *a* is constant, the derivative of that will give us just 0. Therefore, if we multiply both numerator and denominator *b* minus *a* over *L*, then nothing will change, let’s do that. *R* will be equal to integral and let’s also take *ρ* and *π* outside of the integral since they are constant, resistivity and *π*. Therefore, we’ll have *ρ* over *π*, and we’re going to multiply both numerator and denominator by *b* minus *a* over *L*.

And now, let’s do it like this. *b* minus *a* over *L*, multiply numerator and divide also by the same quantity. And I’m going to write that part outside of the integral since this is a constant quantity. I will have *b* minus *a* over *L*, *x* plus *a* quantity squared in the denominator, integrated from 0 to *L*. Therefore, in this notation, the quantity in the numerator is equal to, in terms of this new variable *du*, and quantity in the denominator is just *u* squared. Which we can easily take this integral *du* over *u* squared, which will be equal to minus 1 over *u*.

So, if we move one further step, here, we can write this part as *ρ* times l over *π* times *b* minus *a*. And the integral of *du* over *u* squared is going to give us -1 over *u*. And that, is therefore, will be minus 1 over, *u* was defined as *b* minus *a* over *L*, *x* plus *a*, closed parenthesis, and we’re going to evaluate this at 0 and *L*.

Now moving on, *R* will be equal to *ρ* times *L* over *π* times *b* minus *a*. We will substitute *L* for *x* first. So we will have -1 over *b* minus *a* over *L*, times *L* for *x*, plus *a*. And then minus — now we will substitute 0 for *x* — minus and this minus will make plus 1 over, if we substitute 0 for x, 0 times this quantity will give us 0. So we’ll only end up with *a*. These *L*‘s will cancel and if you look at the denominator then, we will have *b* minus *a* plus *a*. Then those *a*‘s will cancel.

Therefore, we will have *R* is equal to *ρL* divided by *π*, times *b* minus *a*. And so we’re going to end up minus 1 over *b* plus 1 over *a*. Having common denominator, we will multiply this ratio by *b*, numerator and denominator. And this ratio by *a*, numerator and denominator, so we will have *b* minus *a* over *a* times *b*. *b* minus *a* in the numerator over here will cancel *b* minus *a*, the one in the denominator. And finally we’re going to end up with *R* is equal to *ρ* times *L* over *π* times *ab*.

So this will give us the resistance of this truncated cone in terms of its length and the radii from both ends, *a* and *b*. Now, in the second part of the problem, if the taper were 0, taper is equal to 0, then the radii at both ends of this cone will be equal to one another. If there’s no taper, then they will both have the same radius, then *a* is going to be equal to *b*.

In that case, *a* times *b* will be equal to *a* times *a*, which will be equal to *a* squared. Our equation, then will take the form of *r* is equal to *ρ* times *L* over *π* *a* squared. But *π* *a* squared is nothing but the cross-sectional area of the resistor. Therefore, *R* will be equal to *ρ* times *L* over *a*. And this is the expression that we obtained earlier for a conductor such that it’s cross-sectional area is the same along its length. Therefore, for this case, our expression reduces to that original form.