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ANALOG DEVICES Placement question.

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asp1987

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Hi,
Today I wrote a placement test for ANALOG DEVICES. Did not qualify though. Here are some of the questions they asked.

1) Find the resistance of the frustum of a cone( its hollow) made with a material of uniform thickness and resistivity 'rho' or lets say 'Z'. The radii of the smaller and larger flat bases are 'L' and '2L' respectively. Height = 'L'. The two flat bases are covered with a zero resistance material. The resistance is measured between two terminals, one lead at the centre of the smaller flat base and the other lead at the centre of the larger flat base.

Also this frustum, a capacitor and a dc voltage source are connected in series. Find an expression for capacitor voltage.

Please help me solve these.
Is this the right forum?
 
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Willbe

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Find the resistance of the frustum of a cone made with a material of uniform thickness and resistivity 'ro'. lets say 'Z'. The radii of the smaller and larger flat bases are 'L' and '2L' respectively. Height = 'L'. The two flat bases are covered with a zero resistance material. The resistance is measured between two terminals, one lead at the centre of the smaller flat base and the other lead at the centre of the larger flat base.
They must be joking. . .

You do have to know the formula for the volume of a conic frustrum or be able to calculate it on the spot.

Z = ro*L/A, where A is the cross-sectional area of a volume of length L and ro is the volume or bulk resistivity, ρ [rho], of the material. Z is resistive, not reactive, since ρ is resistive.

The volume of this conic frustrum, V, equals (PI*L/3)*(4*[L^2] + 2*[L^2] + [L^2])

Assume this conic frustrum is a cylinder of length L and average cross-sectional area A, where A*L = V. So, A = V/L.

Therefore, I believe that Z = ro*[L^2]/V.


Once you have Z, if V is a DC voltage source and C is the value in farads of a capacitor and Z is the value in ohms of the R from above the voltage across the capacitor Vc = V*{1-e^(-t/[Z*C])}; again, Z is resistive.

These are probably questions from a PE exam for electrical engineers.
 

asp1987

New Member
The volume of this conic frustrum, V, equals (PI*L/3)*(4*[L^2] + 2*[L^2] + [L^2])
Could u explain how u got this value? The approach i took was to apply the same expression R=ρL/A. But instead of calculating volume, i tried to calculate its surface area by subtracting that of a small cone from a larger cone so as to obtain the given frustum. Whats the problem in this method?
The second part- i derived the answer taking resistance simply as 'R'. :eek:
N do u feel it was a silly question? :confused:
 
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Willbe

New Member
Could u explain how u got this value?
From Reference Data for Radio Engineers, Howard Sams & Co, 1981, page 46-3.
It's probably on the Internet, somewhere.

To check it I'd put the formula into a spreadsheet and plug in integer values; if you get the same answer you must have done all the steps correctly.

You could also calculate the Z for a cylinder of radius L and check that it is larger than for the conic frustrum.

I thought at first it was an off-the-wall question, but the more I thought about it AD probably does a lot with bulk resistivity and geometric shapes and interelectrode capacitance [which determines risetime which detemines speed].

In retrospect, I would have liked to work for these guys. . .
 
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