Current Division...... URGENT!

[MMK]

Q:- The resistances of commercially-available discrete resistors are restricted to particular sets. For example, the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from 100 through 105. The E12 set is:

E12={10,12,15,18,22,27,33,39,47,56,68,82…}

Thus, you can buy 10% resistors with a nominal resistance of 330Ω or 33kΩ, but not 350Ω. Furthermore, the "tolerance" means that if you buy a 10% 390Ω resistor you can be sure that its resistance is between 351Ω and 429Ω.

In this problem we need to choose 10% resistors to make a current divider that meets a given specification.

 

Image (Circuit Diagram) could be found out at: http://tinyurl.com/d7yv99y

We are given an input current Iin=100μA and a linear resistive load represented by the resistor RL . Our resistive load is not regulated for the high current provided by our current source, so we add a resistor RS in parallel to divide the current such that IL≈40μA. An additional requirement is that the Thevenin resistance as seen from the load terminals is between 60kΩ and 80kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors RS and RL such that the divider ratio IL/Iin is within 10% of the requirement.

Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal value, what is the largest and smallest value that VL may have?

Find: Rs, RL, VL(max) and VL (min)........

[SandeepReddy]

Im too stuck at this problem. please help needed !!!

Actually, Plugging in the values independently is not helping there, First carry out your procedure and calculate the values Rs, RL, VL(max) and VL (min) and plug them all at once to see actual output.

If you plug values Rs, RL then you would get cross even if they are correct!.

Dont know why!

[imatfaal]

!

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You need to show us what you have done towards the problem - then perhaps some of the membership will be willing to help.

 

[MMK]

Mod I havent seen a single post which will help us in solving this question.........