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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 330O or 33kO, but not 350O. Furthermore, the "tolerance" means that if you buy a 10% 390O resistor you can be sure that its resistance is between 351O and 429O. In this problem we need to choose 10% resistors to make a current divider that meets a given specification. 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 60kO and 80kO. 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 IL may have? Enter your values below. (NOTE: Both resistor values are needed for a correct answer.) RS (in Ohms): RL (in Ohms): ILmax (in Amps): ILmin (in Amps):
Find the resultant of two vectors defined as A=6km east B=8km north
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Q. Collector current I c =20 mA and base current I b = 50 A. Find current gain β of a transistor. β = I C / I B = 20 X 10 -3 / 50 X 10 -6 = 400
give applications of resonance
If small test charge is taken as positive than why measured value of electric intensity be somewhat less than the actual value of elctric intensity and vica versa
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