The voltage at terminal B of the potentiometer can be set toany value between 0V

Question

The voltage at terminal B of the potentiometer can be set toany value between 0V and 10V. We now want to modify this circuit sothat the voltage at terminal B of the potentiometer will rangebetween 3V and 8V (using the same 10V supply). This is done byadding appropriate fixed resistors in series between terminal A andthe power supply, and in series between terminal C and ground.Using knowledge of voltage division, determine the values of R1 andR2 to obtain the modified range. Sorry, i didnt want to have to remake thepicture, but terminal B and C need to be switched in thepicture.

Explanation / Answer

Okay, In the given Diagram there are 3 resistors.(2 fixed, 1variable between 0 and 20k). The measured voltage drop occurs due to 2 resistors : R2 andthe potentiometer. In the given circuit, let V0 be the output voltage and Rp bethe resistance due to the potentiometer.. V0 = 10*(R2+Rp)/(R1+R2+Rp)    = 10* [1 - R1/(R1+R2+Rp)] From the above expression, it is clear that higher Rp meanshigher V0.
Hence., 3V is output when Rp= 0 and 8V is output when Rp=20k
3= 10*[1 - R1/(R1+R2)] 3*R1 +3*R2 = 10*R2 3*R1 = 7*R2 ..... equation 1.
8 = 10* [1 - R1/(R1+R2+20)] 4*R1 +4*R2 +80 = 5*R2+100 4*R1 = R2 +20 using equation 1: R2+20 = (4*7/3) R2 R2 = 2.4k Thus, R1 = 5.6k
I hope you have got your answer.

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