The circuit is made of a variable resistor of 10kΩ value and two fixed resistors R1 and R2 .The output voltage V0 should be variable from 2V to 9V, i have to calculate the values of the two fixed resistors R1 and R2. I don't know how to start.
Since this is a homework question, so the battery should be perfect with no internal resistor and the variable risistor should range from 0 to 10k. This come to these solution steps:
1, The V0 = 2V when the variable resistor is at zero, hence the equation: 2V = 12V*R2/(R1+R2+10k)
2, The V0 = 9V when the variable resistor is at its max, hence the second equation: 9V = 12V*(R2+10k)/(R1+R2+10k)
Hope this will help.
Start by assuming that the output current is approx zero, and think of the circuit as a pure voltage divider.
Since the extreme values are 2V and 9V, you can analyse each of those cases individually. Don't think of the pot as a pot, but just as a resistor where Vo is connected to one end or the other, and figure out the voltage-divider behaviour:
2 = 12*R2/(R1+R2+10k)
9 = 12*(R2+10k)/(R1+R2+10k)
That gives you a pair of simultaneous equations you can solve for R1 and R2.