Calculating Resistance of Y with X & Variable Resistor

[CAH]

I've got two resistors in parallel (X and Y) and a variable resistor, ammeter and voltmeter. SEE PHOTO!

I've calculated the value of X by measuring V and I when the switch is open. I can close the switch and measure V and I but how do I calculate the resistance of Y?

I know 1/x + 1/y = 1/R, this may be a stupid question? When I do 1/R - 1/x (where x is known from previous calculation) I get a negative number.

Thanks

 

[Alettix]

Are you sure 1/R - 1/x gives a negative number? The total resistans of two parallel resistors should always be less than the resistans of the individual ones.

How do you use the variable resistor in your measurments and calculations?

 

[CAH]

I varied the resistor to get values of I from 10mA, 20, 30, 40, 50mA and recorded voltage for each one and then across the parallel I did 10mA, 30, 50, 70, 90mA and recorded V then drew the graphs and used the gradient for resistance of each! :)

 

[Drakkith]

Since R is less than X, 1/R will be greater than 1/X and you shouldn't be getting a negative number. Set your variable resistor so that you have 10 mA of current with the switch open. Then, without changing the variable resistor close the switch and check your ammeter. What is your current at now?

 

[Alettix]

I varied the resistor to get values of I from 10mA, 20, 30, 40, 50mA and recorded voltage for each one and then across the parallel I did 10mA, 30, 50, 70, 90mA and recorded V then drew the graphs and used the gradient for resistance of each! :)

I can't see what's wrong here, but I agree with Drakkith: the total resistance of two resistors in parallel is always less than any resistance of the individual ones, thus 1/R>1/x .
Are you sure you haven't done some little misstake, such as plotting I(U) instead of U(I) or exchanging the two series of measurements? :)

 

[CAH]

Thanks I made a ridiculously stupid mistake: I was doing 1/Rx - 1/Rt insead of 1/Rt - 1/Rx!