Engineering Resolve Parallel Resistor Error: I Set Rtot = 0.168

[amwil]

Homework Statement

Two resistors, R1 and R2, are connected in parallel to a power supply that has voltage V and negligible internal resistance. R2 = 9.00 Ω and the resistance of R1 is not known. For several values of V, you measure the current I flowing through the voltage source. You plot the data as I versus V and find that they lie close to a straight line that has slope 0.168 Ω^−1.

Relevant Equations

V = IR
1/Rtot = 1/R1 +1/R2

Since the slope of the I versus V graph is equal to 1/R, I set Rtot equal to 0.168. So then I subtracted 1/0.168 by 1/9 and got 1/R1 =5.841. I then found R1 equal to 1/5.841 or 0.171 but it said I was wrong. Can someone tell me what I did wrong?

 

[Baluncore]

1/Rtot = 1/R1 +1/R2
Can someone tell me what I did wrong?

0.168 Ω^−1. is a conductance = I / V.
V / I = 1 / 0.168 = 5.952 Ω

1 / ( 1 / 5.952 - 1 / 9 ) = 17.575 Ω
1 / ( 0.168 - 1 / 9 ) = 17.578 Ω

 

[DaveE]

Try again, but everytime you write down a number include it's units. As @Baluncore said it's not 0.168, it's 0.168⋅(1/Ω) = 0.168⋅(A/V). This practice will help in future problems too.

So, in your solution you suggest 1/(9V/A) - 1/(0.168A/V), this doesn't make sense because the units don't match.