Why Is the Experimental R_eq Value Higher Than Theoretical?

Click For Summary
The experimental equivalent resistance (R_eq) value of 8.94 ohms is higher than the theoretical value of 8.33 ohms due to potential sources of error in the measurement process. Instrumentation inaccuracies and the inherent tolerance of the resistors, which can vary by ±7%, may contribute to this deviation. Additionally, the resistance of other circuit components, often assumed negligible, could introduce extra resistance. The close y-intercept of the regression line at 0.004 V suggests minimal systematic error. Overall, these factors can explain the discrepancy between experimental and theoretical R_eq values.
Soaring Crane
Messages
461
Reaction score
0

Homework Statement



On my potential vs. current graph for resistors in parallel, the slope of the regression line is 8.94 ohm (V/I = R) for a 10 ohm resistor and a 50 ohm resistor. The theoretical value of R_eq is R_eq = 1/[(1/50 ohm) + (1/10 ohm)] = 8.33 ohm. Upon comparing the two R_eq values, I was wondering why the experimental R_eq value is higher than theoretical value. What sources of error could account for this deviation? (Oh, the regression line's y-intercept value is close to 0; it is 0.004 V.)


Homework Equations



See above.

The Attempt at a Solution



See above.

Thanks.
 
Physics news on Phys.org
Whats the error on your instruments? Whats the error on your resistors? You answer is with \pm 7%, which is pretty good if your using high school equipment.
 
The other wires in the circuit have a small resistance which is usually assumed to be negligible. This could account for some extra resistance. Also, the resistors are labeled 10 and 50 ohms, but could be higher or lower based on the percentage of error (bottom line on the resistor)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 2 ·
Replies
2
Views
27K
Replies
3
Views
13K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
11K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K