What is the resistance of the wire and its uncertainty?

In summary, an experiment was conducted to measure the resistance of a wire by calculating the percentage uncertainties of the current and potential difference. The total percentage uncertainty was found to be 25%. The resistance of the wire can be determined using Ohm's law, and the choices for the uncertainty of the resistance are (A) (8.0 \pm 0.2)\Omega, (B) (8.0 \pm 0.6)\Omega, (C) (8 \pm 1)\Omega, and (D) (8 \pm 2)\Omega.
  • #1
looi76
81
0

Homework Statement


An experiment is done to measure the resistance of a wire.
The current in the wire is [tex]1.0 \pm 0.2A[/tex] and the potential difference across the wire is [tex]8.0 \pm 0.4V[/tex].
What is the resistance of the wire and its uncertainty?

Homework Equations


-------------

The Attempt at a Solution



[tex]\frac{0.2}{1.0} \times 100 = 20\%[/tex]

[tex]\frac{0.4}{8.0} \times 100 = 5\%[/tex]

Total Percentage Uncertainty = [tex]25\%[/tex]

How can I get the uncertainty?
 
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  • #2
looi76 said:

Homework Statement


An experiment is done to measure the resistance of a wire.
The current in the wire is [tex]1.0 \pm 0.2A[/tex] and the potential difference across the wire is [tex]8.0 \pm 0.4V[/tex].
What is the resistance of the wire and its uncertainty?

Homework Equations


-------------

The Attempt at a Solution



[tex]\frac{0.2}{1.0} \times 100 = 20\%[/tex]

[tex]\frac{0.4}{8.0} \times 100 = 5\%[/tex]

Total Percentage Uncertainty = [tex]25\%[/tex]

How can I get the uncertainty?

Do you have a lab manual for your course? Sometimes they cover this in the introduction. Also, in this course, do they use calculus, or is it precalc?
 
  • #3
I don't think this question need calculus! It is a multiple choice question and the choices are:
(A) [tex](8.0 \pm 0.2)\omega[/tex]
(B) [tex](8.0 \pm 0.6)\omega[/tex]
(C) [tex](8 \pm 1)\omega[/tex]
(D) [tex](8 \pm 2)\omega[/tex]
 
  • #4
[itex]\omega[/itex]?...I think You mean [itex]\Omega[/itex]. You have been given the Voltage (V) and the Current (I). Use Ohm's law R = V/I to find the resistance. You can't add the percentages like that. Hint: You have to include the tolerance value of the Current and voltage in the formula.
 
  • #5
looi76 said:
I don't think this question need calculus! It is a multiple choice question and the choices are:
(A) [tex](8.0 \pm 0.2)\omega[/tex]
(B) [tex](8.0 \pm 0.6)\omega[/tex]
(C) [tex](8 \pm 1)\omega[/tex]
(D) [tex](8 \pm 2)\omega[/tex]

Okay, which of those answers has a 25% uncertainty?
 

1. What is resistance?

Resistance is a measure of how much a material or object resists the flow of electricity. It is the opposition to the movement of electrons through a conductor, such as a wire.

2. How is resistance measured?

Resistance is measured in units called ohms (Ω). It can be measured using a device called an ohmmeter, which sends a small current through the wire and measures the voltage drop across it.

3. What factors affect the resistance of a wire?

The resistance of a wire is affected by its length, cross-sectional area, and the material it is made of. Longer wires have higher resistance, while thicker wires have lower resistance. Different materials have different resistances, with some being better conductors than others.

4. What is the uncertainty of resistance?

The uncertainty of resistance refers to the range of values that the resistance measurement could potentially fall within. This can be affected by factors such as the precision of the measuring device and any external factors that may affect the resistance of the wire, such as temperature.

5. How can the uncertainty of resistance be reduced?

The uncertainty of resistance can be reduced by using more precise measuring equipment, controlling external factors that may affect resistance, and taking multiple measurements to calculate an average value. It is also important to use wires of consistent length and thickness for more accurate results.

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