Solving Resistors in Series: 1.50 V Battery

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SUMMARY

The discussion focuses on calculating the potential difference across three silver wires connected in series to a 1.50 V battery, with cross-sectional areas of 1.00 cm², 5.00 cm², and 10.0 cm². The resistance of each wire is inversely proportional to its cross-sectional area, leading to the conclusion that R1 < R2 < R3. The current is not provided, but users are encouraged to assign a resistance value to the first wire to facilitate calculations for the others.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of resistance calculations for series circuits
  • Familiarity with the relationship between resistivity, length, and cross-sectional area of conductors
  • Basic algebra for manipulating equations
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  • Research how to calculate resistance in series circuits with varying cross-sectional areas
  • Learn about the concept of resistivity and its dependence on material properties
  • Explore practical applications of Ohm's Law in electrical circuits
  • Investigate the effects of wire gauge on electrical resistance and current flow
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Students studying electrical engineering, physics enthusiasts, and anyone looking to understand the principles of resistance in electrical circuits.

franky1994
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Homework Statement



Three silver wires of equal length are connected in series with a 1.50 V battery. Their cross-sectional areas are 1.00, 5.00 and 10.0 cm2. (a) What is the potential difference across the narrowest wire? (b) The medium wire? (c) The widest wire?

Homework Equations


V= IR

1/Req = 1/R1 + 1/R2 + 1/R3 + … + 1/Rn



The Attempt at a Solution


Since the cross sectional area increases for each wire in the series, it can be assumed that the resistivity decreases from left to right or right to left (depending on how you draw your diagram).
That is:
R1 < R2 < R3.

I know the length of each wire is the same. However, I do not know the current. All the question provides me is the cross-sectional area of each wire.


Any help would be appreciated. (Also I will return the favor if I can. That depends on whether or not I have studied the problem you need assistance with.)

Thank you!
 
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Your relevant equation is for parallel resistors.
 
franky1994 said:

Homework Statement



Three silver wires of equal length are connected in series with a 1.50 V battery. Their cross-sectional areas are 1.00, 5.00 and 10.0 cm2. (a) What is the potential difference across the narrowest wire? (b) The medium wire? (c) The widest wire?

Homework Equations


V= IR

1/Req = 1/R1 + 1/R2 + 1/R3 + … + 1/Rn



The Attempt at a Solution


Since the cross sectional area increases for each wire in the series, it can be assumed that the resistivity decreases from left to right or right to left (depending on how you draw your diagram).
That is:
R1 < R2 < R3.

I know the length of each wire is the same. However, I do not know the current. All the question provides me is the cross-sectional area of each wire.


Any help would be appreciated. (Also I will return the favor if I can. That depends on whether or not I have studied the problem you need assistance with.)

Thank you!

Hi franky1994, Welcome to Physics Forums.

While current, lengths, and resistivity of the metal are not supplied, you should be able to compare the resistances of the segments. Consider how the resistance of a wire varies with respect to length and cross sectional area. Hint: What happens if you arbitrarily assign a resistance of 1Ω to the first wire segment? What would the other resistances be?
 

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