Solving Resistors in Series: 1.50 V Battery

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Three silver wires of equal length are connected in series with a 1.50 V battery, each having different cross-sectional areas of 1.00, 5.00, and 10.0 cm². The resistance of each wire is inversely related to its cross-sectional area, leading to the conclusion that the narrowest wire has the highest resistance and the widest has the lowest. Without the current value, the potential difference across each wire can be determined by understanding the relationship between resistance and voltage in series circuits. Assigning a resistance value to one wire can help calculate the others, facilitating the determination of voltage drops across each segment. This approach allows for solving the problem despite missing current information.
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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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