Solving Resistors in Series: 1.50 V Battery

In summary, the question asks for the potential difference across each wire in a series circuit with a 1.50 V battery and three wires of equal length but different cross-sectional areas. Given that the cross-sectional areas increase, it can be assumed that the resistivity decreases. With the relevant equation for parallel resistors, the resistances of the segments can be compared. By arbitrarily assigning a resistance of 1Ω to the first segment, the other resistances can be determined and used to calculate the potential differences across each wire.
  • #1
franky1994
1
0

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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  • #2
Your relevant equation is for parallel resistors.
 
  • #3
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?
 

Related to Solving Resistors in Series: 1.50 V Battery

What is the formula for calculating the total resistance in a series circuit with a 1.50 V battery?

The formula for calculating the total resistance in a series circuit is R = R1 + R2 + R3 + ..., where R is the total resistance and R1, R2, R3, etc. are the resistances of each component in the circuit. This formula applies to circuits with a 1.50 V battery or any other voltage source.

How do I calculate the current in a series circuit with a 1.50 V battery?

To calculate the current in a series circuit, you can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In a series circuit, the current is the same at all points, so you can use the voltage of the 1.50 V battery and the total resistance of the circuit to calculate the current.

Do I need to consider the internal resistance of the 1.50 V battery when calculating the total resistance in a series circuit?

Yes, the internal resistance of the battery should be included when calculating the total resistance in a series circuit. This is because the internal resistance affects the overall voltage and current in the circuit, and can impact the accuracy of your calculations.

Can I use the same formula for solving resistors in series with a 1.50 V battery for circuits with different voltage sources?

Yes, the formula for solving resistors in series can be applied to circuits with different voltage sources. However, the value of the voltage source should be substituted into the formula for V, and the total resistance should be recalculated based on the resistances of the individual components in the circuit.

What is the purpose of solving resistors in series with a 1.50 V battery?

The purpose of solving resistors in series with a 1.50 V battery is to determine the total resistance and current in the circuit. This information is essential for understanding the behavior of the circuit and ensuring that the components are properly connected and functioning as intended.

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