What is the new resistance of a wire when it is stretched till its length is increased three times its original length?
2 Answers
When a wire is stretched to increase its length, its resistance changes due to the change in dimensions. The resistance \(R\) of a wire depends on three factors:
1. The resistivity \(\rho\) of the material, which remains constant if the material remains the same.
2. The length \(L\) of the wire.
3. The cross-sectional area \(A\) of the wire.
The relationship for resistance is given by:
\[
R = \rho \frac{L}{A}
\]
### Step-by-Step Explanation:
Let’s assume the following:
- The original length of the wire is \(L_0\).
- The original cross-sectional area is \(A_0\).
- The original resistance is \(R_0\).
The initial resistance can be written as:
\[
R_0 = \rho \frac{L_0}{A_0}
\]
#### Now, the wire is stretched until its length becomes three times its original length.
Thus, the new length of the wire is:
\[
L_{\text{new}} = 3L_0
\]
When a wire is stretched, its volume remains constant. This means the product of the length and cross-sectional area must remain the same. Mathematically:
\[
L_0 A_0 = L_{\text{new}} A_{\text{new}}
\]
Substituting \(L_{\text{new}} = 3L_0\):
\[
L_0 A_0 = 3L_0 A_{\text{new}}
\]
Simplifying, we get the new cross-sectional area:
\[
A_{\text{new}} = \frac{A_0}{3}
\]
#### Now, substitute the new length and area into the resistance formula:
The new resistance \(R_{\text{new}}\) is:
\[
R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}}
\]
Substituting the values of \(L_{\text{new}} = 3L_0\) and \(A_{\text{new}} = \frac{A_0}{3}\):
\[
R_{\text{new}} = \rho \frac{3L_0}{\frac{A_0}{3}} = \rho \frac{3L_0 \times 3}{A_0} = \rho \frac{9L_0}{A_0}
\]
But we know that \(R_0 = \rho \frac{L_0}{A_0}\), so:
\[
R_{\text{new}} = 9R_0
\]
### Conclusion:
When the wire is stretched to three times its original length, its resistance increases by a factor of 9. Therefore, the new resistance of the wire is:
\[
R_{\text{new}} = 9R_0
\]
1. The resistivity \(\rho\) of the material, which remains constant if the material remains the same.
2. The length \(L\) of the wire.
3. The cross-sectional area \(A\) of the wire.
The relationship for resistance is given by:
\[
R = \rho \frac{L}{A}
\]
### Step-by-Step Explanation:
Let’s assume the following:
- The original length of the wire is \(L_0\).
- The original cross-sectional area is \(A_0\).
- The original resistance is \(R_0\).
The initial resistance can be written as:
\[
R_0 = \rho \frac{L_0}{A_0}
\]
#### Now, the wire is stretched until its length becomes three times its original length.
Thus, the new length of the wire is:
\[
L_{\text{new}} = 3L_0
\]
When a wire is stretched, its volume remains constant. This means the product of the length and cross-sectional area must remain the same. Mathematically:
\[
L_0 A_0 = L_{\text{new}} A_{\text{new}}
\]
Substituting \(L_{\text{new}} = 3L_0\):
\[
L_0 A_0 = 3L_0 A_{\text{new}}
\]
Simplifying, we get the new cross-sectional area:
\[
A_{\text{new}} = \frac{A_0}{3}
\]
#### Now, substitute the new length and area into the resistance formula:
The new resistance \(R_{\text{new}}\) is:
\[
R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}}
\]
Substituting the values of \(L_{\text{new}} = 3L_0\) and \(A_{\text{new}} = \frac{A_0}{3}\):
\[
R_{\text{new}} = \rho \frac{3L_0}{\frac{A_0}{3}} = \rho \frac{3L_0 \times 3}{A_0} = \rho \frac{9L_0}{A_0}
\]
But we know that \(R_0 = \rho \frac{L_0}{A_0}\), so:
\[
R_{\text{new}} = 9R_0
\]
### Conclusion:
When the wire is stretched to three times its original length, its resistance increases by a factor of 9. Therefore, the new resistance of the wire is:
\[
R_{\text{new}} = 9R_0
\]
To determine the new resistance of a wire when it is stretched to three times its original length, we need to understand how stretching affects the wire’s resistance.
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material (which remains constant if the material and temperature are unchanged),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Stretching the Wire
When the wire is stretched, its length increases, and its cross-sectional area decreases. Let’s denote the original length of the wire as \( L_0 \) and its original cross-sectional area as \( A_0 \).
After stretching, the new length \( L \) becomes:
\[ L = 3 L_0 \]
The volume of the wire remains constant during stretching. Hence, the initial volume \( V_0 \) is equal to the final volume \( V \):
\[ V_0 = A_0 L_0 \]
\[ V = A L \]
Since \( V_0 = V \):
\[ A_0 L_0 = A (3 L_0) \]
We can solve for the new cross-sectional area \( A \):
\[ A = \frac{A_0 L_0}{3 L_0} \]
\[ A = \frac{A_0}{3} \]
Thus, the new cross-sectional area \( A \) is one-third of the original cross-sectional area \( A_0 \).
### Calculating the New Resistance
We can now calculate the new resistance \( R' \) using the formula:
\[ R' = \rho \frac{L}{A} \]
Substitute the new length and new cross-sectional area:
\[ R' = \rho \frac{3 L_0}{A_0 / 3} \]
\[ R' = \rho \frac{3 L_0 \cdot 3}{A_0} \]
\[ R' = 9 \left( \rho \frac{L_0}{A_0} \right) \]
\[ R' = 9 R_0 \]
where \( R_0 \) is the original resistance of the wire.
### Conclusion
The new resistance of the wire, when stretched to three times its original length, is nine times the original resistance.
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material (which remains constant if the material and temperature are unchanged),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Stretching the Wire
When the wire is stretched, its length increases, and its cross-sectional area decreases. Let’s denote the original length of the wire as \( L_0 \) and its original cross-sectional area as \( A_0 \).
After stretching, the new length \( L \) becomes:
\[ L = 3 L_0 \]
The volume of the wire remains constant during stretching. Hence, the initial volume \( V_0 \) is equal to the final volume \( V \):
\[ V_0 = A_0 L_0 \]
\[ V = A L \]
Since \( V_0 = V \):
\[ A_0 L_0 = A (3 L_0) \]
We can solve for the new cross-sectional area \( A \):
\[ A = \frac{A_0 L_0}{3 L_0} \]
\[ A = \frac{A_0}{3} \]
Thus, the new cross-sectional area \( A \) is one-third of the original cross-sectional area \( A_0 \).
### Calculating the New Resistance
We can now calculate the new resistance \( R' \) using the formula:
\[ R' = \rho \frac{L}{A} \]
Substitute the new length and new cross-sectional area:
\[ R' = \rho \frac{3 L_0}{A_0 / 3} \]
\[ R' = \rho \frac{3 L_0 \cdot 3}{A_0} \]
\[ R' = 9 \left( \rho \frac{L_0}{A_0} \right) \]
\[ R' = 9 R_0 \]
where \( R_0 \) is the original resistance of the wire.
### Conclusion
The new resistance of the wire, when stretched to three times its original length, is nine times the original resistance.