When a wire of resistance r is cut into two?
2 Answers
When a wire of resistance \( R \) is cut into two pieces, the resistances of the resulting pieces will be different, depending on how the wire is cut. Let’s explore how the resistance of each piece is affected.
### Basic Concept
Resistance \( R \) of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Cutting the Wire
Assume the wire is cut into two pieces, one with length \( L_1 \) and the other with length \( L_2 \). The total length of the wire before cutting is \( L = L_1 + L_2 \).
The cross-sectional area \( A \) of both pieces remains the same as the original wire, so the resistivity \( \rho \) is also unchanged.
#### Resistance of Each Piece
1. **Resistance of the First Piece (Length \( L_1 \))**
The resistance \( R_1 \) of the first piece is:
\[ R_1 = \rho \frac{L_1}{A} \]
2. **Resistance of the Second Piece (Length \( L_2 \))**
The resistance \( R_2 \) of the second piece is:
\[ R_2 = \rho \frac{L_2}{A} \]
Since the total length of the wire is the sum of \( L_1 \) and \( L_2 \), the total resistance of the original wire \( R \) is:
\[ R = \rho \frac{L}{A} \]
We can express \( R \) in terms of \( L_1 \) and \( L_2 \) as:
\[ R = \rho \frac{L_1 + L_2}{A} \]
Therefore, the resistance of each piece after cutting is proportional to its length. If \( L_1 \) is half of \( L \), then \( R_1 \) will be half of \( R \), and similarly for \( L_2 \).
### Summary
When a wire is cut into two pieces:
- The resistance of each piece is directly proportional to its length.
- If the wire is cut into two pieces with lengths \( L_1 \) and \( L_2 \), the resistances \( R_1 \) and \( R_2 \) of the pieces will be:
\[ R_1 = \frac{L_1}{L} \times R \]
\[ R_2 = \frac{L_2}{L} \times R \]
The total resistance of the two pieces in series would be the sum of \( R_1 \) and \( R_2 \), which is equal to the original resistance \( R \).
### Basic Concept
Resistance \( R \) of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Cutting the Wire
Assume the wire is cut into two pieces, one with length \( L_1 \) and the other with length \( L_2 \). The total length of the wire before cutting is \( L = L_1 + L_2 \).
The cross-sectional area \( A \) of both pieces remains the same as the original wire, so the resistivity \( \rho \) is also unchanged.
#### Resistance of Each Piece
1. **Resistance of the First Piece (Length \( L_1 \))**
The resistance \( R_1 \) of the first piece is:
\[ R_1 = \rho \frac{L_1}{A} \]
2. **Resistance of the Second Piece (Length \( L_2 \))**
The resistance \( R_2 \) of the second piece is:
\[ R_2 = \rho \frac{L_2}{A} \]
Since the total length of the wire is the sum of \( L_1 \) and \( L_2 \), the total resistance of the original wire \( R \) is:
\[ R = \rho \frac{L}{A} \]
We can express \( R \) in terms of \( L_1 \) and \( L_2 \) as:
\[ R = \rho \frac{L_1 + L_2}{A} \]
Therefore, the resistance of each piece after cutting is proportional to its length. If \( L_1 \) is half of \( L \), then \( R_1 \) will be half of \( R \), and similarly for \( L_2 \).
### Summary
When a wire is cut into two pieces:
- The resistance of each piece is directly proportional to its length.
- If the wire is cut into two pieces with lengths \( L_1 \) and \( L_2 \), the resistances \( R_1 \) and \( R_2 \) of the pieces will be:
\[ R_1 = \frac{L_1}{L} \times R \]
\[ R_2 = \frac{L_2}{L} \times R \]
The total resistance of the two pieces in series would be the sum of \( R_1 \) and \( R_2 \), which is equal to the original resistance \( R \).
When a wire of resistance \( R \) is cut into two parts, the resistance of each part depends on how the wire is divided. Here's a detailed explanation:
### **1. Cutting the Wire into Two Equal Parts**
If the wire is cut into two equal-length pieces, each piece will have half the length of the original wire. Resistance \( R \) of a wire is directly proportional to its length \( L \), so if the length is halved, the resistance will also be halved.
Let’s denote the resistance of each equal part as \( R_1 \). Since each part is half the length of the original wire, its resistance will be:
\[ R_1 = \frac{R}{2} \]
### **2. Cutting the Wire into Two Unequal Parts**
If the wire is cut into two parts with lengths \( L_1 \) and \( L_2 \) such that \( L_1 + L_2 = L \), the resistances of the two parts will be different, and can be calculated as follows:
The resistance of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.
Let’s denote:
- \( R_1 \) as the resistance of the first part with length \( L_1 \)
- \( R_2 \) as the resistance of the second part with length \( L_2 \)
Given that:
\[ R_1 = \rho \frac{L_1}{A} \]
\[ R_2 = \rho \frac{L_2}{A} \]
Since \( L_1 + L_2 = L \), we can express \( L_2 \) as \( L - L_1 \). Therefore:
\[ R_2 = \rho \frac{L - L_1}{A} \]
If the original wire’s resistance is \( R \), then:
\[ R = \rho \frac{L}{A} \]
Thus:
\[ R_1 = R \frac{L_1}{L} \]
\[ R_2 = R \frac{L_2}{L} \]
### **Summary**
- When a wire of resistance \( R \) is cut into two equal-length parts, each part will have a resistance of \( \frac{R}{2} \).
- When a wire is cut into two parts of different lengths, the resistance of each part is proportional to its length relative to the original length of the wire.
### **1. Cutting the Wire into Two Equal Parts**
If the wire is cut into two equal-length pieces, each piece will have half the length of the original wire. Resistance \( R \) of a wire is directly proportional to its length \( L \), so if the length is halved, the resistance will also be halved.
Let’s denote the resistance of each equal part as \( R_1 \). Since each part is half the length of the original wire, its resistance will be:
\[ R_1 = \frac{R}{2} \]
### **2. Cutting the Wire into Two Unequal Parts**
If the wire is cut into two parts with lengths \( L_1 \) and \( L_2 \) such that \( L_1 + L_2 = L \), the resistances of the two parts will be different, and can be calculated as follows:
The resistance of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.
Let’s denote:
- \( R_1 \) as the resistance of the first part with length \( L_1 \)
- \( R_2 \) as the resistance of the second part with length \( L_2 \)
Given that:
\[ R_1 = \rho \frac{L_1}{A} \]
\[ R_2 = \rho \frac{L_2}{A} \]
Since \( L_1 + L_2 = L \), we can express \( L_2 \) as \( L - L_1 \). Therefore:
\[ R_2 = \rho \frac{L - L_1}{A} \]
If the original wire’s resistance is \( R \), then:
\[ R = \rho \frac{L}{A} \]
Thus:
\[ R_1 = R \frac{L_1}{L} \]
\[ R_2 = R \frac{L_2}{L} \]
### **Summary**
- When a wire of resistance \( R \) is cut into two equal-length parts, each part will have a resistance of \( \frac{R}{2} \).
- When a wire is cut into two parts of different lengths, the resistance of each part is proportional to its length relative to the original length of the wire.