A wire of 5Ω resistance is drawn out so that its new length is two times the original length. If the resistivity of the wire remains the same and the cross-sectional area is halved, the new resistance is
Given:
- Original resistance \( R_1 = 5 \, \Omega \)
- The new length \( L_2 = 2L_1 \) (where \( L_1 \) is the original length)
- The new cross-sectional area \( A_2 = \frac{1}{2} A_1 \) (where \( A_1 \) is the original area)
Step 1: Recall the formula for resistance
The resistance \( R \) of a wire is given by:
\(R = \rho \frac{L}{A}\)
where:
- \( \rho \) is the resistivity,
- \( L \) is the length,
- \( A \) is the cross-sectional area.
Step 2: Calculate the new resistance
The new resistance \( R_2 \) can be calculated using the new length and new area:
\(R_2 = \rho \frac{L_2}{A_2}\)
Substituting for \( L_2 \) and \( A_2 \):
\(R_2 = \rho \frac{2L_1}{\frac{1}{2}A_1} = \rho \frac{2L_1 \cdot 2}{A_1} = \rho \frac{4L_1}{A_1}\)
Step 3: Relate \( R_1 \) and \( R_2 \)
From the original resistance \( R_1 \):
\(R_1 = \rho \frac{L_1}{A_1}\)
Now, we can express \( R_2 \) in terms of \( R_1 \):
\(R_2 = 4 \left(\rho \frac{L_1}{A_1}\right) = 4R_1\)
Step 4: Substitute the value of \( R_1 \)
Now substitute \( R_1 = 5 \, \Omega \):
\(R_2 = 4 \times 5 \, \Omega = 20 \, \Omega\)
Therefore, the new resistance is: \(\boxed{20 \, \Omega}\)
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