A 6 ohms resistance wire is doubled up by folding. Calculate the new resistance in wire.
How area is get doubled in the solution of this question?

As given resistance of the wire $R=6\ ohms$

We know the resistance formula $R=\rho\frac{l}{A}$ ..... $( i)$

When $\rho\rightarrow$ a constant known as resistivity of the wire

$l\rightarrow$ length of the wire

$A\rightarrow$ cross section area of the wire

When the wire is doubled, Length becomes $\frac{l}{2}$ and cross section area becomes doubled $2A$ as shoun in the figure below:

Therefore, new resistance of the wire $R_{new}=\rho\frac{\frac{l}{2}}{2A}$

$=\rho\frac{l}{4A}$

$=\rho\frac{l}{A}\times\frac{1}{4}$

$=\frac{R}{4}$           [from $(i)$]

$=\frac{6}{4}$      [given resistance of the wire $R=6\ ohms$]

$=1.5\ ohms$

Therefore, new resistance of the wire is $1.5\ ohms$ after folding the wire double.

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Updated on: 10-Oct-2022

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