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A 6 ohms resistance wire is doubled up by folding. Calculate the new resistance in wire.\nHow 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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