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# A $6\Omega$ resistance wire is double up by folding. Calculate the new resistance of the wire.

We know that the resistance of a conductor and length of the conductor is directly proportional to each other $(L \propto R)$ which means as the length of the conductor increases, the resistance of the conductor also increases. Whereas, if the length is decreased then resistance also gets decreased.

Similarly, the area of cross-section of the conductor and the resistance are inversely proportional to each other $(A\propto \frac {1}{R})$, which means as the area of cross-section increases, the resistance decreases. Whereas, if the area of the cross section is decreased then resistance gets increased.

Given:

Resistance, $R$ = $6\Omega$

To find: Resistance of the wire, $(R')$ when it is double up by folding.

Solution:

The relation between the length, area of cross-section, and resistance of a conductor is given as-

$R=ρ{\frac {L}{A}}$

where,

$R$ = Resistance of the conductor,

$ρ$ = Resistivity of the conductor,

$L$ = Length of the conductor,

$A$ = Area of the cross-section.

Putting the value of $R$ in the equation we get-

$6=ρ{\frac {L}{A}}$ ------------(i)

Here, in this case, a piece of wire is double up by folding. So, the length gets halved $(\frac {l}{2})$ and the area of cross-section gets doubled $(2A)$.

Then the equation is written as-

$R'=ρ{\frac {\frac {1}{2}L}{2A}}$

$R'=ρ{\frac {L}{2A\times 2}}$

$R'=ρ{\frac {L}{4A}}$

$4R'=ρ{\frac {L}{A}}$ ------------(ii)

Now putting the value of $ρ{\frac {L}{A}}=6$ from equation (i) in equation (ii) we get-

$4R'=6$

$R'=\frac {6}{4}$

$R'=1.5\Omega$

Therefore, the resistance, $R'$ becomes 1.5 Ohm.