Question

A wire is cut in to $4$ pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was $R$, the resistance of the bundle will be:

• A. $\frac{R}{4}$
• B. $\frac{R}{8}$
  
• C. $\frac{R}{16}$
  
• D. $\frac{R}{32}$
  

Answer 1

The correct option is C $\frac{R}{16}$

Resistance of wire $R=\rho \frac{L}{A}$

where $L$ is the length of wire and $A$ is its cross-section area.

$⟹$ $R\propto L$

Thus if a wire is cut into $4$ equal pieces, then the resistance of each piece is equal to the $(1/4{)}^{th}$ of the resistance of whole wire i.e. $R^{\prime }=\frac{R}{4}$

Now these $4$ parts are connected in parallel to each other.

Equivalent resistance of parallel combination $\frac{1}{R_p}=\frac{1}{R^{\prime }}$ $+\frac{1}{R^{\prime }}+\frac{1}{R^{\prime }}\frac{1}{R^{\prime }}$

$\therefore$ $R_p=\frac{R^{\prime }}{4}$

$⟹$ $R_p=\frac{R}{16}$