Question

A wire is broken in four equal parts. A packet is formed by keeping the four wires together in parallel combination. The resistance of the packet in comparison to the resistance of the wire will be

• A. $equal$
• B. ${\left(\frac{1}{4}\right)}^{th}$
• C. ${\left(\frac{1}{8}\right)}^{th}$
• D. ${\left(\frac{1}{16}\right)}^{th}$

Answer 1

The correct option is D ${\left(\frac{1}{16}\right)}^{th}$

Let. $R$ be the resistance of the wire.
When it is broken into four parts the resistance of each part is $\frac{R}{4}$
These four wires are in parallel combination hence the equivalent resistance is
$\frac{1}{R_{eq}}=\frac{1}{\frac{R}{4}}+\frac{1}{\frac{R}{4}}+\frac{1}{\frac{R}{4}}+\frac{1}{\frac{R}{4}}$
$\frac{1}{R_{eq}}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R}$
$\frac{1}{R_{eq}}=\frac{16}{R}$
$R_{eq}=\frac{R}{16}$
Therefore, the resistance of the packet in comparison to the resistance of the wire is ${\frac{1}{16}}^{th}$ times.