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Current Electricity

Section: Physics

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Question : 13 of 37

Marks: +1, -0

A wire of resistance

100 \Omega

is stretched so that its length increases by

20 \%

. The stretched wire is then bent in the form of a rectangle whose length and breadth are in the ratio

3 : 2

. The effective resistance between the ends of any diagonal of the rectangle is

[AP EAPCET 21 May 2025 S1]

$36 \Omega$
$72 \Omega$
$28.8 \Omega$
$43.2 \Omega$

Solution:

\;R_1=100 \Omega l_1=l

\; \;\;\;\; R_2\;=?, l_2=l_1+20 \% \;\text{ of }\; l_1

\;=l+0.2 l=1.2 l

\therefore \;\; R^{'}\;=n^2 R=(1.2)^2 R

\;=1.44 R=1.44 \times 100=144 \Omega

The wire of resistance

144 \Omega

is bent into a rectangle with sides in the ratio

3 : 2

. Let the side be

3 x

and

2 x

.

Total length,

l=2(3 x+2 x)=10 x

\therefore \;\; R_{\;\text{length }}\;=\;\frac{3 x}{10 x} \times 144=43.2 \Omega

R_{\;\text{width }}\;=\;\frac{2 x}{10 x} \times 144=28.8 \Omega

The effective resistance across the diagonal is the equivalent resistance of two parallel branches. On branch consists of the other two sides of the same resistance. Each branch has a resistance of

\;R_{\;\text{length }}+R_{\;\text{width }}=43.2+28.8=72 \Omega

\; \therefore \;\; \;\frac{1}{R_{\;\text{eq }}}=\;\frac{1}{72}+\;\frac{1}{72}=\;\frac{1}{36}

\;\Rightarrow \;\; R_{\;\text{eq }}=36 \Omega

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